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I would like to simplify

(a == c && c != 0) || (a == 0 && c == 0) 

to

a == c 

How can I do that?

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2 Answers 2

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Reduce[(a==c&&c!=0)||(a==0&&c==0),Reals] 

Mathematica graphics

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  • $\begingroup$ Interesting. Why Reduce[(a==c&&c!=0)||(a==0&&c==0),Complexes] doesn't work? $\endgroup$ Commented Nov 5, 2023 at 19:41
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Not assuming a,c \[Element] Reals,

Solve[(a == c && c != 0) || (a == 0 && c == 0), c] // FullSimplify 

{{c -> a}}

and

 Solve[(a == c && c != 0) || (a == 0 && c == 0), a] // FullSimplify 

{{a -> c}}

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