Answer to Question #172871 in Discrete Mathematics for sushii

Question #172871

Prove the equivalence of the following in three different ways (truth table, simplification,

each is a logical consequence of the other): p → (q ∨ r) ≡ (p ∧ ~q) → r.


1
Expert's answer
2021-03-19T12:40:54-0400

We prove the equivalence using the truth table.





We see that the formulas p → (q ∨ r) and (p ∧ ~q) → r take the same truth value therefore the formulas are equivalent.Proven.


Prove by simplifying:

p(qr)pqrp \to \left( {q \vee r} \right) \equiv \sim p \vee q \vee r

(pq)r(pq)rpqr\left( {p \wedge \sim q} \right) \to r \equiv \sim \left( {p \wedge \sim q} \right) \vee r \equiv \sim p \vee q \vee r

It's obvious that pqrpqrp(qr)(pq)r\sim p \vee q \vee r \equiv \sim p \vee q \vee r \Rightarrow p \to \left( {q \vee r} \right) \equiv \left( {p \wedge \sim q} \right) \to r . Proven.


Show that (p ∧ ~q) → r is a logical consequence of p → (q ∨ r).

Let p → (q ∨ r) ≡1. Let's say that (p ∧ ~q) → r ≡0. Then (p ∧ ~q) ≡1 and r ≡0, but then p ≡1 and q≡0. Then p → (q ∨ r) ≡1→(0 ∨ 0)≡0 - we have a contradiction. Therefore, if p → (q ∨ r) ≡1 then (p ∧ ~q) → r ≡1, so p(qr)=(pq)r\left. {p \to \left( {q \vee r} \right)} \right| = \left( {p \wedge \sim q} \right) \to r .

Show that p → (q ∨ r) is a logical consequence of (p ∧ ~q) → r.

Let (p ∧ ~q) → r≡1. Let's say that p → (q ∨ r) ≡0. Then p ≡1 and (q ∨ r) ≡0. Then q ≡r≡0. But then (p ∧ ~q) → r≡(1 ∧1)→0≡0 - we have a contradiction. Therefore, if (p ∧ ~q) → r≡1 then p → (q ∨ r) ≡1, so (pq)r=p(qr)\left. {\left( {p \wedge \sim q} \right) \to r} \right| = p \to \left( {q \vee r} \right) .

We have p(qr)=(pq)r\left. {p \to \left( {q \vee r} \right)} \right| = \left( {p \wedge \sim q} \right) \to r and (pq)r=p(qr)\left. {\left( {p \wedge \sim q} \right) \to r} \right| = p \to \left( {q \vee r} \right) , therefore p → (q ∨ r) ≡ (p ∧ ~q) → r. Proven.


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