Is it possible for Mathematica to do simplifications for expressions where the variables are binary, such as:
a + b = 1 + 2*c => a + b = 1
Here c must be 0 because if it is 1, the RHS is 3 but the LHS can be at most 2.
a + 2*b*a + 2*c = 2*d => c=d
Here the LHS is: a(1+2*b) + 2*c. The RHS must be even, so the LHS must be even. But (1 + 2*b) can never be even, so 'a' must be 0.
I looked at Simplify[ ] and FullSimplify[ ] using assumptions, and various other stackexchange questions, but to my surprise this doesn't seem possible in Mathematica.
I want to comment on the answer by Algohi, but can't seem to add an image to the comment. What Algohi is doing solves the binary equations, but what I need is the simplified equations (a + b = 1, and c=d). I would have to convert the boolean expressions in that answer to equations:
However I can't find a way to do this in Mathematica. If I could convert back & forth easily, I would have converted the equations to a boolean expression and then used BooleanMinimize[ ] to simplify.