5 points Show that p + (q + r) and q + (pvr) are logically equivalent...
2. (a) Show that (PVQ) + R is not logically equivalent to (P + R) V(Q + R) using a truth table. (b) Is (PAQ) → R logically equivalent to (P + R) A( Q R )? If so, use a truth table to establish this. If not, show that it is false.
Assume that p NAND q is logically equivalent to ¬(p ∧ q). Then, (a) prove that {NAND} is functionally complete, i.e., any propositional formula is equivalent to one whose only connective is NAND. Now, (b) prove that any propositional formula is equivalent to one whose only connectives are XOR and AND, along with the constant TRUE. Prove these using a series of logical equivalences.
WITHOUT constructing TT Show whether or not p-, q ^ (q-r)-p-, r is logically equivalent to
6. Maximum score 3 ( 1 per part).Show that:(b) (p → q) → r and p →(q → r) are not logically equivalent.(c) p ↔ q and ¬ p ↔ ¬ q are logically equivalent.
Problem 12.1: Let p and be logical statements. By using a truth table determine if the following compound statements are logically equivalent. Show work! Circle one: A: The statements are equivalent. B: The statements are not equivalent. Problem 12.2: Let P, Q, and be be logical statements. By using a truth table determine if the following compound statements are logically equivalent. Show work! Circle one: A: The statements are equivalent. B: The statements are not equivalent.
Show that negation \neg (p xor q) and p if and only if q are logically equivalent without using a truth
3. (10 pts.) Use logical equivalences to show that (p r)v(q r) and (pAq) r ane logically equivalent.
Show that ~p -> (q -> r) and q-> (p v r) are logically equivalent
Discrete Math: Decide whether (p^q)r and (pr)^(qr) are logically equivalent using boolean algebra. Show work! Do NOT use truth table. We were unable to transcribe this imageWe were unable to transcribe this imageWe were unable to transcribe this image
How do you show the following propositions are logically equivalent? (a) [(p → q) → r] ⊕ (p ∧ q ∧ r) and (p ∨ r) ⊕ (p ∧ q) (b) ¬∃x {P(x) → ∃y [Q(x, y) ⊕ R(x, y)] } and (∀x P(x)) ∧ [∀x ∀y(Q(x, y) ↔ R(x, y))] (c) Does [(p → q) ∧ (q → r)] → r implies (p → r) → r?