Reducing Boolean Expressions: Axioms and Simplification Techniques

AI Thread Summary
The discussion focuses on simplifying the Boolean expression A(B+C)+B'D+C'D'. The initial poster identifies that the term (B+C) is redundant when A is true, leading to a simplified expression of A+B'D+C'D'. Participants provide hints on proving that B'C' is included in the simplified expression B'D+C'D'. The conversation highlights the importance of axioms in Boolean simplification and concludes with the poster expressing gratitude for the guidance received.
deathprog23
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I have been stumped by a simplification problem - well, I can solve it, but I'm not sure how to do it axiomatically!

The expression is A(B+C)+B'D+C'D'

I can see that the (B+C) is redundant in the first term - if A is true, the whole is true regardless of (B+C)'s value. So it reduces to A+B'D+C'D'

What axioms are used in the proof of this? Thanks!
 
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Hi deathprog23! :smile:

Hint: you need to prove that B'C' lies in B'D+C'D' :wink:
 
tiny-tim said:
Hi deathprog23! :smile:

Hint: you need to prove that B'C' lies in B'D+C'D' :wink:

Nice hint! I got it now, thanks very much :)
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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