Show that $(p ⇒ q) ⇒ (r ⇒ s) ⇐⇒ (p ⇒ r) ⇒ (q ⇒ s)$ is a tautology?
I am having a little trouble in proving this proofs without truth tables
the idea for solve this question is to work out left side and right side separately, trying to match the result.but i can't match the result :(
I have tried:
left side:
1. $(p ⇒ q) ⇒ (r ⇒ s)$
2. $(¬p ∨ q) ⇒ (¬r ∨ s)$
3. $¬(¬p ∨ q) ∨ (¬r ∨ s)$
4. $(p ∧ ¬q) ∨ (¬r ∨ s)$
5. $[(p ∧ ¬q) ∨ ¬r] ∨ s$
6. $[(p ∨ ¬r) ∧ (¬q ∨ ¬r)] ∨ s$
right side:
1. $(p ⇒ r) ⇒ (q ⇒ s)$
2. $(¬p ∨ r) ⇒ (¬q ∨ s)$
3. $¬(¬p ∨ r) ∨ (¬q ∨ s)$
4. $(p ∧ ¬r) ∨ (¬q ∨ s)$
5. $[(p ∧ ¬r) ∨ ¬q] ∨ s$
6. $[(p ∨ ¬q) ∧ (¬r ∨ ¬q)] ∨ s$
however don't see where this will help me. I don't really know how to go on from her.
any assistance would be greatly appreciated.
Suppose $p, r, s$ are F, and $q$ is T.
Then $(p \to q) \to (r \to s)$ is T
And $(p \to r) \to (q \to s)$ is F
So on that valuation, $((p \to q) \to (r \to s)) \leftrightarrow ((p \to r) \to (q \to s))$ is F, and isn't a tautology.
[That's assuming, as your answer attempt assumes too, that $\to$ takes precedence over $\leftrightarrow$, so we should bracket as shown.]