I'm having trouble trying to simplify one boolean equation into another (and visa versa).

Why are these Boolean expressions,

A.B + C.(A⊕B) = A.B + C.B + A.C


They have the same truth table but I cannot figure out how one simplifies into the other. Here is what I have attempted so far:

A.B + C.(A⊕B)

A.B + C.(A'.B + A.B') ...XOR equivelent

A.B + C.A'.B + C.A.B' ...distributive

I don't know what else I can do. If I try from the other direction, it still does not work:

A.B + C.B + A.C

A.B + C.(B + A) ...distributive

A.B + C.(A'.B')' ...de Morgan's

I don't know if I'm heading in the write direction. Any advice would be appreciated.


1 Answer 1


Starting from where you left off: $$ A \cdot B + C \cdot A' \cdot B + C \cdot A \cdot B' = \\ (C + C')A \cdot B + C \cdot A' \cdot B + C \cdot A \cdot B' =\\ C\cdot A \cdot B + C'\cdot A \cdot B + C \cdot A' \cdot B + C \cdot A \cdot B' =\\ C \cdot A \cdot (B + B') + A \cdot B\cdot (C + C') = \\ A \cdot C + A \cdot B $$ This kind of simplification is more typically done using Karnaugh maps. The right-hand-side simplifies similarly: $$ AB + CB + AC = \\ AC + AB + (A + A')CB = \\ AC + AB + ABC + A'BC = \\ AC(1 + B) + AB(1 + C) = \\ AC + AB $$


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