how to prove boolean identities

Question

I'm working on 2 boolean proofs

1. (¬p⊕q)=(p⊕¬q=¬(p⊕q) <- I assume its equality law i'm not sure how to do this problem(I verified using truth table but I need to do algebraically)
2. (¬p^¬q)∨(q^¬r)∨(¬p^¬r) = (q^¬r) ∨ (¬p^¬q)

what I did so far (for b)

(¬p^¬q)∨(¬p^¬r)∨(q^¬r) (just commutative law)

¬p^(¬q∨¬r)∨(q^¬r)

I assume I should leave (¬p^¬q) as it is and make ¬p^(¬q∨¬r) equal to (q^¬r) but I'm not sure how to go further from here..

Anyone can please clarify this??

Answer 1

For problem 1. use the XOR definition:

X⊕Y = ¬XY ∨ X¬Y

Then:

¬p⊕q = ¬(¬p)q ∨ (¬p)¬q = ¬¬pq ∨ ¬p¬q = pq ∨ ¬p¬q = [fork here]

= p¬¬q ∨ ¬p¬q = p¬(¬q) ∨ ¬p(¬q) = p⊕¬q

= ¬¬(pq ∨ ¬p¬q) = ¬(¬(pq) ∧ ¬(¬p¬q)) = ¬((¬p∨¬q)(p∨q))
= ¬(¬pp ∨ ¬pq ∨ ¬qp ∨ ¬qq) = ¬(0 ∨ ¬pq ∨ ¬qp ∨ 0)
= ¬(¬pq ∨ ¬qp) = ¬(p⊕q)

Answer 2

For problem 2:

(¬p¬q)∨(q¬r)∨(¬p¬r)
= ¬p¬q ∨ q¬r ∨ ¬p¬r   // simplified notation
= ¬p¬q ∨ q¬r ∨ (1)¬p¬r
= ¬p¬q ∨ q¬r ∨ (q ∨ ¬q)¬p¬r
= ¬p¬q ∨ q¬r ∨ q¬p¬r ∨ ¬q¬p¬r
    |
    `--------------.
                   |
= q¬r ∨ q¬p¬r ∨ ¬p¬q ∨ ¬q¬p¬r
= q¬r(1 ∨ ¬p) ∨ ¬p¬q(1 ∨ ¬r)
= q¬r(1) ∨ ¬p¬q(1)
= q¬r ∨ ¬p¬q

q.e.d