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Equiv: Answer test results

This page exposes the results of running answer tests on STACK test cases. This page is automatically generated from the STACK unit tests and is designed to show question authors what answer tests actually do. This includes cases where answer tests currentl fail, which gives a negative expected mark. Comments and further test cases are very welcome.

Equiv

Test
?
Student response
Teacher answer
Opt
Mark
Answer note
Equiv 
x
 
[x^2=4,x=2 or x=-2]
 -1 ATEquiv_SA_not_list.
The first argument to the Equiv answer test should be a list, but the test failed. Please contact your teacher.
Equiv 
[x^2=4,x=2 or x=-2]
 
x
 -1 ATEquiv_SB_not_list.
The second argument to the Equiv answer test should be a list, but the test failed. Please contact your teacher.
Equiv 
[1/0]
 
[x^2=4,x=2 or x=-2]
 -1 ATEquiv_STACKERROR_SAns.
Equiv 
[x^2=4,x=2 or x=-2]
 
[1/0]
 -1 ATEquiv_STACKERROR_TAns.
Equiv 
[x^2=4,x=2 or x=-2]
 
[x^2=4,x=2 or x=-2]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &x^2=4& \cr \color{green}{\Leftrightarrow}&x=2\,{\text{ or }}\, x=-2& \cr \end{array}\]
Equiv 
[x^2=4,x=#pm#2,x=2 and x=-2]
 
[x^2=4,x=2 or x=-2]
 0 (EMPTYCHAR, EQUIVCHAR,ANDOR)
\[\begin{array}{lll} &x^2=4& \cr \color{green}{\Leftrightarrow}&x= \pm 2& \cr \color{red}{\text{and/or confusion!}}&\left\{\begin{array}{l}x=2\cr x=-2\cr \end{array}\right.& \cr \end{array}\]
Equiv 
[x^2=4,x=2]
 
[x^2=4,x=2 or x=-2]
 0 (EMPTYCHAR,IMPLIEDCHAR)
\[\begin{array}{lll} &x^2=4& \cr \color{red}{\Leftarrow}&x=2& \cr \end{array}\]
Equiv 
[x^2=4,x=2]
 
[x^2=4,x=2]
 
[assumepos]
 1 (ASSUMEPOSVARS, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{\text{Assume +ve vars}}&x^2=4& \cr \color{green}{\Leftrightarrow}&x=2& \cr \end{array}\]
Equiv 
[x^2=4,x^2-4=0,(x-2)*(x+2)=0,x
=2 or x=-2]
 
[x^2=4,x=2 or x=-2]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &x^2=4& \cr \color{green}{\Leftrightarrow}&x^2-4=0& \cr \color{green}{\Leftrightarrow}&\left(x-2\right)\cdot \left(x+2\right)=0& \cr \color{green}{\Leftrightarrow}&x=2\,{\text{ or }}\, x=-2& \cr \end{array}\]
Equiv 
[x^2=4,x= #pm#2, x=2 or x=-2]
 
[x^2=4,x=2 or x=-2]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &x^2=4& \cr \color{green}{\Leftrightarrow}&x= \pm 2& \cr \color{green}{\Leftrightarrow}&x=2\,{\text{ or }}\, x=-2& \cr \end{array}\]
Equiv 
[x^2-6*x+9=0,x=3]
 
[x^2-6*x+9=0,x=3]
 1 (EMPTYCHAR,SAMEROOTS)
\[\begin{array}{lll} &x^2-6\cdot x+9=0& \cr \color{green}{\text{(Same roots)}}&x=3& \cr \end{array}\]
Equiv 
[]
 
[]
 1 (EMPTYCHAR)
\[\begin{array}{lll} &\left[ \right] & \cr \end{array}\]
Equiv 
[x^2=-1]
 
[]
 1 (EMPTYCHAR)
\[\begin{array}{lll} &x^2=-1& \cr \end{array}\]
Equiv 
[x=x,all]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &x=x& \cr \color{green}{\Leftrightarrow}&\mathbb{R}& \cr \end{array}\]
Equiv 
[x=x,true]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &x=x& \cr \color{green}{\Leftrightarrow}&\mathbf{True}& \cr \end{array}\]
Equiv 
[x=x,false]
 
[]
 0 (EMPTYCHAR,QMCHAR)
\[\begin{array}{lll} &x=x& \cr \color{red}{?}&\mathbf{False}& \cr \end{array}\]
Equiv 
[1=1,all]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &1=1& \cr \color{green}{\Leftrightarrow}&\mathbb{R}& \cr \end{array}\]
Equiv 
[1=1,true]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &1=1& \cr \color{green}{\Leftrightarrow}&\mathbf{True}& \cr \end{array}\]
Equiv 
[0=0,all]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &0=0& \cr \color{green}{\Leftrightarrow}&\mathbb{R}& \cr \end{array}\]
Equiv 
[0=0,true]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &0=0& \cr \color{green}{\Leftrightarrow}&\mathbf{True}& \cr \end{array}\]
Equiv 
[1=2,false]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &1=2& \cr \color{green}{\Leftrightarrow}&\mathbf{False}& \cr \end{array}\]
Equiv 
[1=2,none]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &1=2& \cr \color{green}{\Leftrightarrow}&\emptyset& \cr \end{array}\]
Equiv 
[1=2,{}]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &1=2& \cr \color{green}{\Leftrightarrow}&\left \{ \right \}& \cr \end{array}\]
Equiv 
[1=2,[]]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &1=2& \cr \color{green}{\Leftrightarrow}&\left[ \right] & \cr \end{array}\]
Equiv 
[3=0,2=sqrt(-5),2=0,2=sqrt(5),
2=0,2=sqrt(-5),3=0]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &3=0& \cr \color{green}{\Leftrightarrow}&2=\sqrt{-5}& \cr \color{green}{\Leftrightarrow}&2=0& \cr \color{green}{\Leftrightarrow}&2=\sqrt{5}& \cr \color{green}{\Leftrightarrow}&2=0& \cr \color{green}{\Leftrightarrow}&2=\sqrt{-5}& \cr \color{green}{\Leftrightarrow}&3=0& \cr \end{array}\]
Equiv 
[3=0,2=sqrt(-5),2=0,2=sqrt(5),
2=0,2=sqrt(-5),3=0]
 
[]
 
[assumereal]
 1 (ASSUMEREALVARS, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{(\mathbb{R})}&3=0& \cr \color{green}{\Leftrightarrow}&2=\sqrt{-5}& \cr \color{green}{\Leftrightarrow}&2=0& \cr \color{green}{\Leftrightarrow}&2=\sqrt{5}& \cr \color{green}{\Leftrightarrow}&2=0& \cr \color{green}{\Leftrightarrow}&2=\sqrt{-5}& \cr \color{green}{\Leftrightarrow}&3=0& \cr \end{array}\]
Equiv 
[x=1,X=1]
 
[]
 0 (EMPTYCHAR,QMCHAR)
\[\begin{array}{lll} &x=1& \cr \color{red}{?}&X=1& \cr \end{array}\]
Equiv 
[1/(x^2+1)=1/((x+%i)*(x-%i)),t
rue]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\frac{1}{x^2+1}=\frac{1}{\left(x+\mathrm{i}\right)\cdot \left(x-\mathrm{i}\right)}& \cr \color{green}{\Leftrightarrow}&\mathbf{True}& \cr \end{array}\]
Equiv 
[2^2,stackeq(4)]
 
[]
 1 (EMPTYCHAR, CHECKMARK)
\[\begin{array}{lll} &2^2& \cr \color{green}{\checkmark}&=4& \cr \end{array}\]
Equiv 
[2^2,stackeq(3)]
 
[]
 0 (EMPTYCHAR,IMPLIESCHAR)
\[\begin{array}{lll} &2^2& \cr \color{red}{\Rightarrow}&=3& \cr \end{array}\]
Equiv 
[2^2,4]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &2^2& \cr \color{green}{\Leftrightarrow}&4& \cr \end{array}\]
Equiv 
[2^2,3]
 
[]
 0 (EMPTYCHAR,IMPLIESCHAR)
\[\begin{array}{lll} &2^2& \cr \color{red}{\Rightarrow}&3& \cr \end{array}\]
Equiv 
[lg(64,4),lg(4^3,4),3*lg(4,4),
3]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\log_{4}\left(64\right)& \cr \color{green}{\Leftrightarrow}&\log_{4}\left(4^3\right)& \cr \color{green}{\Leftrightarrow}&3\cdot \log_{4}\left(4\right)& \cr \color{green}{\Leftrightarrow}&3& \cr \end{array}\]
Equiv 
[lg(64,4),stackeq(lg(4^3,4)),s
tackeq(3*lg(4,4)),stackeq(3)]
 
[]
 1 (EMPTYCHAR, CHECKMARK, CHECKMARK, CHECKMARK)
\[\begin{array}{lll} &\log_{4}\left(64\right)& \cr \color{green}{\checkmark}&=\log_{4}\left(4^3\right)& \cr \color{green}{\checkmark}&=3\cdot \log_{4}\left(4\right)& \cr \color{green}{\checkmark}&=3& \cr \end{array}\]
Equiv 
[x=1 or x=2,x=1 or 2]
 
[]
 0 (EMPTYCHAR,MISSINGVAR)
\[\begin{array}{lll} &x=1\,{\text{ or }}\, x=2& \cr \color{red}{\text{Missing assignments}}&x=1\,{\text{ or }}\, 2& \cr \end{array}\]
Equiv 
[x=1 or x=2,x=1 and x=2]
 
[]
 0 (EMPTYCHAR,ANDOR)
\[\begin{array}{lll} &x=1\,{\text{ or }}\, x=2& \cr \color{red}{\text{and/or confusion!}}&\left\{\begin{array}{l}x=1\cr x=2\cr \end{array}\right.& \cr \end{array}\]
Equiv 
[x=1 and y=2,x=1 or y=2]
 
[]
 0 (EMPTYCHAR,ANDOR)
\[\begin{array}{lll} &\left\{\begin{array}{l}x=1\cr y=2\cr \end{array}\right.& \cr \color{red}{\text{and/or confusion!}}&x=1\,{\text{ or }}\, y=2& \cr \end{array}\]
Equiv 
[a=b,a^2=b^2]
 
[]
 0 (EMPTYCHAR,IMPLIESCHAR)
\[\begin{array}{lll} &a=b& \cr \color{red}{\Rightarrow}&a^2=b^2& \cr \end{array}\]
Equiv 
[a=b,sqrt(a)=sqrt(b)]
 
[]
 0 (EMPTYCHAR,IMPLIEDCHAR)
\[\begin{array}{lll} &a=b& \cr \color{red}{\Leftarrow}&\sqrt{a}=\sqrt{b}& \cr \end{array}\]
Equiv 
[a^2=b^2,a=b]
 
[]
 0 (EMPTYCHAR,IMPLIEDCHAR)
\[\begin{array}{lll} &a^2=b^2& \cr \color{red}{\Leftarrow}&a=b& \cr \end{array}\]
Equiv 
[a^2=b^2,a=b or a=-b]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &a^2=b^2& \cr \color{green}{\Leftrightarrow}&a=b\,{\text{ or }}\, a=-b& \cr \end{array}\]
Equiv 
[a^2=b^2,a= #pm#b,a= b or a=-b
]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &a^2=b^2& \cr \color{green}{\Leftrightarrow}&a= \pm b& \cr \color{green}{\Leftrightarrow}&a=b\,{\text{ or }}\, a=-b& \cr \end{array}\]
Equiv 
[9*x^2/2-81*x/2+90=5*x^2/2-5*x
-20 nounor 9*x^2/2-81*x/2+90=-
(5*x^2/2-5*x-20),9*x^2-81*x+18
0=5*x^2-10*x-40 nounor 9*x^2-8
1*x+180=-5*x^2+10*x+40,4*x^2-7
1*x+220=0 nounor 14*x^2-91*x+1
40=0,x=(71 #pm# sqrt(71^2-4*4*
220))/(2*4) nounor x=(91 #pm# 
sqrt(91^2-4*14*140))/(2*14),x=
55/4 nounor x=4 nounor x=5/2]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR,SAMEROOTS)
\[\begin{array}{lll} &\frac{9\cdot x^2}{2}-\frac{81\cdot x}{2}+90=\frac{5\cdot x^2}{2}-5\cdot x-20\,{\text{ or }}\, \frac{9\cdot x^2}{2}-\frac{81\cdot x}{2}+90=-\left(\frac{5\cdot x^2}{2}-5\cdot x-20\right)& \cr \color{green}{\Leftrightarrow}&9\cdot x^2-81\cdot x+180=5\cdot x^2-10\cdot x-40\,{\text{ or }}\, 9\cdot x^2-81\cdot x+180=-5\cdot x^2+10\cdot x+40& \cr \color{green}{\Leftrightarrow}&4\cdot x^2-71\cdot x+220=0\,{\text{ or }}\, 14\cdot x^2-91\cdot x+140=0& \cr \color{green}{\Leftrightarrow}&x=\frac{{71 \pm \sqrt{71^2-4\cdot 4\cdot 220}}}{2\cdot 4}\,{\text{ or }}\, x=\frac{{91 \pm \sqrt{91^2-4\cdot 14\cdot 140}}}{2\cdot 14}& \cr \color{green}{\text{(Same roots)}}&x=\frac{55}{4}\,{\text{ or }}\, x=4\,{\text{ or }}\, x=\frac{5}{2}& \cr \end{array}\]
Equiv 
[a=b,abs(a)=abs(b),a=b]
 
[]
 0 (EMPTYCHAR,IMPLIESCHAR,IMPLIEDCHAR)
\[\begin{array}{lll} &a=b& \cr \color{red}{\Rightarrow}&\left| a\right| =\left| b\right| & \cr \color{red}{\Leftarrow}&a=b& \cr \end{array}\]
Equiv 
[abs(a)=abs(b),a=b or a=-b]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\left| a\right| =\left| b\right| & \cr \color{green}{\Leftrightarrow}&a=b\,{\text{ or }}\, a=-b& \cr \end{array}\]
Equiv 
[abs(a)=abs(b),a^2=b^2]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\left| a\right| =\left| b\right| & \cr \color{green}{\Leftrightarrow}&a^2=b^2& \cr \end{array}\]
Equiv 
[x^3=8,x=2]
 
[]
 0 (EMPTYCHAR,IMPLIEDCHAR)
\[\begin{array}{lll} &x^3=8& \cr \color{red}{\Leftarrow}&x=2& \cr \end{array}\]
Equiv 
[x^3=8,x=2]
 
[]
 
