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.
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:
It's obvious that . 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 .
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 .
We have and , therefore p → (q ∨ r) ≡ (p ∧ ~q) → r. Proven.
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