Prove that (((p v ~q) ⊕ p) v ~p) ⊕ (p v ~q) ⊕ (p ⊕...

Prove that (((p v ~q) ⊕ p) v ~p) ⊕ (p v ~q) ⊕ (p ⊕ q) is equivalent to p ^ q. Please show your work and name all the logical equivalence laws for each step. ( v = or, ~ = not, ⊕ = XOR)

Thank you

Expert Solution

Find the solution given in uploaded images (2 images are there)

venereology answered 1 year ago

Using logical equivalence laws, show that (((p v ~ q) ⊕ p) v ~p) ⊕ (p...

Using logical equivalence laws, show that (((p v ~ q) ⊕ p) v ~p) ⊕ (p v ~q) is equivalent to p v q. v = or, ~ = not, ⊕ = exclusive or (XOR). Please show the steps with the name of the law beside each step, thanks so much!

Prove p → (q ∨ r), q → s, r → s ⊢ p → s

prove or disprove using logical equivalences (a) p ∧ (q → r) ⇐⇒ (p → q)...

prove or disprove using logical equivalences (a) p ∧ (q → r) ⇐⇒ (p → q) → r (b) x ∧ (¬y ↔ z) ⇐⇒ ((x → y) ∨ ¬z) → (x ∧ ¬(y → z)) (c) (x ∨ y ∨ ¬z) ∧ (¬x ∨ y ∨ z) ⇐⇒ ¬y → (x ↔ z)

Prove or disprove using a Truth Table( De Morgan's Law) ¬(p∧q) ≡ ¬p∨¬q

Prove or disprove using a Truth Table( De Morgan's Law) ¬(p∧q) ≡ ¬p∨¬q Show the Truth Table for (p∨r) (r→¬q)

Discrete math question Prove that ¬(q→p)∧(p∧q∧s→r)∧p is a contradiction without using truth table

If the pth term of an AP is q and the qth term is p, prove that its nth term is (p + q - n).

Using the inner product on 〈p, q〉 = ∫(0 to1) p(x)q(x)dx on P2, write v as the...

Using the inner product on 〈p, q〉 = ∫(0 to1) p(x)q(x)dx on P2, write v as the sum of a vector in U and a vector in U⊥, where v=x^2, U =span{x+1,9x−5}.

1. m • (n • p) 2. (q ⊃ ~t) • (~m v q) 3. ~t...

1. m • (n • p) 2. (q ⊃ ~t) • (~m v q) 3. ~t ⊃ z : . z

(c) (¬p ∨ q) → (p ∧ q) and p (d) (p → q) ∨ p...

(c) (¬p ∨ q) → (p ∧ q) and p (d) (p → q) ∨ p and T I was wondering if I could get help proving these expressions are logically equivalent by applying laws of logic. Also these 2 last questions im having trouble with. Rewrite the negation of each of the following logical expressions so that all negations immediately precede predicates. (a) ¬∀x(¬P(x) → Q(x)) (b) ¬∃x(P(x) → ¬Q(x))

Let p and q be propositions. (i) Show (p →q) ≡ (p ∧ ¬q) →F (ii.)...

Let p and q be propositions. (i) Show (p →q) ≡ (p ∧ ¬q) →F (ii.) Why does this equivalency allow us to use the proof by contradiction technique?

Question