[assumereal]
 1 (ASSUMEREALVARS, EQUIVCHARREAL)
\[\begin{array}{lll}\color{blue}{(\mathbb{R})}&x^3=8& \cr \color{green}{\Leftrightarrow}\, \color{blue}{(\mathbb{R})}&x=2& \cr \end{array}\]
Equiv 
[abs(x-1/2)+abs(x+1/2)=2,abs(x
)=1]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\left| x-\frac{1}{2}\right| +\left| x+\frac{1}{2}\right| =2& \cr \color{green}{\Leftrightarrow}&\left| x\right| =1& \cr \end{array}\]
Equiv 
[a^2=9 and a>0,a=3]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\left\{\begin{array}{l}a^2=9\cr a > 0\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&a=3& \cr \end{array}\]
Equiv 
[T=2*pi*sqrt(L/g),T^2=4*pi^2*L
/g,g=4*pi^2*L/T^2]
 
[]
 
[assumepos]
 1 (ASSUMEPOSVARS, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{\text{Assume +ve vars}}&T=2\cdot \pi\cdot \sqrt{\frac{L}{g}}& \cr \color{green}{\Leftrightarrow}&T^2=\frac{4\cdot \pi^2\cdot L}{g}& \cr \color{green}{\Leftrightarrow}&g=\frac{4\cdot \pi^2\cdot L}{T^2}& \cr \end{array}\]
Equiv 
[a=b,a^2=b^2]
 
[]
 
[assumepos]
 1 (ASSUMEPOSVARS, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{\text{Assume +ve vars}}&a=b& \cr \color{green}{\Leftrightarrow}&a^2=b^2& \cr \end{array}\]
Equiv 
[a=b,sqrt(a)=sqrt(b)]
 
[]
 
[assumepos]
 1 (ASSUMEPOSVARS, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{\text{Assume +ve vars}}&a=b& \cr \color{green}{\Leftrightarrow}&\sqrt{a}=\sqrt{b}& \cr \end{array}\]
Equiv 
[a^2=b^2,a=b]
 
[]
 
[assumepos]
 1 (ASSUMEPOSVARS, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{\text{Assume +ve vars}}&a^2=b^2& \cr \color{green}{\Leftrightarrow}&a=b& \cr \end{array}\]
Equiv 
[a^2=b^2,a=b or a=-b]
 
[]
 
[assumepos]
 1 (ASSUMEPOSVARS, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{\text{Assume +ve vars}}&a^2=b^2& \cr \color{green}{\Leftrightarrow}&a=b\,{\text{ or }}\, a=-b& \cr \end{array}\]
Equiv 
[a=b,abs(a)=abs(b)]
 
[]
 
[assumepos]
 1 (ASSUMEPOSVARS, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{\text{Assume +ve vars}}&a=b& \cr \color{green}{\Leftrightarrow}&\left| a\right| =\left| b\right| & \cr \end{array}\]
Equiv 
[abs(a)=abs(b),a=b]
 
[]
 
[assumepos]
 1 (ASSUMEPOSVARS, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{\text{Assume +ve vars}}&\left| a\right| =\left| b\right| & \cr \color{green}{\Leftrightarrow}&a=b& \cr \end{array}\]
Equiv 
[abs(a)=abs(b),a=-b]
 
[]
 
[assumepos]
 1 (ASSUMEPOSVARS, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{\text{Assume +ve vars}}&\left| a\right| =\left| b\right| & \cr \color{green}{\Leftrightarrow}&a=-b& \cr \end{array}\]
Equiv 
[abs(a)=abs(b),a=b or a=-b]
 
[]
 
[assumepos]
 1 (ASSUMEPOSVARS, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{\text{Assume +ve vars}}&\left| a\right| =\left| b\right| & \cr \color{green}{\Leftrightarrow}&a=b\,{\text{ or }}\, a=-b& \cr \end{array}\]
Equiv 
[x=abs(-2),x=2]
 
[]
 
[assumepos]
 1 (ASSUMEPOSVARS, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{\text{Assume +ve vars}}&x=\left| -2\right| & \cr \color{green}{\Leftrightarrow}&x=2& \cr \end{array}\]
Equiv 
[abs(a)=abs(b),a^2=b^2]
 
[]
 
[assumepos]
 1 (ASSUMEPOSVARS, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{\text{Assume +ve vars}}&\left| a\right| =\left| b\right| & \cr \color{green}{\Leftrightarrow}&a^2=b^2& \cr \end{array}\]
Equiv 
[x^2=9,x=#pm#3,x=3 or x=-3,x=3
]
 
[]
 
[assumepos]
 1 (ASSUMEPOSVARS, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{\text{Assume +ve vars}}&x^2=9& \cr \color{green}{\Leftrightarrow}&x= \pm 3& \cr \color{green}{\Leftrightarrow}&x=3\,{\text{ or }}\, x=-3& \cr \color{green}{\Leftrightarrow}&x=3& \cr \end{array}\]
Equiv 
[x^2=9,x=3]
 
[]
 
[assumepos]
 1 (ASSUMEPOSVARS, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{\text{Assume +ve vars}}&x^2=9& \cr \color{green}{\Leftrightarrow}&x=3& \cr \end{array}\]
Equiv 
[x^2=2,x=#pm#sqrt(2),x=sqrt(2)
 or x=-sqrt(2)]
 
[]
 
[assumepos]
 1 (ASSUMEPOSVARS, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{\text{Assume +ve vars}}&x^2=2& \cr \color{green}{\Leftrightarrow}&x= \pm \sqrt{2}& \cr \color{green}{\Leftrightarrow}&x=\sqrt{2}\,{\text{ or }}\, x=-\sqrt{2}& \cr \end{array}\]
Equiv 
[x^2=2,x=sqrt(2)]
 
[]
 
[assumepos]
 1 (ASSUMEPOSVARS, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{\text{Assume +ve vars}}&x^2=2& \cr \color{green}{\Leftrightarrow}&x=\sqrt{2}& \cr \end{array}\]
Equiv 
[x^2 = a^2-b,x = sqrt(a^2-b)]
 
[]
 
[assumepos]
 1 (ASSUMEPOSVARS, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{\text{Assume +ve vars}}&x^2=a^2-b& \cr \color{green}{\Leftrightarrow}&x=\sqrt{a^2-b}& \cr \end{array}\]
Equiv 
[2*(x-3) = 4*x-3*(x+2),2*x-6=x
-6,x=0]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &2\cdot \left(x-3\right)=4\cdot x-3\cdot \left(x+2\right)& \cr \color{green}{\Leftrightarrow}&2\cdot x-6=x-6& \cr \color{green}{\Leftrightarrow}&x=0& \cr \end{array}\]
Equiv 
[2*(x-3) = 5*x-3*(x+2),2*x-6=2
*x-6,0=0,all]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &2\cdot \left(x-3\right)=5\cdot x-3\cdot \left(x+2\right)& \cr \color{green}{\Leftrightarrow}&2\cdot x-6=2\cdot x-6& \cr \color{green}{\Leftrightarrow}&0=0& \cr \color{green}{\Leftrightarrow}&\mathbb{R}& \cr \end{array}\]
Equiv 
[2*(x-3) = 5*x-3*(x+1),2*x-6=2
*x-3,0=3,{}]
 
[]
 1 (EMPTYCHAR,SAMEROOTS, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &2\cdot \left(x-3\right)=5\cdot x-3\cdot \left(x+1\right)& \cr \color{green}{\text{(Same roots)}}&2\cdot x-6=2\cdot x-3& \cr \color{green}{\Leftrightarrow}&0=3& \cr \color{green}{\Leftrightarrow}&\left \{ \right \}& \cr \end{array}\]
Equiv 
[a^2=b^2,a^2-b^2=0,(a-b)*(a+b)
=0,a=b or a=-b]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &a^2=b^2& \cr \color{green}{\Leftrightarrow}&a^2-b^2=0& \cr \color{green}{\Leftrightarrow}&\left(a-b\right)\cdot \left(a+b\right)=0& \cr \color{green}{\Leftrightarrow}&a=b\,{\text{ or }}\, a=-b& \cr \end{array}\]
Equiv 
[a^3=b^3,a^3-b^3=0,(a-b)*(a^2+
a*b+b^2)=0,(a-b)=0,a=b]
 
[]
 0 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR,IMPLIEDCHAR, EQUIVCHAR)
\[\begin{array}{lll} &a^3=b^3& \cr \color{green}{\Leftrightarrow}&a^3-b^3=0& \cr \color{green}{\Leftrightarrow}&\left(a-b\right)\cdot \left(a^2+a\cdot b+b^2\right)=0& \cr \color{red}{\Leftarrow}&a-b=0& \cr \color{green}{\Leftrightarrow}&a=b& \cr \end{array}\]
Equiv 
[a^3=b^3,a^3-b^3=0,(a-b)*(a^2+
a*b+b^2)=0,(a-b)=0 or (a^2+a*b
+b^2)=0, a=b or (a+(1+%i*sqrt(
3))/2*b)*(a+(1-%i*sqrt(3))/2*b
)=0, a=b or a=-(1+%i*sqrt(3))/
2*b or a=-(1-%i*sqrt(3))/2*b]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &a^3=b^3& \cr \color{green}{\Leftrightarrow}&a^3-b^3=0& \cr \color{green}{\Leftrightarrow}&\left(a-b\right)\cdot \left(a^2+a\cdot b+b^2\right)=0& \cr \color{green}{\Leftrightarrow}&a-b=0\,{\text{ or }}\, a^2+a\cdot b+b^2=0& \cr \color{green}{\Leftrightarrow}&a=b\,{\text{ or }}\, \left(a+\frac{1+\mathrm{i}\cdot \sqrt{3}}{2}\cdot b\right)\cdot \left(a+\frac{1-\mathrm{i}\cdot \sqrt{3}}{2}\cdot b\right)=0& \cr \color{green}{\Leftrightarrow}&a=b\,{\text{ or }}\, a=\frac{-\left(1+\mathrm{i}\cdot \sqrt{3}\right)}{2}\cdot b\,{\text{ or }}\, a=\frac{-\left(1-\mathrm{i}\cdot \sqrt{3}\right)}{2}\cdot b& \cr \end{array}\]
Equiv 
[x^2-x=30,x^2-x-30=0,(x-6)*(x+
5)=0,x-6=0 or x+5=0,x=6 or x=-
5]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &x^2-x=30& \cr \color{green}{\Leftrightarrow}&x^2-x-30=0& \cr \color{green}{\Leftrightarrow}&\left(x-6\right)\cdot \left(x+5\right)=0& \cr \color{green}{\Leftrightarrow}&x-6=0\,{\text{ or }}\, x+5=0& \cr \color{green}{\Leftrightarrow}&x=6\,{\text{ or }}\, x=-5& \cr \end{array}\]
Equiv 
[x^2=2,x^2-2=0,(x-sqrt(2))*(x+
sqrt(2))=0,x=sqrt(2) or x=-sqr
t(2)]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &x^2=2& \cr \color{green}{\Leftrightarrow}&x^2-2=0& \cr \color{green}{\Leftrightarrow}&\left(x-\sqrt{2}\right)\cdot \left(x+\sqrt{2}\right)=0& \cr \color{green}{\Leftrightarrow}&x=\sqrt{2}\,{\text{ or }}\, x=-\sqrt{2}& \cr \end{array}\]
Equiv 
[x^2=2,x=#pm#sqrt(2),x=sqrt(2)
 or x=-sqrt(2)]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &x^2=2& \cr \color{green}{\Leftrightarrow}&x= \pm \sqrt{2}& \cr \color{green}{\Leftrightarrow}&x=\sqrt{2}\,{\text{ or }}\, x=-\sqrt{2}& \cr \end{array}\]
Equiv 
[(2*x-7)^2=(x+1)^2,(2*x-7)^2 -
(x+1)^2=0,(2*x-7+x+1)*(2*x-7-x
-1)=0,(3*x-6)*(x-8)=0,x=2 or x
=8]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &{\left(2\cdot x-7\right)}^2={\left(x+1\right)}^2& \cr \color{green}{\Leftrightarrow}&{\left(2\cdot x-7\right)}^2-{\left(x+1\right)}^2=0& \cr \color{green}{\Leftrightarrow}&\left(2\cdot x-7+x+1\right)\cdot \left(2\cdot x-7-x-1\right)=0& \cr \color{green}{\Leftrightarrow}&\left(3\cdot x-6\right)\cdot \left(x-8\right)=0& \cr \color{green}{\Leftrightarrow}&x=2\,{\text{ or }}\, x=8& \cr \end{array}\]
Equiv 
[x^2-6*x=-9,(x-3)^2=0,x-3=0,x=
3]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR,SAMEROOTS, EQUIVCHAR)
\[\begin{array}{lll} &x^2-6\cdot x=-9& \cr \color{green}{\Leftrightarrow}&{\left(x-3\right)}^2=0& \cr \color{green}{\text{(Same roots)}}&x-3=0& \cr \color{green}{\Leftrightarrow}&x=3& \cr \end{array}\]
Equiv 
[(2*x-7)^2=(x+1)^2,sqrt((2*x-7
)^2)=sqrt((x+1)^2),2*x-7=x+1,x
=8]
 
