Show that $p →(q ∨ r)$ and $(p ∧ ∼q) →r$ and $(p ∧ ∼r) →q$ are all logically equivalent.

Question

I need to show that the following are logical equivalent:

• $p → (q ∨ r) $
• $(p ∧ ∼q) →r$
• $(p ∧ ∼r) →q$

Here are the steps I took. I started with $p → (q ∨ r) $.

1. $p → (q ∨ r) $
2. $∼p ∨ (q ∨ r)$
3. $∼ (∼p ∨ (q ∨ r))$
4. $p ∧ ∼ (q ∨ r)$
5. $p ∧ (∼q) ∧ (∼r)$

My problem is how to justify going from $p ∧ (∼q) ∧ (∼r)$ to $(p ∧ ∼q) →r$ or $(p ∧ ∼r)$ →q.

Answer 1

What you've done between step two and three is wrong. Obviously the 2nd $∼p ∨ (q ∨ r)$ is not equal to its complement the 3rd $∼ (∼p ∨ (q ∨ r))$

You could've continued as following to complete the proof:

3. $(∼p ∨ q ) ∨ r $ (associativity and commutativity)
4. $∼(∼p ∨ q ) \implies r$ (Conditional identity)
5. $ (p ∧ ∼ q ) \implies r$ (De Morgan's)

Notice that each step is connected with if and only if condition. So for the backward implication, you can reverse the order of the steps.

proving $ p \implies (q ∨ r)$ if and only if $ (p ∧ ∼ r ) \implies q$ is almost same by writing $(∼p ∨ r ) ∨ q $ on 3rd step onward.