Question

Prove that for a triple element x, y, z the distributive law is always satisfied if one of the elements equals 0 or 1, or if two of the elements are equal.

Answer 1

We need to prove x+yz=(x+y)(x+z) for all permutations of Boolean variables x, y, z.

Multiplication is understood to be logical AND, and addition to be logical OR (so 1+1=1, not 2).

We can do this by setting up a table to cover all possibilities. There are 8 variants of x, y, z.

x	y	z	x+yz	x+y	x+z	(x+y)(x+z)
0	0	0	0+0=0	0	0	0
0	0	1	0+0=0	0	1	0
0	1	0	0+0=0	1	0	0
0	1	1	0+1=1	1	1	1
1	0	0	1+0=1	1	1	1
1	0	1	1+0=1	1	1	1
1	1	0	1+0=1	1	1	1
1	1	1	1+1=1	1	1	1

From the table it's clear that, since the results in columns 4 and 7 are the same, that the Distributive Law applies.