[]
 0 (EMPTYCHAR, EQUIVCHAR,IMPLIEDCHAR, EQUIVCHAR)
\[\begin{array}{lll} &{\left(2\cdot x-7\right)}^2={\left(x+1\right)}^2& \cr \color{green}{\Leftrightarrow}&\sqrt{{\left(2\cdot x-7\right)}^2}=\sqrt{{\left(x+1\right)}^2}& \cr \color{red}{\Leftarrow}&2\cdot x-7=x+1& \cr \color{green}{\Leftrightarrow}&x=8& \cr \end{array}\]
Equiv 
[x^2-10*x+9 = 0, (x-5)^2-16 = 
0, (x-5)^2 =16, x-5 =#pm#4, x-
5 =4 or x-5=-4, x = 1 or x = 9
]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &x^2-10\cdot x+9=0& \cr \color{green}{\Leftrightarrow}&{\left(x-5\right)}^2-16=0& \cr \color{green}{\Leftrightarrow}&{\left(x-5\right)}^2=16& \cr \color{green}{\Leftrightarrow}&x-5= \pm 4& \cr \color{green}{\Leftrightarrow}&x-5=4\,{\text{ or }}\, x-5=-4& \cr \color{green}{\Leftrightarrow}&x=1\,{\text{ or }}\, x=9& \cr \end{array}\]
Equiv 
[x^2-2*p*x-q=0,x^2-2*p*x=q,x^2
-2*p*x+p^2=q+p^2,(x-p)^2=q+p^2
,x-p=#pm#sqrt(q+p^2),x-p=sqrt(
q+p^2) or x-p=-sqrt(q+p^2),x=p
+sqrt(q+p^2) or x=p-sqrt(q+p^2
)]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &x^2-2\cdot p\cdot x-q=0& \cr \color{green}{\Leftrightarrow}&x^2-2\cdot p\cdot x=q& \cr \color{green}{\Leftrightarrow}&x^2-2\cdot p\cdot x+p^2=q+p^2& \cr \color{green}{\Leftrightarrow}&{\left(x-p\right)}^2=q+p^2& \cr \color{green}{\Leftrightarrow}&x-p= \pm \sqrt{q+p^2}& \cr \color{green}{\Leftrightarrow}&x-p=\sqrt{q+p^2}\,{\text{ or }}\, x-p=-\sqrt{q+p^2}& \cr \color{green}{\Leftrightarrow}&x=p+\sqrt{q+p^2}\,{\text{ or }}\, x=p-\sqrt{q+p^2}& \cr \end{array}\]
Equiv 
[x^2-10*x+7=0,(x-5)^2-18=0,(x-
5)^2=sqrt(18)^2,(x-5)^2-sqrt(1
8)^2=0,(x-5-sqrt(18))*(x-5+sqr
t(18))=0,x=5-sqrt(18) or x=5+s
qrt(18)]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &x^2-10\cdot x+7=0& \cr \color{green}{\Leftrightarrow}&{\left(x-5\right)}^2-18=0& \cr \color{green}{\Leftrightarrow}&{\left(x-5\right)}^2={\sqrt{18}}^2& \cr \color{green}{\Leftrightarrow}&{\left(x-5\right)}^2-{\sqrt{18}}^2=0& \cr \color{green}{\Leftrightarrow}&\left(x-5-\sqrt{18}\right)\cdot \left(x-5+\sqrt{18}\right)=0& \cr \color{green}{\Leftrightarrow}&x=5-\sqrt{18}\,{\text{ or }}\, x=5+\sqrt{18}& \cr \end{array}\]
Equiv 
[9*x^2/2-81*x/2+90=5*x^2/2-5*x
-20,4*x^2-71*x+220 = 0,x = (71
 #pm# 39)/8,x=55/4 nounor x=4]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\frac{9\cdot x^2}{2}-\frac{81\cdot x}{2}+90=\frac{5\cdot x^2}{2}-5\cdot x-20& \cr \color{green}{\Leftrightarrow}&4\cdot x^2-71\cdot x+220=0& \cr \color{green}{\Leftrightarrow}&x=\frac{{71 \pm 39}}{8}& \cr \color{green}{\Leftrightarrow}&x=\frac{55}{4}\,{\text{ or }}\, x=4& \cr \end{array}\]
Equiv 
[(x-4)*(x-7)=-3*(x-4),x-7=-3,x
=4]
 
[]
 1 (EMPTYCHAR,SAMEROOTS, EQUIVCHAR)
\[\begin{array}{lll} &\left(x-4\right)\cdot \left(x-7\right)=-3\cdot \left(x-4\right)& \cr \color{green}{\text{(Same roots)}}&x-7=-3& \cr \color{green}{\Leftrightarrow}&x=4& \cr \end{array}\]
Equiv 
[x^2+2*a*x = 0, x*(x+2*a)=0, (
x+a-a)*(x+a+a)=0, (x+a)^2-a^2=
0]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &x^2+2\cdot a\cdot x=0& \cr \color{green}{\Leftrightarrow}&x\cdot \left(x+2\cdot a\right)=0& \cr \color{green}{\Leftrightarrow}&\left(x+a-a\right)\cdot \left(x+a+a\right)=0& \cr \color{green}{\Leftrightarrow}&{\left(x+a\right)}^2-a^2=0& \cr \end{array}\]
Equiv 
[x^3-1=0,(x-1)*(x^2+x+1)=0,x=1
]
 
[]
 0 (EMPTYCHAR, EQUIVCHAR,IMPLIEDCHAR)
\[\begin{array}{lll} &x^3-1=0& \cr \color{green}{\Leftrightarrow}&\left(x-1\right)\cdot \left(x^2+x+1\right)=0& \cr \color{red}{\Leftarrow}&x=1& \cr \end{array}\]
Equiv 
[x^3-1=0,(x-1)*(x^2+x+1)=0,x=1
 or x^2+x+1=0,x=1 or x = -(sqr
t(3)*%i+1)/2 or x=(sqrt(3)*%i-
1)/2]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &x^3-1=0& \cr \color{green}{\Leftrightarrow}&\left(x-1\right)\cdot \left(x^2+x+1\right)=0& \cr \color{green}{\Leftrightarrow}&x=1\,{\text{ or }}\, x^2+x+1=0& \cr \color{green}{\Leftrightarrow}&x=1\,{\text{ or }}\, x=\frac{-\left(\sqrt{3}\cdot \mathrm{i}+1\right)}{2}\,{\text{ or }}\, x=\frac{\sqrt{3}\cdot \mathrm{i}-1}{2}& \cr \end{array}\]
Equiv 
[a*x^2+b*x+c=0 or a=0,a^2*x^2+
a*b*x+a*c=0,(a*x)^2+b*(a*x)+a*
c=0, (a*x)^2+b*(a*x)+b^2/4-b^2
/4+a*c=0,(a*x+b/2)^2-b^2/4+a*c
=0,(a*x+b/2)^2=b^2/4-a*c, a*x+
b/2= #pm#sqrt(b^2/4-a*c),a*x=-
b/2+sqrt(b^2/4-a*c) or a*x=-b/
2-sqrt(b^2/4-a*c), (a=0 or x=(
-b+sqrt(b^2-4*a*c))/(2*a)) or 
(a=0 or x=(-b-sqrt(b^2-4*a*c))
/(2*a)), a^2=0 or x=(-b+sqrt(b
^2-4*a*c))/(2*a) or x=(-b-sqrt
(b^2-4*a*c))/(2*a)]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &a\cdot x^2+b\cdot x+c=0\,{\text{ or }}\, a=0& \cr \color{green}{\Leftrightarrow}&a^2\cdot x^2+a\cdot b\cdot x+a\cdot c=0& \cr \color{green}{\Leftrightarrow}&{\left(a\cdot x\right)}^2+b\cdot \left(a\cdot x\right)+a\cdot c=0& \cr \color{green}{\Leftrightarrow}&{\left(a\cdot x\right)}^2+b\cdot \left(a\cdot x\right)+\frac{b^2}{4}-\frac{b^2}{4}+a\cdot c=0& \cr \color{green}{\Leftrightarrow}&{\left(a\cdot x+\frac{b}{2}\right)}^2-\frac{b^2}{4}+a\cdot c=0& \cr \color{green}{\Leftrightarrow}&{\left(a\cdot x+\frac{b}{2}\right)}^2=\frac{b^2}{4}-a\cdot c& \cr \color{green}{\Leftrightarrow}&a\cdot x+\frac{b}{2}= \pm \sqrt{\frac{b^2}{4}-a\cdot c}& \cr \color{green}{\Leftrightarrow}&a\cdot x=-\frac{b}{2}+\sqrt{\frac{b^2}{4}-a\cdot c}\,{\text{ or }}\, a\cdot x=-\frac{b}{2}-\sqrt{\frac{b^2}{4}-a\cdot c}& \cr \color{green}{\Leftrightarrow}&a=0\,{\text{ or }}\, x=\frac{-b+\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}\,{\text{ or }}\, \left(a=0\,{\text{ or }}\, x=\frac{-b-\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}\right)& \cr \color{green}{\Leftrightarrow}&a^2=0\,{\text{ or }}\, x=\frac{-b+\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}\,{\text{ or }}\, x=\frac{-b-\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}& \cr \end{array}\]
Equiv 
[a*x^2+b*x=-c,4*a^2*x^2+4*a*b*
x+b^2=b^2-4*a*c,(2*a*x+b)^2=b^
2-4*a*c,2*a*x+b=#pm#sqrt(b^2-4
*a*c),2*a*x=-b#pm#sqrt(b^2-4*a
*c),x=(-b#pm#sqrt(b^2-4*a*c))/
(2*a)]
 
[]
 0 (EMPTYCHAR,IMPLIESCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR,QMCHAR)
\[\begin{array}{lll} &a\cdot x^2+b\cdot x=-c& \cr \color{red}{\Rightarrow}&4\cdot a^2\cdot x^2+4\cdot a\cdot b\cdot x+b^2=b^2-4\cdot a\cdot c& \cr \color{green}{\Leftrightarrow}&{\left(2\cdot a\cdot x+b\right)}^2=b^2-4\cdot a\cdot c& \cr \color{green}{\Leftrightarrow}&2\cdot a\cdot x+b= \pm \sqrt{b^2-4\cdot a\cdot c}& \cr \color{green}{\Leftrightarrow}&2\cdot a\cdot x={-b \pm \sqrt{b^2-4\cdot a\cdot c}}& \cr \color{red}{?}&x=\frac{{-b \pm \sqrt{b^2-4\cdot a\cdot c}}}{2\cdot a}& \cr \end{array}\]
Equiv 
[a*x^2+b*x=-c or a=0,4*a^2*x^2
+4*a*b*x+b^2=b^2-4*a*c,(2*a*x+
b)^2=b^2-4*a*c,2*a*x+b=#pm#sqr
t(b^2-4*a*c),2*a*x=-b#pm#sqrt(
b^2-4*a*c),x=(-b#pm#sqrt(b^2-4
*a*c))/(2*a) or a=0]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &a\cdot x^2+b\cdot x=-c\,{\text{ or }}\, a=0& \cr \color{green}{\Leftrightarrow}&4\cdot a^2\cdot x^2+4\cdot a\cdot b\cdot x+b^2=b^2-4\cdot a\cdot c& \cr \color{green}{\Leftrightarrow}&{\left(2\cdot a\cdot x+b\right)}^2=b^2-4\cdot a\cdot c& \cr \color{green}{\Leftrightarrow}&2\cdot a\cdot x+b= \pm \sqrt{b^2-4\cdot a\cdot c}& \cr \color{green}{\Leftrightarrow}&2\cdot a\cdot x={-b \pm \sqrt{b^2-4\cdot a\cdot c}}& \cr \color{green}{\Leftrightarrow}&x=\frac{{-b \pm \sqrt{b^2-4\cdot a\cdot c}}}{2\cdot a}\,{\text{ or }}\, a=0& \cr \end{array}\]
Equiv 
[sqrt(3*x+4) = 2+sqrt(x+2), 3*
x+4=4+4*sqrt(x+2)+(x+2),x-1=2*
sqrt(x+2),x^2-2*x+1 = 4*x+8,x^
2-6*x-7 = 0,(x-7)*(x+1) = 0,x=
7 or x=-1]
 
[]
 0 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR,IMPLIESCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\sqrt{3\cdot x+4}=2+\sqrt{x+2}&{\color{blue}{{x \in {\left[ -\frac{4}{3},\, \infty \right)}}}}\cr \color{green}{\Leftrightarrow}&3\cdot x+4=4+4\cdot \sqrt{x+2}+\left(x+2\right)&{\color{blue}{{x \in {\left[ -2,\, \infty \right)}}}}\cr \color{green}{\Leftrightarrow}&x-1=2\cdot \sqrt{x+2}&{\color{blue}{{x \in {\left[ -2,\, \infty \right)}}}}\cr \color{red}{\Rightarrow}&x^2-2\cdot x+1=4\cdot x+8& \cr \color{green}{\Leftrightarrow}&x^2-6\cdot x-7=0& \cr \color{green}{\Leftrightarrow}&\left(x-7\right)\cdot \left(x+1\right)=0& \cr \color{green}{\Leftrightarrow}&x=7\,{\text{ or }}\, x=-1& \cr \end{array}\]
Equiv 
[sqrt(3*x+4) = 2+sqrt(x+2), 3*
x+4=4+4*sqrt(x+2)+(x+2),x-1=2*
sqrt(x+2),x^2-2*x+1 = 4*x+8,x^
2-6*x-7 = 0,(x-7)*(x+1) = 0,x=
7 or x=-1,x=7]
 
[]
 
[assumepos]
 1 (ASSUMEPOSVARS, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{\text{Assume +ve vars}}&\sqrt{3\cdot x+4}=2+\sqrt{x+2}&{\color{blue}{{x \in {\left[ 0,\, \infty \right)}}}}\cr \color{green}{\Leftrightarrow}&3\cdot x+4=4+4\cdot \sqrt{x+2}+\left(x+2\right)&{\color{blue}{{x \in {\left[ 0,\, \infty \right)}}}}\cr \color{green}{\Leftrightarrow}&x-1=2\cdot \sqrt{x+2}&{\color{blue}{{x \in {\left[ 0,\, \infty \right)}}}}\cr \color{green}{\Leftrightarrow}&x^2-2\cdot x+1=4\cdot x+8& \cr \color{green}{\Leftrightarrow}&x^2-6\cdot x-7=0& \cr \color{green}{\Leftrightarrow}&\left(x-7\right)\cdot \left(x+1\right)=0& \cr \color{green}{\Leftrightarrow}&x=7\,{\text{ or }}\, x=-1& \cr \color{green}{\Leftrightarrow}&x=7& \cr \end{array}\]
Equiv 
[x*(x-1)*(x-2)=0,x*(x-1)=0,x*(
x-1)*(x-2)=0,x*(x^2-2)=0]
 
[]
 0 (EMPTYCHAR,IMPLIEDCHAR,IMPLIESCHAR,QMCHAR)
\[\begin{array}{lll} &x\cdot \left(x-1\right)\cdot \left(x-2\right)=0& \cr \color{red}{\Leftarrow}&x\cdot \left(x-1\right)=0& \cr \color{red}{\Rightarrow}&x\cdot \left(x-1\right)\cdot \left(x-2\right)=0& \cr \color{red}{?}&x\cdot \left(x^2-2\right)=0& \cr \end{array}\]
Equiv 
[x^2-6*x=-9,x=3]
 
[]
 1 (EMPTYCHAR,SAMEROOTS)
\[\begin{array}{lll} &x^2-6\cdot x=-9& \cr \color{green}{\text{(Same roots)}}&x=3& \cr \end{array}\]
Equiv 
[x=1 nounor x=-2 nounor x=1,x^
3-3*x=-2,x=1 nounor x=-2]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR,SAMEROOTS)
\[\begin{array}{lll} &x=1\,{\text{ or }}\, x=-2\,{\text{ or }}\, x=1& \cr \color{green}{\Leftrightarrow}&x^3-3\cdot x=-2& \cr \color{green}{\text{(Same roots)}}&x=1\,{\text{ or }}\, x=-2& \cr \end{array}\]
Equiv 
[9*x^3-24*x^2+13*x=2,x=1/3 nou
nor x=2]
 
[]
 1 (EMPTYCHAR,SAMEROOTS)
\[\begin{array}{lll} &9\cdot x^3-24\cdot x^2+13\cdot x=2& \cr \color{green}{\text{(Same roots)}}&x=\frac{1}{3}\,{\text{ or }}\, x=2& \cr \end{array}\]
Equiv 
[(x-2)^43*(x+1/3)^60=0,(3*x+1)
^4*(x-2)^2=0,x=-1/3 nounor x=2
]
 
