Solution to Prove the equivalence of the following in three different ways (truth table, simplification, each is … - Sikademy
Author Image

Archangel Macsika

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.

The Answer to the Question
is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

Get the Answers Now!

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question

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

\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 \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 \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 \left. {\left( {p \wedge \sim q} \right) \to r} \right| = p \to \left( {q \vee r} \right) .

We have \left. {p \to \left( {q \vee r} \right)} \right| = \left( {p \wedge \sim q} \right) \to r and \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.

Related Answers

Was this answer helpful?

Join our Community to stay in the know

Get updates for similar and other helpful Answers

Question ID: mtid-5-stid-8-sqid-3277-qpid-1976