minimization

Question

Let $‘1’$ and $‘0’$ denote the binary constants. Consider the following Boolean expression for $F$ over two variables $P$ and $Q$:

$F(P, Q) = ( ( 1 \oplus P) \oplus (P \oplus Q) ) \oplus ( (P \oplus Q) \oplus (Q \oplus 0) ) $

The equivalent expression for $F$ is

Answer 1

f= p xnor q

Comments

Is expression of F will change if i solve this with another way?

Answer 2

We know that 

$P \oplus 1 = P'$  And

$Q \oplus 0 = Q$

And $A \oplus A = 0$

So, Using the above properties, We can easily solve the given expression :

F(P, Q) = ( ( 1 X P) X (P X Q) ) X ( (P X Q) X (Q X 0) ) 

= (  $P'$ $\oplus$  ($P \oplus$ Q)  )  $\oplus$ (  ($P \oplus$ Q) $\oplus$ $Q$  )

And Since $\oplus$ is Commutative and associative... We can write the above expression as follows :

=  $P'$ $\oplus$ (  ($P \oplus$ $Q$)   $\oplus$  ($P \oplus$ $Q$) ) $\oplus$ $Q$ 

= $P'$ $\oplus$ $0$ $\oplus$ $Q$   // (Using the Property $A \oplus A = 0$ )

= $P'$  $\oplus$ $Q$   // ( Using the Property  $Q \oplus 0 = Q$ )

 =  $P \odot Q$

Answer 3

BASICS :-

1) A⊕0 = A and A⊕1=A' and A⊕A =0

2) EX-OR IS ASSOCIATIVE

Given that

F(P,Q) =  ( ( 1 ⊕ P) ⊕ (P ⊕ Q) ) ⊕ ( (P ⊕ Q) ⊕ (Q ⊕ 0) )

= ((P' ⊕ ( P ⊕ Q ) ) ⊕ ((P⊕ Q) ⊕ Q))

= P' ⊕((P⊕Q) ⊕ ( P ⊕ Q)) ⊕ Q  --- Due to Associative Property

=P' ⊕ 0 ⊕ Q    

= P' ⊕ Q (WHEN 0 merge with Q )