[]
 1 (EMPTYCHAR,SAMEROOTS,SAMEROOTS)
\[\begin{array}{lll} &{\left(x-2\right)}^{43}\cdot {\left(x+\frac{1}{3}\right)}^{60}=0& \cr \color{green}{\text{(Same roots)}}&{\left(3\cdot x+1\right)}^4\cdot {\left(x-2\right)}^2=0& \cr \color{green}{\text{(Same roots)}}&x=\frac{-1}{3}\,{\text{ or }}\, x=2& \cr \end{array}\]
Equiv 
[2^x=4,x*log(2)=log(4),x=log(2
^2)/log(2),x=2*log(2)/log(2),x
=2]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &2^{x}=4& \cr \color{green}{\Leftrightarrow}&x\cdot \ln \left( 2 \right)=\ln \left( 4 \right)& \cr \color{green}{\Leftrightarrow}&x=\frac{\ln \left( 2^2 \right)}{\ln \left( 2 \right)}& \cr \color{green}{\Leftrightarrow}&x=\frac{2\cdot \ln \left( 2 \right)}{\ln \left( 2 \right)}& \cr \color{green}{\Leftrightarrow}&x=2& \cr \end{array}\]
Equiv 
[x^log(y),stackeq(e^(log(x)*lo
g(y))),stackeq(e^(log(y)*log(x
))),stackeq(y^log(x))]
 
[]
 1 (EMPTYCHAR, CHECKMARK, CHECKMARK, CHECKMARK)
\[\begin{array}{lll} &x^{\ln \left( y \right)}& \cr \color{green}{\checkmark}&=e^{\ln \left( x \right)\cdot \ln \left( y \right)}& \cr \color{green}{\checkmark}&=e^{\ln \left( y \right)\cdot \ln \left( x \right)}& \cr \color{green}{\checkmark}&=y^{\ln \left( x \right)}& \cr \end{array}\]
Equiv 
[lg(x+17,3)-2=lg(2*x,3),lg(x+1
7,3)-lg(2*x,3)=2,lg((x+17)/(2*
x),3)=2,(x+17)/(2*x)=3^2,(x+17
)=18*x,17*x=17,x=1]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR,EQUIVLOG, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\log_{3}\left(x+17\right)-2=\log_{3}\left(2\cdot x\right)&{\color{blue}{{x \in {\left( 0,\, \infty \right)}}}}\cr \color{green}{\Leftrightarrow}&\log_{3}\left(x+17\right)-\log_{3}\left(2\cdot x\right)=2&{\color{blue}{{x \in {\left( 0,\, \infty \right)}}}}\cr \color{green}{\Leftrightarrow}&\log_{3}\left(\frac{x+17}{2\cdot x}\right)=2& \cr \color{green}{\log(?)}&\frac{x+17}{2\cdot x}=3^2&{\color{blue}{{x \not\in {\left \{0 \right \}}}}}\cr \color{green}{\Leftrightarrow}&x+17=18\cdot x& \cr \color{green}{\Leftrightarrow}&17\cdot x=17& \cr \color{green}{\Leftrightarrow}&x=1& \cr \end{array}\]
Equiv 
[a=logbase(9,3),3^a=9,3^a=3^2,
a=2]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &a=\log_{3}\left(9\right)& \cr \color{green}{\Leftrightarrow}&3^{a}=9& \cr \color{green}{\Leftrightarrow}&3^{a}=3^2& \cr \color{green}{\Leftrightarrow}&a=2& \cr \end{array}\]
Equiv 
[x=(1+y/n)^n,x^(1/n)=(1+y/n),y
/n=x^(1/n)-1,y=n*(x^(1/n)-1)]
 
[]
 0 (EMPTYCHAR,QMCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &x={\left(1+\frac{y}{n}\right)}^{n}& \cr \color{red}{?}&x^{\frac{1}{n}}=1+\frac{y}{n}& \cr \color{green}{\Leftrightarrow}&\frac{y}{n}=x^{\frac{1}{n}}-1& \cr \color{green}{\Leftrightarrow}&y=n\cdot \left(x^{\frac{1}{n}}-1\right)& \cr \end{array}\]
Equiv 
[a^3=b^3,a^3-b^3=0,(a-b)*(a^2+
a*b+b^2)=0,(a-b)=0,a=b]
 
[]
 
[assumereal]
 0 (ASSUMEREALVARS, EQUIVCHAR, EQUIVCHAR,IMPLIEDCHAR, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{(\mathbb{R})}&a^3=b^3& \cr \color{green}{\Leftrightarrow}&a^3-b^3=0& \cr \color{green}{\Leftrightarrow}&\left(a-b\right)\cdot \left(a^2+a\cdot b+b^2\right)=0& \cr \color{red}{\Leftarrow}&a-b=0& \cr \color{green}{\Leftrightarrow}&a=b& \cr \end{array}\]
Equiv 
[x^3-1=0,(x-1)*(x^2+x+1)=0,x=1
]
 
[]
 
[assumereal]
 1 (ASSUMEREALVARS, EQUIVCHAR, EQUIVCHARREAL)
\[\begin{array}{lll}\color{blue}{(\mathbb{R})}&x^3-1=0& \cr \color{green}{\Leftrightarrow}&\left(x-1\right)\cdot \left(x^2+x+1\right)=0& \cr \color{green}{\Leftrightarrow}\, \color{blue}{(\mathbb{R})}&x=1& \cr \end{array}\]
Equiv 
[x^4=2,x^4-2=0,(x^2-sqrt(2))*(
x^2+sqrt(2))=0,x^2=sqrt(2),x=#
pm# 2^(1/4)]
 
[]
 
[assumereal]
 1 (ASSUMEREALVARS, EQUIVCHAR, EQUIVCHAR, EQUIVCHARREAL, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{(\mathbb{R})}&x^4=2& \cr \color{green}{\Leftrightarrow}&x^4-2=0& \cr \color{green}{\Leftrightarrow}&\left(x^2-\sqrt{2}\right)\cdot \left(x^2+\sqrt{2}\right)=0& \cr \color{green}{\Leftrightarrow}\, \color{blue}{(\mathbb{R})}&x^2=\sqrt{2}& \cr \color{green}{\Leftrightarrow}&x= \pm 2^{\frac{1}{4}}& \cr \end{array}\]
Equiv 
[6*x-12=3*(x-2),6*x-12+3*(x-2)
=0,9*x-18=0,x=2]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &6\cdot x-12=3\cdot \left(x-2\right)& \cr \color{green}{\Leftrightarrow}&6\cdot x-12+3\cdot \left(x-2\right)=0& \cr \color{green}{\Leftrightarrow}&9\cdot x-18=0& \cr \color{green}{\Leftrightarrow}&x=2& \cr \end{array}\]
Equiv 
[x^2-6*x+9=0,x^2-6*x=-9,x*(x-6
)=3*-3,x=3 or x-6=-3,x=3]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR,SAMEROOTS)
\[\begin{array}{lll} &x^2-6\cdot x+9=0& \cr \color{green}{\Leftrightarrow}&x^2-6\cdot x=-9& \cr \color{green}{\Leftrightarrow}&x\cdot \left(x-6\right)=3\cdot \left(-3\right)& \cr \color{green}{\Leftrightarrow}&x=3\,{\text{ or }}\, x-6=-3& \cr \color{green}{\text{(Same roots)}}&x=3& \cr \end{array}\]
Equiv 
[(x+3)*(2-x)=4,x+3=4 or (2-x)=
4,x=1 or x=-2]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\left(x+3\right)\cdot \left(2-x\right)=4& \cr \color{green}{\Leftrightarrow}&x+3=4\,{\text{ or }}\, 2-x=4& \cr \color{green}{\Leftrightarrow}&x=1\,{\text{ or }}\, x=-2& \cr \end{array}\]
Equiv 
[(x-p)*(x-q)=0,x^2-p*x-q*x+p*q
=0,1+q-x-p-p*q+p*x+x+q*x-x^2=1
-p+q,(1+q-x)*(1-p+x)=1-p+q,(1+
q-x)=1-p+q or (1-p+x)=1-p+q,x=
p or x=q]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\left(x-p\right)\cdot \left(x-q\right)=0& \cr \color{green}{\Leftrightarrow}&x^2-p\cdot x+\left(-q\right)\cdot x+p\cdot q=0& \cr \color{green}{\Leftrightarrow}&1+q-x-p+\left(-p\right)\cdot q+p\cdot x+x+q\cdot x-x^2=1-p+q& \cr \color{green}{\Leftrightarrow}&\left(1+q-x\right)\cdot \left(1-p+x\right)=1-p+q& \cr \color{green}{\Leftrightarrow}&1+q-x=1-p+q\,{\text{ or }}\, 1-p+x=1-p+q& \cr \color{green}{\Leftrightarrow}&x=p\,{\text{ or }}\, x=q& \cr \end{array}\]
Equiv 
[a=b, a^2=a*b, a^2-b^2=a*b-b^2
, (a-b)*(a+b)=b*(a-b), a+b=b, 
2*a=a, 1=2]
 
[]
 0 (EMPTYCHAR,IMPLIESCHAR, EQUIVCHAR, EQUIVCHAR,IMPLIEDCHAR, EQUIVCHAR,IMPLIEDCHAR)
\[\begin{array}{lll} &a=b& \cr \color{red}{\Rightarrow}&a^2=a\cdot b& \cr \color{green}{\Leftrightarrow}&a^2-b^2=a\cdot b-b^2& \cr \color{green}{\Leftrightarrow}&\left(a-b\right)\cdot \left(a+b\right)=b\cdot \left(a-b\right)& \cr \color{red}{\Leftarrow}&a+b=b& \cr \color{green}{\Leftrightarrow}&2\cdot a=a& \cr \color{red}{\Leftarrow}&1=2& \cr \end{array}\]
Equiv 
[a=b or a=0, a^2=a*b, a^2-b^2=
a*b-b^2, (a-b)*(a+b)=b*(a-b), 
a+b=b or a-b=0, 2*a=a or a=b, 
2=1 or a=0 or a=b, a=0 or a=b]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &a=b\,{\text{ or }}\, a=0& \cr \color{green}{\Leftrightarrow}&a^2=a\cdot b& \cr \color{green}{\Leftrightarrow}&a^2-b^2=a\cdot b-b^2& \cr \color{green}{\Leftrightarrow}&\left(a-b\right)\cdot \left(a+b\right)=b\cdot \left(a-b\right)& \cr \color{green}{\Leftrightarrow}&a+b=b\,{\text{ or }}\, a-b=0& \cr \color{green}{\Leftrightarrow}&2\cdot a=a\,{\text{ or }}\, a=b& \cr \color{green}{\Leftrightarrow}&2=1\,{\text{ or }}\, a=0\,{\text{ or }}\, a=b& \cr \color{green}{\Leftrightarrow}&a=0\,{\text{ or }}\, a=b& \cr \end{array}\]
Equiv 
[(x^2-4)/(x-2)=0,(x-2)*(x+2)/(
x-2)=0,x+2=0,x=-2]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\frac{x^2-4}{x-2}=0&{\color{blue}{{x \not\in {\left \{2 \right \}}}}}\cr \color{green}{\Leftrightarrow}&\frac{\left(x-2\right)\cdot \left(x+2\right)}{x-2}=0& \cr \color{green}{\Leftrightarrow}&x+2=0& \cr \color{green}{\Leftrightarrow}&x=-2& \cr \end{array}\]
Equiv 
[(x^2-4)/(x-2)=0,(x^2-4)=0,(x-
2)*(x+2)=0,x=-2 or x=2]
 
[]
 0 (EMPTYCHAR,IMPLIESCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\frac{x^2-4}{x-2}=0&{\color{blue}{{x \not\in {\left \{2 \right \}}}}}\cr \color{red}{\Rightarrow}&x^2-4=0& \cr \color{green}{\Leftrightarrow}&\left(x-2\right)\cdot \left(x+2\right)=0& \cr \color{green}{\Leftrightarrow}&x=-2\,{\text{ or }}\, x=2& \cr \end{array}\]
Equiv 
[5*x/(2*x+1)-3/(x+1) = 1,5*x*(
x+1)-3*(2*x+1)=(x+1)*(2*x+1),5
*x^2+5*x-6*x-3=2*x^2+3*x+1,3*x
^2-4*x-4=0,(x-2)*(3*x+2)=0,x=2
 or x=-2/3]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\frac{5\cdot x}{2\cdot x+1}-\frac{3}{x+1}=1&{\color{blue}{{x \not\in {\left \{-1 , -\frac{1}{2} \right \}}}}}\cr \color{green}{\Leftrightarrow}&5\cdot x\cdot \left(x+1\right)-3\cdot \left(2\cdot x+1\right)=\left(x+1\right)\cdot \left(2\cdot x+1\right)& \cr \color{green}{\Leftrightarrow}&5\cdot x^2+5\cdot x-6\cdot x-3=2\cdot x^2+3\cdot x+1& \cr \color{green}{\Leftrightarrow}&3\cdot x^2-4\cdot x-4=0& \cr \color{green}{\Leftrightarrow}&\left(x-2\right)\cdot \left(3\cdot x+2\right)=0& \cr \color{green}{\Leftrightarrow}&x=2\,{\text{ or }}\, x=\frac{-2}{3}& \cr \end{array}\]
Equiv 
[(x+10)/(x-6)-5= (4*x-40)/(13-
x),(x+10-5*(x-6))/(x-6)= (4*x-
40)/(13-x), (4*x-40)/(6-x)= (4
*x-40)/(13-x),6-x= 13-x,6= 13]
 
[]
 0 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR,QMCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\frac{x+10}{x-6}-5=\frac{4\cdot x-40}{13-x}&{\color{blue}{{x \not\in {\left \{6 , 13 \right \}}}}}\cr \color{green}{\Leftrightarrow}&\frac{x+10-5\cdot \left(x-6\right)}{x-6}=\frac{4\cdot x-40}{13-x}&{\color{blue}{{x \not\in {\left \{6 , 13 \right \}}}}}\cr \color{green}{\Leftrightarrow}&\frac{4\cdot x-40}{6-x}=\frac{4\cdot x-40}{13-x}&{\color{blue}{{x \not\in {\left \{6 , 13 \right \}}}}}\cr \color{red}{?}&6-x=13-x& \cr \color{green}{\Leftrightarrow}&6=13& \cr \end{array}\]
Equiv 
[(x+5)/(x-7)-5= (4*x-40)/(13-x
),(x+5-5*(x-7))/(x-7)= (4*x-40
)/(13-x), (4*x-40)/(7-x)= (4*x
-40)/(13-x),7-x= 13-x,7= 13]
 
