Question

Derive backward from S = (x XOR y) XOR z to ~x~yz + ~xy~z + xyz + x~y~z

Answer 1

Here is the solution to the problem.

x XOR y XOR z
= (x XOR y) XOR z
= (x XOR y)’z + (x XOR y)z’
= (x’y+xy’)’z + (x’y+xy’)z’
= ((x’y)’·(xy’)’)z + x’yz’ + xy’z’
= ((x+y’)·(x’+y))z + x’yz’ + xy’z’
= (xx’+xy+x’y’+yy’)z+x’yz + xy’z’
= xx’z + xyz + x’y’z +yy’z + x’yz’ + xy’z’
= 0 + xyz + x’y’z + 0 +x’yz’ + xy’z’
= xyz + x’y’z + x’yz’ + xy’z’

Here is a screenshot if the above derivation is ot clear.

Hope this helps.

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