Method I:

 Is the following distributive law valid or invalid? Prove your answer.

x XOR (y AND z) = (x XOR y) AND (x XOR z)

 The given expression is X XOR(y AND z) = (x XOR y) AND( x XOR z)

To prove the given expression we have two methods by using Boolean algebra method and by using truth table method

When both the inputs are different then XOR gate give ‘0’ and when both the inputs are same the XOR gate will give ‘1’

When both the inputs are 1 then only the AND gate will give ‘1’ for remaining all conditions it will give ‘0’.

 From the above two truth tables we can conclude that the following distribute law is invalid x XOR (y AND z) = (x XOR y) AND (x XOR z)

 Method II:

The given expression can be rewritten by using the following algebraic identities  :- x AND y= x.y and x XOR y= x’y+xy’

In the expression the Left hand side(L.H.S) expression is :

= x XOR (y AND z)

= x XOR(yz)

=x(yz)’+x’(yz)           {here we have to apply DeMorgans law (yz)’=y’+z’}

=x(y’+z’)+x’yz

=xy’+xz’+x’yz

The Right hand side expression is :

= (x XOR y) AND (x XOR z)

= (x’y+xy’) AND (x’z+xz’)

=(x’y+xy’) . (x’z+xz’)

=x’yz+xy’z’      {here we have to apply x.x’=0}

Hence the LHS and RHS are not equal we can conclude that the following distribute law is invalid x XOR (y AND z) = (x XOR y) AND (x XOR z)