[]
 0 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR,IMPLIEDCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\frac{x+5}{x-7}-5=\frac{4\cdot x-40}{13-x}&{\color{blue}{{x \not\in {\left \{7 , 13 \right \}}}}}\cr \color{green}{\Leftrightarrow}&\frac{x+5-5\cdot \left(x-7\right)}{x-7}=\frac{4\cdot x-40}{13-x}&{\color{blue}{{x \not\in {\left \{7 , 13 \right \}}}}}\cr \color{green}{\Leftrightarrow}&\frac{4\cdot x-40}{7-x}=\frac{4\cdot x-40}{13-x}&{\color{blue}{{x \not\in {\left \{7 , 13 \right \}}}}}\cr \color{red}{\Leftarrow}&7-x=13-x& \cr \color{green}{\Leftrightarrow}&7=13& \cr \end{array}\]
Equiv 
[(x+5)/(x-7)-5= (4*x-40)/(13-x
),(x+5-5*(x-7))/(x-7)= (4*x-40
)/(13-x), (4*x-40)/(7-x)= (4*x
-40)/(13-x),7-x= 13-x or 4*x-4
0=0,7= 13 or 4*x=40,x=10]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\frac{x+5}{x-7}-5=\frac{4\cdot x-40}{13-x}&{\color{blue}{{x \not\in {\left \{7 , 13 \right \}}}}}\cr \color{green}{\Leftrightarrow}&\frac{x+5-5\cdot \left(x-7\right)}{x-7}=\frac{4\cdot x-40}{13-x}&{\color{blue}{{x \not\in {\left \{7 , 13 \right \}}}}}\cr \color{green}{\Leftrightarrow}&\frac{4\cdot x-40}{7-x}=\frac{4\cdot x-40}{13-x}&{\color{blue}{{x \not\in {\left \{7 , 13 \right \}}}}}\cr \color{green}{\Leftrightarrow}&7-x=13-x\,{\text{ or }}\, 4\cdot x-40=0& \cr \color{green}{\Leftrightarrow}&7=13\,{\text{ or }}\, 4\cdot x=40& \cr \color{green}{\Leftrightarrow}&x=10& \cr \end{array}\]
Equiv 
[1/(a-b)-1/(b-a),stackeq(1/(a-
b)+1/(b-a))]
 
[]
 0 (EMPTYCHAR,QMCHAR)
\[\begin{array}{lll} &\frac{1}{a-b}-\frac{1}{b-a}& \cr \color{red}{?}&=\frac{1}{a-b}+\frac{1}{b-a}& \cr \end{array}\]
Equiv 
[a*x^2+b*x+c=0,a=0 nounand b=0
 nounand c=0,a*x^2+b*x+c=0]
 
[]
 1 (EMPTYCHAR,EQUATECOEFFLOSS(x),EQUATECOEFFGAIN(x))
\[\begin{array}{lll} &a\cdot x^2+b\cdot x+c=0& \cr \color{green}{\equiv (\cdots ? x)}&\left\{\begin{array}{l}a=0\cr b=0\cr c=0\cr \end{array}\right.& \cr \color{green}{(\cdots ? x)\equiv}&a\cdot x^2+b\cdot x+c=0& \cr \end{array}\]
Equiv 
[a*x^2+b*x+c=A*x^2+B*x+C,a=A n
ounand b=B nounand c=C,a*x^2+b
*x+c=A*x^2+B*x+C]
 
[]
 1 (EMPTYCHAR,EQUATECOEFFLOSS(x),EQUATECOEFFGAIN(x))
\[\begin{array}{lll} &a\cdot x^2+b\cdot x+c=A\cdot x^2+B\cdot x+C& \cr \color{green}{\equiv (\cdots ? x)}&\left\{\begin{array}{l}a=A\cr b=B\cr c=C\cr \end{array}\right.& \cr \color{green}{(\cdots ? x)\equiv}&a\cdot x^2+b\cdot x+c=A\cdot x^2+B\cdot x+C& \cr \end{array}\]
Equiv 
[(x-1)*(x+4), stackeq(x^2-x+4*
x-4),stackeq(x^2+3*x-4)]
 
[]
 1 (EMPTYCHAR, CHECKMARK, CHECKMARK)
\[\begin{array}{lll} &\left(x-1\right)\cdot \left(x+4\right)& \cr \color{green}{\checkmark}&=x^2-x+4\cdot x-4& \cr \color{green}{\checkmark}&=x^2+3\cdot x-4& \cr \end{array}\]
Equiv 
[(x-1)*(x+4), stackeq(x^2-x+4*
x-4),stackeq(x^2+3*x-4)]
 
[]
 1 (EMPTYCHAR, CHECKMARK, CHECKMARK)
\[\begin{array}{lll} &\left(x-1\right)\cdot \left(x+4\right)& \cr \color{green}{\checkmark}&=x^2-x+4\cdot x-4& \cr \color{green}{\checkmark}&=x^2+3\cdot x-4& \cr \end{array}\]
Equiv 
[x^2-2,stackeq((x-sqrt(2))*(x+
sqrt(2)))]
 
[]
 1 (EMPTYCHAR, CHECKMARK)
\[\begin{array}{lll} &x^2-2& \cr \color{green}{\checkmark}&=\left(x-\sqrt{2}\right)\cdot \left(x+\sqrt{2}\right)& \cr \end{array}\]
Equiv 
[x^2+4,stackeq((x-2*i)*(x+2*i)
)]
 
[]
 1 (EMPTYCHAR, CHECKMARK)
\[\begin{array}{lll} &x^2+4& \cr \color{green}{\checkmark}&=\left(x-2\cdot \mathrm{i}\right)\cdot \left(x+2\cdot \mathrm{i}\right)& \cr \end{array}\]
Equiv 
[x^2+2*a*x,x^2+2*a*x+a^2-a^2,(
x+a)^2-a^2]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &x^2+2\cdot a\cdot x& \cr \color{green}{\Leftrightarrow}&x^2+2\cdot a\cdot x+a^2-a^2& \cr \color{green}{\Leftrightarrow}&{\left(x+a\right)}^2-a^2& \cr \end{array}\]
Equiv 
[x^2+2*a*x,stackeq(x^2+2*a*x+a
^2-a^2),stackeq((x+a)^2-a^2)]
 
[]
 1 (EMPTYCHAR, CHECKMARK, CHECKMARK)
\[\begin{array}{lll} &x^2+2\cdot a\cdot x& \cr \color{green}{\checkmark}&=x^2+2\cdot a\cdot x+a^2-a^2& \cr \color{green}{\checkmark}&={\left(x+a\right)}^2-a^2& \cr \end{array}\]
Equiv 
[(y-z)/(y*z)+(z-x)/(z*x)+(x-y)
/(x*y),(x*(y-z)+y*(z-x)+z*(x-y
))/(x*y*z),0]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\frac{y-z}{y\cdot z}+\frac{z-x}{z\cdot x}+\frac{x-y}{x\cdot y}& \cr \color{green}{\Leftrightarrow}&\frac{x\cdot \left(y-z\right)+y\cdot \left(z-x\right)+z\cdot \left(x-y\right)}{x\cdot y\cdot z}& \cr \color{green}{\Leftrightarrow}&0& \cr \end{array}\]
Equiv 
[(y-z)/(y*z)+(z-x)/(z*x)+(x-y)
/(x*y),stackeq((x*(y-z)+y*(z-x
)+z*(x-y))/(x*y*z)),stackeq(0)
]
 
[]
 1 (EMPTYCHAR, CHECKMARK, CHECKMARK)
\[\begin{array}{lll} &\frac{y-z}{y\cdot z}+\frac{z-x}{z\cdot x}+\frac{x-y}{x\cdot y}& \cr \color{green}{\checkmark}&=\frac{x\cdot \left(y-z\right)+y\cdot \left(z-x\right)+z\cdot \left(x-y\right)}{x\cdot y\cdot z}& \cr \color{green}{\checkmark}&=0& \cr \end{array}\]
Equiv 
[2*(a^2*b^2+b^2*c^2+c^2*a^2)-(
a^4+b^4+c^4),stackeq(4*a^2*b^2
-(a^4+b^4+c^4+2*a^2*b^2-2*b^2*
c^2-2*c^2*a^2)),stackeq((2*a*b
)^2-(b^2+a^2-c^2)^2,(2*a*b+b^2
+a^2-c^2)*(2*a*b-b^2-a^2+c^2))
,stackeq(((a+b)^2-c^2)*(c^2-(a
-b)^2)),stackeq((a+b+c)*(a+b-c
)*(c+a-b)*(c-a+b))]
 
[]
 1 (EMPTYCHAR, CHECKMARK, CHECKMARK, CHECKMARK, CHECKMARK)
\[\begin{array}{lll} &2\cdot \left(a^2\cdot b^2+b^2\cdot c^2+c^2\cdot a^2\right)-\left(a^4+b^4+c^4\right)& \cr \color{green}{\checkmark}&=4\cdot a^2\cdot b^2-\left(a^4+b^4+c^4+2\cdot a^2\cdot b^2-2\cdot b^2\cdot c^2-2\cdot c^2\cdot a^2\right)& \cr \color{green}{\checkmark}&={\left(2\cdot a\cdot b\right)}^2-{\left(b^2+a^2-c^2\right)}^2& \cr \color{green}{\checkmark}&=\left({\left(a+b\right)}^2-c^2\right)\cdot \left(c^2-{\left(a-b\right)}^2\right)& \cr \color{green}{\checkmark}&=\left(a+b+c\right)\cdot \left(a+b-c\right)\cdot \left(c+a-b\right)\cdot \left(c-a+b\right)& \cr \end{array}\]
Equiv 
[abs(x-1/2)+abs(x+1/2)-2,stack
eq(abs(x)-1)]
 
[]
 0 (EMPTYCHAR,QMCHAR)
\[\begin{array}{lll} &\left| x-\frac{1}{2}\right| +\left| x+\frac{1}{2}\right| -2& \cr \color{red}{?}&=\left| x\right| -1& \cr \end{array}\]
Equiv 
[11*sqrt(abs(x)+1)=25-x,11^2*(
abs(x)+1)=(25-x)^2,11^2*abs(x)
=(25-x)^2-11^2,11^4*x^2=((25-x
)^2-11^2)^2, ((25-x)^2-11^2)^2
-11^4*x^2=0,((25-x)^2-11^2-11^
2*x)*((25-x)^2-11^2+11^2*x)=0,
(x^2-50*x+504-121*x)*(x^2-50*x
+504+121*x)=0, (x-168)*(x-3)*(
x+8)*(x+63)=0]
 
[]
 0 (EMPTYCHAR,QMCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &11\cdot \sqrt{\left| x\right| +1}=25-x& \cr \color{red}{?}&11^2\cdot \left(\left| x\right| +1\right)={\left(25-x\right)}^2& \cr \color{green}{\Leftrightarrow}&11^2\cdot \left| x\right| ={\left(25-x\right)}^2-11^2& \cr \color{green}{\Leftrightarrow}&11^4\cdot x^2={\left({\left(25-x\right)}^2-11^2\right)}^2& \cr \color{green}{\Leftrightarrow}&{\left({\left(25-x\right)}^2-11^2\right)}^2-11^4\cdot x^2=0& \cr \color{green}{\Leftrightarrow}&\left({\left(25-x\right)}^2-11^2+\left(-11^2\right)\cdot x\right)\cdot \left({\left(25-x\right)}^2-11^2+11^2\cdot x\right)=0& \cr \color{green}{\Leftrightarrow}&\left(x^2-50\cdot x+504-121\cdot x\right)\cdot \left(x^2-50\cdot x+504+121\cdot x\right)=0& \cr \color{green}{\Leftrightarrow}&\left(x-168\right)\cdot \left(x-3\right)\cdot \left(x+8\right)\cdot \left(x+63\right)=0& \cr \end{array}\]
Equiv 
[1/(x^2+1)=1/((x+%i)*(x-%i)), 
stackeq(1/(2*%i)*(1/(x-%i)-1/(
x+%i)))]
 
[]
 1 (CHECKMARK, CHECKMARK)
\[\begin{array}{lll}\color{green}{\checkmark}&\frac{1}{x^2+1}=\frac{1}{\left(x+\mathrm{i}\right)\cdot \left(x-\mathrm{i}\right)}& \cr \color{green}{\checkmark}&=\frac{1}{2\cdot \mathrm{i}}\cdot \left(\frac{1}{x-\mathrm{i}}-\frac{1}{x+\mathrm{i}}\right)& \cr \end{array}\]
Equiv 
[((a-b)/(a^2+a*b))/((a^2-2*a*b
+b^2)/(a^4-b^4)),stackeq(((a-b
)*(a-b)*(a+b)*(a^2+b^2))/(a*(a
+b)*(a-b)^2)),stackeq((a^2+b^2
)/a),stackeq(a+b^2/a)]
 
[]
 1 (EMPTYCHAR, CHECKMARK, CHECKMARK, CHECKMARK)
\[\begin{array}{lll} &\frac{\frac{a-b}{a^2+a\cdot b}}{\frac{a^2-2\cdot a\cdot b+b^2}{a^4-b^4}}& \cr \color{green}{\checkmark}&=\frac{\left(a-b\right)\cdot \left(a-b\right)\cdot \left(a+b\right)\cdot \left(a^2+b^2\right)}{a\cdot \left(a+b\right)\cdot {\left(a-b\right)}^2}& \cr \color{green}{\checkmark}&=\frac{a^2+b^2}{a}& \cr \color{green}{\checkmark}&=a+\frac{b^2}{a}& \cr \end{array}\]
Equiv 
[a^4+4*b^4,stackeq((a^2)^2+4*a
^2*b^2+(2*b^2)^2-4*a^2*b^2),st
ackeq((a^2+2*b^2)^2-(2*a*b)^2)
,stackeq((2*b^2-2*a*b+a^2)*(2*
b^2+2*a*b+a^2))]
 
[]
 1 (EMPTYCHAR, CHECKMARK, CHECKMARK, CHECKMARK)
\[\begin{array}{lll} &a^4+4\cdot b^4& \cr \color{green}{\checkmark}&={\left(a^2\right)}^2+4\cdot a^2\cdot b^2+{\left(2\cdot b^2\right)}^2-4\cdot a^2\cdot b^2& \cr \color{green}{\checkmark}&={\left(a^2+2\cdot b^2\right)}^2-{\left(2\cdot a\cdot b\right)}^2& \cr \color{green}{\checkmark}&=\left(2\cdot b^2-2\cdot a\cdot b+a^2\right)\cdot \left(2\cdot b^2+2\cdot a\cdot b+a^2\right)& \cr \end{array}\]
Equiv 
[sum(k,k,1,n+1),stackeq(sum(k,
k,1,n)+(n+1)),stackeq(n*(n+1)/
2 +n+1),stackeq((n+1)*(n+1+1)/
2),stackeq((n+1)*(n+2)/2)]
 
[]
 1 (EMPTYCHAR, CHECKMARK, CHECKMARK, CHECKMARK, CHECKMARK)
\[\begin{array}{lll} &\sum_{k=1}^{n+1}{k}& \cr \color{green}{\checkmark}&=\sum_{k=1}^{n}{k}+\left(n+1\right)& \cr \color{green}{\checkmark}&=\frac{n\cdot \left(n+1\right)}{2}+n+1& \cr \color{green}{\checkmark}&=\frac{\left(n+1\right)\cdot \left(n+1+1\right)}{2}& \cr \color{green}{\checkmark}&=\frac{\left(n+1\right)\cdot \left(n+2\right)}{2}& \cr \end{array}\]
Equiv 
[log((a-1)^n*product(x_i^(-a),
i,1,n)),stackeq(n*log(a-1)-a*s
um(log(x_i),i,1,n))]
 
[]
 1 (EMPTYCHAR, CHECKMARK)
\[\begin{array}{lll} &\ln \left( {\left(a-1\right)}^{n}\cdot \prod_{i=1}^{n}{\frac{1}{{{x}_{i}}^{a}}} \right)& \cr \color{green}{\checkmark}&=n\cdot \ln \left( a-1 \right)-a\cdot \sum_{i=1}^{n}{\ln \left( {x}_{i} \right)}& \cr \end{array}\]
Equiv 
[binomial(n,k)+binomial(n,k+1)
,stackeq(n!/(k!*(n-k)!)+n!/((k
+1)!*(n-k-1)!)),stackeq(n!/(k!
*(n-k)*(n-k-1)!)+n!/((k+1)!*(n
-k-1)!)),stackeq(n!/(k!*(n-k-1
)!)*(1/(n-k)+1/(k+1))),stackeq
(n!/(k!*(n-k-1)!)*((n+1)/((n-k
)*(k+1)))),stackeq((n+1)*n!/(k
!*(n-k-1)!)*(1/((k+1)*(n-k))))
,stackeq((n+1)*n!/((k+1)*k!*(n
-k)*(n-k-1)!)),stackeq(((n+1)!
/((k+1)!)*(1/((n-k)*(n-k-1)!))
)),stackeq((n+1)!/((k+1)!*(n-k
)!)),stackeq(binomial(n+1,k+1)
)]
 
[]
 1 (EMPTYCHAR, CHECKMARK, CHECKMARK, CHECKMARK, CHECKMARK, CHECKMARK, CHECKMARK, CHECKMARK, CHECKMARK, CHECKMARK)
\[\begin{array}{lll} &{{n}\choose{k}}+{{n}\choose{k+1}}& \cr \color{green}{\checkmark}&=\frac{n!}{k!\cdot \left(n-k\right)!}+\frac{n!}{\left(k+1\right)!\cdot \left(n-k-1\right)!}& \cr \color{green}{\checkmark}&=\frac{n!}{k!\cdot \left(n-k\right)\cdot \left(n-k-1\right)!}+\frac{n!}{\left(k+1\right)!\cdot \left(n-k-1\right)!}& \cr \color{green}{\checkmark}&=\frac{n!}{k!\cdot \left(n-k-1\right)!}\cdot \left(\frac{1}{n-k}+\frac{1}{k+1}\right)&{\color{blue}{{n \not\in {\left \{4 \right \}}}}}\cr \color{green}{\checkmark}&=\frac{n!}{k!\cdot \left(n-k-1\right)!}\cdot \left(\frac{n+1}{\left(n-k\right)\cdot \left(k+1\right)}\right)& \cr \color{green}{\checkmark}&=\frac{\left(n+1\right)\cdot n!}{k!\cdot \left(n-k-1\right)!}\cdot \left(\frac{1}{\left(k+1\right)\cdot \left(n-k\right)}\right)& \cr \color{green}{\checkmark}&=\frac{\left(n+1\right)\cdot n!}{\left(k+1\right)\cdot k!\cdot \left(n-k\right)\cdot \left(n-k-1\right)!}& \cr \color{green}{\checkmark}&=\frac{\left(n+1\right)!}{\left(k+1\right)!}\cdot \left(\frac{1}{\left(n-k\right)\cdot \left(n-k-1\right)!}\right)& \cr \color{green}{\checkmark}&=\frac{\left(n+1\right)!}{\left(k+1\right)!\cdot \left(n-k\right)!}& \cr \color{green}{\checkmark}&={{n+1}\choose{k+1}}& \cr \end{array}\]
Equiv 
[binomial(n,k)+binomial(n,k-1)
,stackeq(n!/((k-1)!*(n-k+1)!)+
n!/(k!*(n-k)!)),stackeq(n!*k/(
k!*(n-k+1)!)+n!*(n-k+1)/(k!*(n
-k+1)!)),stackeq(n!*k/(k!*(n-k
+1)!)+n!/(k!*(n-k)!)),stackeq(
((n-k+1)*n!+k*n!)/(k!*(n-k+1)!
)),stackeq(((n+1)*n!)/(k!*(n-k
+1)!))]
 
[]
 1 (EMPTYCHAR, CHECKMARK, CHECKMARK, CHECKMARK, CHECKMARK, CHECKMARK)
\[\begin{array}{lll} &{{n}\choose{k}}+{{n}\choose{k-1}}& \cr \color{green}{\checkmark}&=\frac{n!}{\left(k-1\right)!\cdot \left(n-k+1\right)!}+\frac{n!}{k!\cdot \left(n-k\right)!}& \cr \color{green}{\checkmark}&=\frac{n!\cdot k}{k!\cdot \left(n-k+1\right)!}+\frac{n!\cdot \left(n-k+1\right)}{k!\cdot \left(n-k+1\right)!}& \cr \color{green}{\checkmark}&=\frac{n!\cdot k}{k!\cdot \left(n-k+1\right)!}+\frac{n!}{k!\cdot \left(n-k\right)!}& \cr \color{green}{\checkmark}&=\frac{\left(n-k+1\right)\cdot n!+k\cdot n!}{k!\cdot \left(n-k+1\right)!}& \cr \color{green}{\checkmark}&=\frac{\left(n+1\right)\cdot n!}{k!\cdot \left(n-k+1\right)!}& \cr \end{array}\]
Equiv 
[(x-1)^2=(x-1)*(x-1), stackeq(
x^2-2*x+1)]
 
[]
 1 (CHECKMARK, CHECKMARK)
\[\begin{array}{lll}\color{green}{\checkmark}&{\left(x-1\right)}^2=\left(x-1\right)\cdot \left(x-1\right)& \cr \color{green}{\checkmark}&=x^2-2\cdot x+1& \cr \end{array}\]
Equiv 
[(x-1)^2=(x-1)*(x-1), stackeq(
x^2-2*x+2)]
 
[]
 0 (CHECKMARK,QMCHAR)
\[\begin{array}{lll}\color{green}{\checkmark}&{\left(x-1\right)}^2=\left(x-1\right)\cdot \left(x-1\right)& \cr \color{red}{?}&=x^2-2\cdot x+2& \cr \end{array}\]
Equiv 
[(x-2)^2=(x-1)*(x-1), stackeq(
x^2-2*x+1)]
 
[]
 0 (QMCHAR, CHECKMARK)
\[\begin{array}{lll}\color{red}{?}&{\left(x-2\right)}^2=\left(x-1\right)\cdot \left(x-1\right)& \cr \color{green}{\checkmark}&=x^2-2\cdot x+1& \cr \end{array}\]
Equiv 
[4^((n+1)+1)-1= 4*4^(n+1)-1,st
ackeq(4*(4^(n+1)-1)+3)]
 
[]
 1 (CHECKMARK, CHECKMARK)
\[\begin{array}{lll}\color{green}{\checkmark}&4^{n+1+1}-1=4\cdot 4^{n+1}-1& \cr \color{green}{\checkmark}&=4\cdot \left(4^{n+1}-1\right)+3& \cr \end{array}\]
Equiv 
[2*x+3*y=6 and 4*x+9*y=15,2*x+
3*y=6 and -2*x=-3,3+3*y=6 and 
2*x=3,y=1 and x=3/2]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\left\{\begin{array}{l}2\cdot x+3\cdot y=6\cr 4\cdot x+9\cdot y=15\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}2\cdot x+3\cdot y=6\cr -2\cdot x=-3\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}3+3\cdot y=6\cr 2\cdot x=3\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}y=1\cr x=\frac{3}{2}\cr \end{array}\right.& \cr \end{array}\]
Equiv 
[2*x+3*y=6 and 4*x+9*y=15,2*x+
3*y=6 and -2*x=-3,3+3*y=6 and 
2*x=3,y=1 and x=3]
 
[]
 0 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR,QMCHAR)
\[\begin{array}{lll} &\left\{\begin{array}{l}2\cdot x+3\cdot y=6\cr 4\cdot x+9\cdot y=15\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}2\cdot x+3\cdot y=6\cr -2\cdot x=-3\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}3+3\cdot y=6\cr 2\cdot x=3\cr \end{array}\right.& \cr \color{red}{?}&\left\{\begin{array}{l}y=1\cr x=3\cr \end{array}\right.& \cr \end{array}\]
Equiv 
[x^2+y^2=8 and x=y, 2*x^2=8 an
d y=x, x^2=4 and y=x, x= #pm#2
 and y=x, (x= 2 and y=x) or (x
=-2 and y=x), (x=2 and y=2) or
 (x=-2 and y=-2)]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\left\{\begin{array}{l}x^2+y^2=8\cr x=y\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}2\cdot x^2=8\cr y=x\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}x^2=4\cr y=x\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}x= \pm 2\cr y=x\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&x=2\,{\text{ and }}\, y=x\,{\text{ or }}\, x=-2\,{\text{ and }}\, y=x& \cr \color{green}{\Leftrightarrow}&x=2\,{\text{ and }}\, y=2\,{\text{ or }}\, x=-2\,{\text{ and }}\, y=-2& \cr \end{array}\]
Equiv 
[x^2+y^2=5 and x*y=2, x^2+y^2-
5=0 and x*y-2=0, x^2-2*x*y+y^2
-1=0 and x^2+2*x*y+y^2-9=0, (x
-y)^2-1=0 and (x+y)^2-3^2=0, (
x-y=1 and x+y=3) or (x-y=-1 an
d x+y=3) or (x-y=1 and x+y=-3)
 or (x-y=-1 and x+y=-3), (x=1 
and y=2) or (x=2 and y=1) or (
x=-2 and y=-1) or (x=-1 and y=
-2)]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\left\{\begin{array}{l}x^2+y^2=5\cr x\cdot y=2\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}x^2+y^2-5=0\cr x\cdot y-2=0\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}x^2-2\cdot x\cdot y+y^2-1=0\cr x^2+2\cdot x\cdot y+y^2-9=0\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}{\left(x-y\right)}^2-1=0\cr {\left(x+y\right)}^2-3^2=0\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&x-y=1\,{\text{ and }}\, x+y=3\,{\text{ or }}\, x-y=-1\,{\text{ and }}\, x+y=3\,{\text{ or }}\, x-y=1\,{\text{ and }}\, x+y=-3\,{\text{ or }}\, x-y=-1\,{\text{ and }}\, x+y=-3& \cr \color{green}{\Leftrightarrow}&x=1\,{\text{ and }}\, y=2\,{\text{ or }}\, x=2\,{\text{ and }}\, y=1\,{\text{ or }}\, x=-2\,{\text{ and }}\, y=-1\,{\text{ or }}\, x=-1\,{\text{ and }}\, y=-2& \cr \end{array}\]
Equiv 
[4*x^2+7*x*y+4*y^2=4 and y=x-4
, 4*x^2+7*x*(x-4)+4*(x-4)^2-4=
0 and y=x-4, 15*x^2-60*x+60=0 
and y=x-4, (x-2)^2=0 and y=x-4
, x=2 and y=x-4, x=2 and y=-2]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\left\{\begin{array}{l}4\cdot x^2+7\cdot x\cdot y+4\cdot y^2=4\cr y=x-4\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}4\cdot x^2+7\cdot x\cdot \left(x-4\right)+4\cdot {\left(x-4\right)}^2-4=0\cr y=x-4\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}15\cdot x^2-60\cdot x+60=0\cr y=x-4\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}{\left(x-2\right)}^2=0\cr y=x-4\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}x=2\cr y=x-4\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}x=2\cr y=-2\cr \end{array}\right.& \cr \end{array}\]
Equiv 
[a^2=b and a^2=1, b=a^2 and (a
=1 or a=-1), (b=1 and a=1) or 
(b=1 and a=-1)]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\left\{\begin{array}{l}a^2=b\cr a^2=1\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}b=a^2\cr a=1\,{\text{ or }}\, a=-1\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&b=1\,{\text{ and }}\, a=1\,{\text{ or }}\, b=1\,{\text{ and }}\, a=-1& \cr \end{array}\]
Equiv 
[a^2=b and x=1, b=a^2 and x=1]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\left\{\begin{array}{l}a^2=b\cr x=1\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}b=a^2\cr x=1\cr \end{array}\right.& \cr \end{array}\]
Equiv 
[a^2=b and b^2=a, b=a^2 and a^
4=a, b=a^2 and a^4-a=0, b=a^2 
and a*(a-1)*(a^2+a+1)=0, b=a^2
 and (a=0 or a=1 or a^2+a+1=0)
, (b=0 and a=0) or (b=1 and a=
1)]
 
[]
 
[assumereal]
 1 (ASSUMEREALVARS, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll}\color{blue}{(\mathbb{R})}&\left\{\begin{array}{l}a^2=b\cr b^2=a\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}b=a^2\cr a^4=a\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}b=a^2\cr a^4-a=0\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}b=a^2\cr a\cdot \left(a-1\right)\cdot \left(a^2+a+1\right)=0\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}b=a^2\cr a=0\,{\text{ or }}\, a=1\,{\text{ or }}\, a^2+a+1=0\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&b=0\,{\text{ and }}\, a=0\,{\text{ or }}\, b=1\,{\text{ and }}\, a=1& \cr \end{array}\]
Equiv 
[2*x^3-9*x^2+10*x-3,stacklet(x
,1),2*1^3-9*1^2+10*1-3,stackeq
(0),"So",2*x^3-9*x^2
+10*x-3,stackeq((x-1)*(2*x^2-7
*x+3)),stackeq((x-1)*(2*x-1)*(
x-3))]
 
[]
 0 (EMPTYCHAR, EMPTYCHAR, EQUIVCHAR, CHECKMARK, EMPTYCHAR, EMPTYCHAR, CHECKMARK, CHECKMARK)
\[\begin{array}{lll} &2\cdot x^3-9\cdot x^2+10\cdot x-3& \cr &\text{Let }x = 1& \cr \color{green}{\Leftrightarrow}&2\cdot 1^3-9\cdot 1^2+10\cdot 1-3& \cr \color{green}{\checkmark}&=0& \cr &\text{So}& \cr &2\cdot x^3-9\cdot x^2+10\cdot x-3& \cr \color{green}{\checkmark}&=\left(x-1\right)\cdot \left(2\cdot x^2-7\cdot x+3\right)& \cr \color{green}{\checkmark}&=\left(x-1\right)\cdot \left(2\cdot x-1\right)\cdot \left(x-3\right)& \cr \end{array}\]
Equiv 
[2*x^2+x>=6, 2*x^2+x-6>=
0, (2*x-3)*(x+2)>= 0,((2*x-
3)>=0 and (x+2)>=0) or (
(2*x-3)<=0 and (x+2)<=0)
, (x>=3/2 and x>=-2) or 
(x<=3/2 and x<=-2), x>
;=3/2 or x <=-2]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &2\cdot x^2+x\geq 6& \cr \color{green}{\Leftrightarrow}&2\cdot x^2+x-6\geq 0& \cr \color{green}{\Leftrightarrow}&\left(2\cdot x-3\right)\cdot \left(x+2\right)\geq 0& \cr \color{green}{\Leftrightarrow}&2\cdot x-3\geq 0\,{\text{ and }}\, x+2\geq 0\,{\text{ or }}\, 2\cdot x-3\leq 0\,{\text{ and }}\, x+2\leq 0& \cr \color{green}{\Leftrightarrow}&x\geq \frac{3}{2}\,{\text{ and }}\, x\geq -2\,{\text{ or }}\, x\leq \frac{3}{2}\,{\text{ and }}\, x\leq -2& \cr \color{green}{\Leftrightarrow}&x\geq \frac{3}{2}\,{\text{ or }}\, x\leq -2& \cr \end{array}\]
Equiv 
[2*x^2+x>=6, 2*x^2+x-6>=
0, (2*x-3)*(x+2)>= 0,((2*x-
3)>=0 and (x+2)>=0) or (
(2*x-3)<=0 and (x+2)<=0)
, (x>=3/2 and x>=-2) or 
(x<=3/2 and x<=-2), x>
;=3/2 or x <=2]
 
[]
 0 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR,QMCHAR)
\[\begin{array}{lll} &2\cdot x^2+x\geq 6& \cr \color{green}{\Leftrightarrow}&2\cdot x^2+x-6\geq 0& \cr \color{green}{\Leftrightarrow}&\left(2\cdot x-3\right)\cdot \left(x+2\right)\geq 0& \cr \color{green}{\Leftrightarrow}&2\cdot x-3\geq 0\,{\text{ and }}\, x+2\geq 0\,{\text{ or }}\, 2\cdot x-3\leq 0\,{\text{ and }}\, x+2\leq 0& \cr \color{green}{\Leftrightarrow}&x\geq \frac{3}{2}\,{\text{ and }}\, x\geq -2\,{\text{ or }}\, x\leq \frac{3}{2}\,{\text{ and }}\, x\leq -2& \cr \color{red}{?}&x\geq \frac{3}{2}\,{\text{ or }}\, x\leq 2& \cr \end{array}\]
Equiv 
[x^2>=9 and x>3, x^2-9&g
t;=0 and x>3, (x>=3 or x
<=-3) and x>3, x>3]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\left\{\begin{array}{l}x^2\geq 9\cr x > 3\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}x^2-9\geq 0\cr x > 3\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}x\geq 3\,{\text{ or }}\, x\leq -3\cr x > 3\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&x > 3& \cr \end{array}\]
Equiv 
[-x^2+a*x+a-3<0, a-3<x^2
-a*x, a-3<(x-a/2)^2-a^2/4, 
a^2/4+a-3<(x-a/2)^2, a^2+4*
a-12<4*(x-a/2)^2, (a-2)*(a+
6)<4*(x-a/2)^2, "This 
inequality is required to be t
rue for all x.", "So
 it must be true when the righ
t hand side takes its minimum 
value.", "This happe
ns for x=a/2.", (a-2)*(a+
6)<0, ((a-2)<0 and (a+6)
>0) or ((a-2)>0 and (a+6
)<0), (a<2 and a>-6) 
or (a>2 and a<-6), (-6&l
t;a and a<2) or false, (-6&
lt;a and a<2)]
 
[]
 0 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EMPTYCHAR, EMPTYCHAR, EMPTYCHAR, EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &-x^2+a\cdot x+a-3 < 0& \cr \color{green}{\Leftrightarrow}&a-3 < x^2-a\cdot x& \cr \color{green}{\Leftrightarrow}&a-3 < {\left(x-\frac{a}{2}\right)}^2-\frac{a^2}{4}& \cr \color{green}{\Leftrightarrow}&\frac{a^2}{4}+a-3 < {\left(x-\frac{a}{2}\right)}^2& \cr \color{green}{\Leftrightarrow}&a^2+4\cdot a-12 < 4\cdot {\left(x-\frac{a}{2}\right)}^2& \cr \color{green}{\Leftrightarrow}&\left(a-2\right)\cdot \left(a+6\right) < 4\cdot {\left(x-\frac{a}{2}\right)}^2& \cr &\text{This inequality is required to be true for all x.}& \cr &\text{So it must be true when the right hand side takes its minimum value.}& \cr &\text{This happens for x=a/2.}& \cr &\left(a-2\right)\cdot \left(a+6\right) < 0& \cr \color{green}{\Leftrightarrow}&a-2 < 0\,{\text{ and }}\, a+6 > 0\,{\text{ or }}\, a-2 > 0\,{\text{ and }}\, a+6 < 0& \cr \color{green}{\Leftrightarrow}&a < 2\,{\text{ and }}\, a > -6\,{\text{ or }}\, a > 2\,{\text{ and }}\, a < -6& \cr \color{green}{\Leftrightarrow}&-6 < a\,{\text{ and }}\, a < 2\,{\text{ or }}\, \mathbf{False}& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}-6 < a\cr a < 2\cr \end{array}\right.& \cr \end{array}\]
Equiv 
[x-2>0 and x*(x-2)<15,x&
gt;2 and x^2-2*x-15<0,x>
2 and (x-5)*(x+3)<0,x>2 
and ((x<5 and x>-3) or (
x>5 and x<-3)),x>2 an
d (x<5 and x>-3),x>2 
and x<5]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\left\{\begin{array}{l}x-2 > 0\cr x\cdot \left(x-2\right) < 15\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}x > 2\cr x^2-2\cdot x-15 < 0\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}x > 2\cr \left(x-5\right)\cdot \left(x+3\right) < 0\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}x > 2\cr x < 5\,{\text{ and }}\, x > -3\,{\text{ or }}\, x > 5\,{\text{ and }}\, x < -3\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}x > 2\cr x < 5\,{\text{ and }}\, x > -3\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}x > 2\cr x < 5\cr \end{array}\right.& \cr \end{array}\]
Equiv 
[x-2>0 and x*(x-2)<15,x&
gt;2 and x^2-2*x-15<0,x>
2 and (x-5)*(x+3)<0,x>2 
and ((x<5 and x>-3) or (
x>5 and x<-3)),x>7 an
d (x<5 and x>-3),x>2 
and x<5]
 
[]
 0 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR,QMCHAR,QMCHAR)
\[\begin{array}{lll} &\left\{\begin{array}{l}x-2 > 0\cr x\cdot \left(x-2\right) < 15\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}x > 2\cr x^2-2\cdot x-15 < 0\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}x > 2\cr \left(x-5\right)\cdot \left(x+3\right) < 0\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}x > 2\cr x < 5\,{\text{ and }}\, x > -3\,{\text{ or }}\, x > 5\,{\text{ and }}\, x < -3\cr \end{array}\right.& \cr \color{red}{?}&\left\{\begin{array}{l}x > 7\cr x < 5\,{\text{ and }}\, x > -3\cr \end{array}\right.& \cr \color{red}{?}&\left\{\begin{array}{l}x > 2\cr x < 5\cr \end{array}\right.& \cr \end{array}\]
Equiv 
[x^2 + (a-2)*x + a = 0,(x + (a
-2)/2)^2 -((a-2)/2)^2 + a = 0,
(x + (a-2)/2)^2 =(a-2)^2/4 - a
,"This has real roots iff
",(a-2)^2/4-a >=0,a^2-
4*a+4-4*a >=0,a^2-8*a+4>
=0,(a-4)^2-16+4>=0,(a-4)^2&
gt;=12,a-4>=sqrt(12) or a-4
<= -sqrt(12),"Ignoring
 the negative solution.",
a>=sqrt(12)+4,"Using e
xternal domain information tha
t a is an integer.",a>
=8]
 
[]
 0 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EMPTYCHAR, EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EMPTYCHAR, EMPTYCHAR, EMPTYCHAR, EMPTYCHAR)
\[\begin{array}{lll} &x^2+\left(a-2\right)\cdot x+a=0& \cr \color{green}{\Leftrightarrow}&{\left(x+\frac{a-2}{2}\right)}^2-{\left(\frac{a-2}{2}\right)}^2+a=0& \cr \color{green}{\Leftrightarrow}&{\left(x+\frac{a-2}{2}\right)}^2=\frac{{\left(a-2\right)}^2}{4}-a& \cr &\text{This has real roots iff}& \cr &\frac{{\left(a-2\right)}^2}{4}-a\geq 0& \cr \color{green}{\Leftrightarrow}&a^2-4\cdot a+4-4\cdot a\geq 0& \cr \color{green}{\Leftrightarrow}&a^2-8\cdot a+4\geq 0& \cr \color{green}{\Leftrightarrow}&{\left(a-4\right)}^2-16+4\geq 0& \cr \color{green}{\Leftrightarrow}&{\left(a-4\right)}^2\geq 12& \cr \color{green}{\Leftrightarrow}&a-4\geq \sqrt{12}\,{\text{ or }}\, a-4\leq -\sqrt{12}& \cr &\text{Ignoring the negative solution.}& \cr &a\geq \sqrt{12}+4& \cr &\text{Using external domain information that a is an integer.}& \cr &a\geq 8& \cr \end{array}\]
Equiv 
[x^2#1,x^2-1#0,(x-1)*(x+1)#0,x
<-1 nounor (-1<x nounand
 x<1) nounor x>1]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &x^2\neq 1& \cr \color{green}{\Leftrightarrow}&x^2-1\neq 0& \cr \color{green}{\Leftrightarrow}&\left(x-1\right)\cdot \left(x+1\right)\neq 0& \cr \color{green}{\Leftrightarrow}&x < -1\,{\text{ or }}\, -1 < x\,{\text{ and }}\, x < 1\,{\text{ or }}\, x > 1& \cr \end{array}\]
Equiv 
["Set P(n) be the stateme
nt that",sum(k^2,k,1,n) =
 n*(n+1)*(2*n+1)/6, "Then
 P(1) is the statement", 
1^2 = 1*(1+1)*(2*1+1)/6, 1 = 1
, "So P(1) holds.  Now as
sume P(n) is true.",sum(k
^2,k,1,n) = n*(n+1)*(2*n+1)/6,
sum(k^2,k,1,n) +(n+1)^2= n*(n+
1)*(2*n+1)/6 +(n+1)^2,sum(k^2,
k,1,n+1)= (n+1)*(n*(2*n+1) +6*
(n+1))/6,sum(k^2,k,1,n+1)= (n+
1)*(2*n^2+7*n+6)/6,sum(k^2,k,1
,n+1)= (n+1)*(n+1+1)*(2*(n+1)+
1)/6]
 
[]
 0 (EMPTYCHAR, EMPTYCHAR, EMPTYCHAR, EMPTYCHAR, EQUIVCHAR, EMPTYCHAR, EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\text{Set P(n) be the statement that}& \cr &\sum_{k=1}^{n}{k^2}=\frac{n\cdot \left(n+1\right)\cdot \left(2\cdot n+1\right)}{6}& \cr &\text{Then P(1) is the statement}& \cr &1^2=\frac{1\cdot \left(1+1\right)\cdot \left(2\cdot 1+1\right)}{6}& \cr \color{green}{\Leftrightarrow}&1=1& \cr &\text{So P(1) holds. Now assume P(n) is true.}& \cr &\sum_{k=1}^{n}{k^2}=\frac{n\cdot \left(n+1\right)\cdot \left(2\cdot n+1\right)}{6}& \cr \color{green}{\Leftrightarrow}&\sum_{k=1}^{n}{k^2}+{\left(n+1\right)}^2=\frac{n\cdot \left(n+1\right)\cdot \left(2\cdot n+1\right)}{6}+{\left(n+1\right)}^2& \cr \color{green}{\Leftrightarrow}&\sum_{k=1}^{n+1}{k^2}=\frac{\left(n+1\right)\cdot \left(n\cdot \left(2\cdot n+1\right)+6\cdot \left(n+1\right)\right)}{6}& \cr \color{green}{\Leftrightarrow}&\sum_{k=1}^{n+1}{k^2}=\frac{\left(n+1\right)\cdot \left(2\cdot n^2+7\cdot n+6\right)}{6}& \cr \color{green}{\Leftrightarrow}&\sum_{k=1}^{n+1}{k^2}=\frac{\left(n+1\right)\cdot \left(n+1+1\right)\cdot \left(2\cdot \left(n+1\right)+1\right)}{6}& \cr \end{array}\]
Equiv 
[(n+1)^2+sum(k^2,k,1,n) = (n+1
)^2+(n*(n+1)*(2*n+1))/6, sum(k
^2,k,1,n+1) = ((n+1)*(n*(2*n+1
)+6*(n+1)))/6, sum(k^2,k,1,n+1
) = ((n+1)*(2*n^2+7*n+6))/6, s
um(k^2,k,1,n+1) = ((n+1)*(n+2)
*(2*(n+1)+1))/6]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &{\left(n+1\right)}^2+\sum_{k=1}^{n}{k^2}={\left(n+1\right)}^2+\frac{n\cdot \left(n+1\right)\cdot \left(2\cdot n+1\right)}{6}& \cr \color{green}{\Leftrightarrow}&\sum_{k=1}^{n+1}{k^2}=\frac{\left(n+1\right)\cdot \left(n\cdot \left(2\cdot n+1\right)+6\cdot \left(n+1\right)\right)}{6}& \cr \color{green}{\Leftrightarrow}&\sum_{k=1}^{n+1}{k^2}=\frac{\left(n+1\right)\cdot \left(2\cdot n^2+7\cdot n+6\right)}{6}& \cr \color{green}{\Leftrightarrow}&\sum_{k=1}^{n+1}{k^2}=\frac{\left(n+1\right)\cdot \left(n+2\right)\cdot \left(2\cdot \left(n+1\right)+1\right)}{6}& \cr \end{array}\]
Equiv 
[conjugate(a)*conjugate(b),sta
cklet(a,x+i*y),stacklet(b,r+i*
s),stackeq(conjugate(x+i*y)*co
njugate(r+i*s)),stackeq((x-i*y
)*(r-i*s)),stackeq((x*r-y*s)-i
*(y*r+x*s)),stackeq(conjugate(
(x*r-y*s)+i*(y*r+x*s))),stacke
q(conjugate((x+i*y)*(r+i*s))),
stacklet(x+i*y,a),stacklet(r+i
*s,b),stackeq(conjugate(a*b))]
 
[]
 1 (EMPTYCHAR, EMPTYCHAR, EMPTYCHAR, CHECKMARK, CHECKMARK, CHECKMARK, CHECKMARK, CHECKMARK, EMPTYCHAR, EMPTYCHAR, CHECKMARK)
\[\begin{array}{lll} &a^\star\cdot b^\star& \cr &\text{Let }a = x+\mathrm{i}\cdot y& \cr &\text{Let }b = r+\mathrm{i}\cdot s& \cr \color{green}{\checkmark}&=\left(x+\mathrm{i}\cdot y\right)^\star\cdot \left(r+\mathrm{i}\cdot s\right)^\star& \cr \color{green}{\checkmark}&=\left(x-\mathrm{i}\cdot y\right)\cdot \left(r-\mathrm{i}\cdot s\right)& \cr \color{green}{\checkmark}&=x\cdot r-y\cdot s-\mathrm{i}\cdot \left(y\cdot r+x\cdot s\right)& \cr \color{green}{\checkmark}&=\left(x\cdot r-y\cdot s+\mathrm{i}\cdot \left(y\cdot r+x\cdot s\right)\right)^\star& \cr \color{green}{\checkmark}&=\left(\left(x+\mathrm{i}\cdot y\right)\cdot \left(r+\mathrm{i}\cdot s\right)\right)^\star& \cr &\text{Let }x+\mathrm{i}\cdot y = a& \cr &\text{Let }r+\mathrm{i}\cdot s = b& \cr \color{green}{\checkmark}&=\left(a\cdot b\right)^\star& \cr \end{array}\]
Equiv 
[nounint(x*e^x,x,-inf,0),nounl
imit(nounint(x*e^x,x,t,0),t,-i
nf),nounlimit(e^t-t*e^t-1,t,-i
nf),nounlimit(e^t,t,-inf)+noun
limit(-t*e^t,t,-inf)+nounlimit
(-1,t,-inf),-1]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\int_{-\infty }^{0}{x\cdot e^{x}\;\mathrm{d}x}& \cr \color{green}{\Leftrightarrow}&\lim_{t\rightarrow -\infty }{\int_{t}^{0}{x\cdot e^{x}\;\mathrm{d}x}}& \cr \color{green}{\Leftrightarrow}&\lim_{t\rightarrow -\infty }{e^{t}-t\cdot e^{t}-1}& \cr \color{green}{\Leftrightarrow}&\lim_{t\rightarrow -\infty }{e^{t}}+\lim_{t\rightarrow -\infty }{\left(-t\right)\cdot e^{t}}+\lim_{t\rightarrow -\infty }{-1}& \cr \color{green}{\Leftrightarrow}&-1& \cr \end{array}\]
Equiv 
[noundiff(x^2,x),stackeq(nounl
imit(((x+h)^2-x^2)/h,h,0)),sta
ckeq(nounlimit(2*x+h,h,0)),sta
ckeq(2*x)]
 
[]
 1 (EMPTYCHAR, CHECKMARK, CHECKMARK, CHECKMARK)
\[\begin{array}{lll} &\frac{\mathrm{d}}{\mathrm{d} x} x^2& \cr \color{green}{\checkmark}&=\lim_{h\rightarrow 0}{\frac{{\left(x+h\right)}^2-x^2}{h}}& \cr \color{green}{\checkmark}&=\lim_{h\rightarrow 0}{2\cdot x+h}& \cr \color{green}{\checkmark}&=2\cdot x& \cr \end{array}\]
Equiv 
[4*x=2,x=0.500,x=1/2]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &4\cdot x=2& \cr \color{green}{\Leftrightarrow}&x=0.5& \cr \color{green}{\Leftrightarrow}&x=\frac{1}{2}& \cr \end{array}\]
Equiv 
[4*x=6,x=0.33333,x=1/3]
 
[]
 0 (EMPTYCHAR,QMCHAR,QMCHAR)
\[\begin{array}{lll} &4\cdot x=6& \cr \color{red}{?}&x=0.33333& \cr \color{red}{?}&x=\frac{1}{3}& \cr \end{array}\]
Equiv 
[-2.5*t-11.25 = 0,-2.5*t = 11.
25,t = 11.25/-2.50,t = -4.500]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &-2.5\cdot t-11.25=0& \cr \color{green}{\Leftrightarrow}&-2.5\cdot t=11.25& \cr \color{green}{\Leftrightarrow}&t=\frac{11.25}{-2.5}& \cr \color{green}{\Leftrightarrow}&t=-4.5& \cr \end{array}\]
Equiv 
[4*x-6*y=-3 and 3*y+4*x=3, 4*x
-6*y=-3 and 9*y=6, 4*x=1 and 3
*y=2, x=0.25 and y=2/3]
 
[]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &\left\{\begin{array}{l}4\cdot x-6\cdot y=-3\cr 3\cdot y+4\cdot x=3\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}4\cdot x-6\cdot y=-3\cr 9\cdot y=6\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}4\cdot x=1\cr 3\cdot y=2\cr \end{array}\right.& \cr \color{green}{\Leftrightarrow}&\left\{\begin{array}{l}x=0.25\cr y=\frac{2}{3}\cr \end{array}\right.& \cr \end{array}\]
Equiv 
[-12+3*noundiff(y(x),x)+8-8*no
undiff(y(x),x)=0,-5*noundiff(y
(x),x)=4,noundiff(y(x),x)=-4/5
]
 
[]
 
[calculus]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &-12+3\cdot \left(\frac{\mathrm{d}}{\mathrm{d} x} y\left(x\right)\right)+8-8\cdot \left(\frac{\mathrm{d}}{\mathrm{d} x} y\left(x\right)\right)=0& \cr \color{green}{\Leftrightarrow}&-5\cdot \left(\frac{\mathrm{d}}{\mathrm{d} x} y\left(x\right)\right)=4& \cr \color{green}{\Leftrightarrow}&\left(\frac{\mathrm{d}}{\mathrm{d} x} y\left(x\right)\right)=\frac{-4}{5}& \cr \end{array}\]
Equiv 
[x^2+1,x^3/3+x,x^2+1,x^3/3+x+c
]
 
[]
 
[calculus]
 1 (EMPTYCHAR,INTCHAR(x),DIFFCHAR(x),INTCHAR(x))
\[\begin{array}{lll} &x^2+1& \cr \color{blue}{\int\ldots\mathrm{d}x}&\frac{x^3}{3}+x& \cr \color{blue}{\frac{\mathrm{d}}{\mathrm{d}x}\ldots}&x^2+1& \cr \color{blue}{\int\ldots\mathrm{d}x}&\frac{x^3}{3}+x+c& \cr \end{array}\]
Equiv 
[3*x^(3/2)-2/x,(9*sqrt(x))/2+2
/x^2,3*x^(3/2)-2/x+c]
 
[]
 
[calculus]
 1 (EMPTYCHAR,DIFFCHAR(x),INTCHAR(x))
\[\begin{array}{lll} &3\cdot x^{\frac{3}{2}}-\frac{2}{x}&{\color{blue}{{x \not\in {\left \{0 \right \}}}}}\cr \color{blue}{\frac{\mathrm{d}}{\mathrm{d}x}\ldots}&\frac{9\cdot \sqrt{x}}{2}+\frac{2}{x^2}&{\color{blue}{{x \in {\left( 0,\, \infty \right)}}}}\cr \color{blue}{\int\ldots\mathrm{d}x}&3\cdot x^{\frac{3}{2}}-\frac{2}{x}+c& \cr \end{array}\]
Equiv 
[x^2+1,stackeq(x^3/3+x),stacke
q(x^2+1),stackeq(x^3/3+x+c)]
 
[]
 
[calculus]
 0 (EMPTYCHAR,QMCHAR,QMCHAR,QMCHAR)
\[\begin{array}{lll} &x^2+1& \cr \color{red}{?}&=\frac{x^3}{3}+x& \cr \color{red}{?}&=x^2+1& \cr \color{red}{?}&=\frac{x^3}{3}+x+c& \cr \end{array}\]
Equiv 
[diff(x^2*sin(x),x),stackeq(x^
2*diff(sin(x),x)+diff(x^2,x)*s
in(x)),stackeq(x^2*cos(x)+2*x*
sin(x))]
 
[]
 
[calculus]
 1 (EMPTYCHAR, CHECKMARK, CHECKMARK)
\[\begin{array}{lll} &+\left(\cos \left( x \right)\right)\cdot x^2+2\cdot x\cdot \sin \left( x \right)& \cr \color{green}{\checkmark}&=x^2\cdot +\left(\cos \left( x \right)\right)+2\cdot x\cdot \sin \left( x \right)& \cr \color{green}{\checkmark}&=x^2\cdot \cos \left( x \right)+2\cdot x\cdot \sin \left( x \right)& \cr \end{array}\]
Equiv 
[y(x)*cos(x)+y(x)^2 = 6*x,cos(
x)*diff(y(x),x)+2*y(x)*diff(y(
x),x)-y(x)*sin(x) = 6,(cos(x)+
2*y(x))*diff(y(x),x) = y(x)*si
n(x)+6,diff(y(x),x) = (y(x)*si
n(x)+6)/(cos(x)+2*y(x))]
 
[]
 
[calculus]
 1 (EMPTYCHAR,DIFFCHAR(x), EQUIVCHAR, EQUIVCHAR)
\[\begin{array}{lll} &y\left(x\right)\cdot \cos \left( x \right)+y^2\left(x\right)=6\cdot x& \cr \color{blue}{\frac{\mathrm{d}}{\mathrm{d}x}\ldots}&\cos \left( x \right)\cdot \left(\frac{\mathrm{d}}{\mathrm{d} x} y\left(x\right)\right)+2\cdot y\left(x\right)\cdot \left(\frac{\mathrm{d}}{\mathrm{d} x} y\left(x\right)\right)+\left(-y\left(x\right)\right)\cdot \sin \left( x \right)=6& \cr \color{green}{\Leftrightarrow}&\left(\cos \left( x \right)+2\cdot y\left(x\right)\right)\cdot \left(\frac{\mathrm{d}}{\mathrm{d} x} y\left(x\right)\right)=y\left(x\right)\cdot \sin \left( x \right)+6& \cr \color{green}{\Leftrightarrow}&\left(\frac{\mathrm{d}}{\mathrm{d} x} y\left(x\right)\right)=\frac{y\left(x\right)\cdot \sin \left( x \right)+6}{\cos \left( x \right)+2\cdot y\left(x\right)}& \cr \end{array}\]
Equiv 
[nounint(s^2+1,s),stackeq(s^3/
3+s+c)]
 
[]
 
[calculus]
 1 (EMPTYCHAR,INTCHAR(s))
\[\begin{array}{lll} &\int {s^2+1}{\;\mathrm{d}s}& \cr \color{blue}{\int\ldots\mathrm{d}s}&=\frac{s^3}{3}+s+c& \cr \end{array}\]
Equiv 
[nounint(x^3*log(x),x),x^4/4*l
og(x)-1/4*nounint(x^3,x),x^4/4
*log(x)-x^4/16]
 
[]
 
[calculus]
 0 (EMPTYCHAR, EQUIVCHAR,PLUSC)
\[\begin{array}{lll} &\int {x^3\cdot \ln \left( x \right)}{\;\mathrm{d}x}& \cr \color{green}{\Leftrightarrow}&\frac{x^4}{4}\cdot \ln \left( x \right)-\frac{1}{4}\cdot \int {x^3}{\;\mathrm{d}x}& \cr \color{red}{\cdots +c\quad ?}&\frac{x^4}{4}\cdot \ln \left( x \right)-\frac{x^4}{16}&{\color{blue}{{x \in {\left( 0,\, \infty \right)}}}}\cr \end{array}\]
Equiv 
[nounint(x^3*log(x),x),x^4/4*l
og(x)-1/4*nounint(x^3,x),x^4/4
*log(x)-x^4/16+c]
 
[]
 
[calculus]
 1 (EMPTYCHAR, EQUIVCHAR,INTCHAR(x))
\[\begin{array}{lll} &\int {x^3\cdot \ln \left( x \right)}{\;\mathrm{d}x}& \cr \color{green}{\Leftrightarrow}&\frac{x^4}{4}\cdot \ln \left( x \right)-\frac{1}{4}\cdot \int {x^3}{\;\mathrm{d}x}& \cr \color{blue}{\int\ldots\mathrm{d}x}&\frac{x^4}{4}\cdot \ln \left( x \right)-\frac{x^4}{16}+c& \cr \end{array}\]
Equiv 
[noundiff(y,x)-2/x*y=x^3*sin(3
*x),1/x^2*noundiff(y,x)-2/x^3*
y=x*sin(3*x),noundiff(y/x^2,x)
=x*sin(3*x),y/x^2 = nounint(x*
sin(3*x),x),y/x^2=(sin(3*x)-3*
x*cos(3*x))/9+c]
 
[]
 
[calculus]
 1 (EMPTYCHAR, EQUIVCHAR, EQUIVCHAR,INTCHAR(x),INTCHAR(x))
\[\begin{array}{lll} &\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2}{x}\cdot y=x^3\cdot \sin \left( 3\cdot x \right)& \cr \color{green}{\Leftrightarrow}&\frac{1}{x^2}\cdot \left(\frac{\mathrm{d} y}{\mathrm{d} x}\right)-\frac{2}{x^3}\cdot y=x\cdot \sin \left( 3\cdot x \right)& \cr \color{green}{\Leftrightarrow}&\left(\frac{\mathrm{d}}{\mathrm{d} x} \frac{y}{x^2}\right)=x\cdot \sin \left( 3\cdot x \right)& \cr \color{blue}{\int\ldots\mathrm{d}x}&\frac{y}{x^2}=\int {x\cdot \sin \left( 3\cdot x \right)}{\;\mathrm{d}x}& \cr \color{blue}{\int\ldots\mathrm{d}x}&\frac{y}{x^2}=\frac{\sin \left( 3\cdot x \right)-3\cdot x\cdot \cos \left( 3\cdot x \right)}{9}+c& \cr \end{array}\]