Is the following expression true? $(q \Rightarrow p) \Rightarrow(p \Rightarrow q)\equiv(p \Rightarrow q)$

Question

Is the following expression true? $$(q \Rightarrow p) \Rightarrow(p \Rightarrow q)\equiv(p \Rightarrow q)$$

It's actually from a bigger problem, and I'm currently at $(q \Rightarrow p) \Rightarrow(p \Rightarrow q)$. I have to get to $p \Rightarrow q$ to prove the theorem.

$$(q \Rightarrow p) \Rightarrow(p \Rightarrow q)$$ $$(q ∧ p) \Rightarrow(p ∧ q)$$ $$(q ∧ p) ∧ (p ∧ q)$$ $$p∧q$$ $$p \Rightarrow q$$

Is there a faster way to prove this? Any help is appreciated!

Answer 1

You can just use a truth-table to confirm that this is indeed true.

Or:

$(q \Rightarrow p) \Rightarrow (p \Rightarrow q) \equiv$ (Implication)

$\neg (q \Rightarrow p) \lor (p \Rightarrow q) \equiv$ (Implication x2)

$(q \land \neg p) \lor \neg p \lor q \equiv$ (Absorption)

$\neg p \lor q \equiv$ (implication)

$p \Rightarrow q$

Answer 2

Yes it is true, but you need to use the fact that $a \Rightarrow b \equiv \neg a \lor b$ as well as the absorption laws (see http://mathworld.wolfram.com/AbsorptionLaw.html) to prove it. Your assumption that $a \Rightarrow b \equiv a \land b$ is not correct.

Answer 3

Your work doesn't look correct. My work is below. I use the usual laws of sentential logic, which includes the law saying that $p \to q$ and $\neg p \vee q$ are equivalent (I'll call this the Conditional law). $$ \begin{aligned}[t] (q \to p) \to (p \to q) &\equiv \neg(\neg q \vee p) \vee (\neg p \vee q)\\ &\equiv (q \wedge \neg p) \vee (\neg p \vee q) \\ &\equiv [(q \wedge \neg p) \vee \neg p] \vee q \\ &\equiv [\neg p] \vee q \\ &\equiv p \to q \end{aligned} \qquad \begin{aligned}[t] &{}\\ &\text{DeMorgan; Double Negation} \\ &\text{Associative} \\ &\text{Absorption} \\ &\text{Conditional} &\end{aligned} $$

Answer 4

I feel compelled to point out that $(q \Rightarrow p) \Rightarrow(p \Rightarrow q)\equiv(p \Rightarrow q)$ is valid in intuitionistic logic (so you don't need to use $a \Rightarrow b \equiv \neg a \lor b$). You also don't need to introduce any connectives (like $\land$ or $\lor$ into the proof). To see this, note that the right-to-left direction of $(q \Rightarrow p) \Rightarrow(p \Rightarrow q)\equiv(p \Rightarrow q)$ is trivial and note that for the left-to-right direction, what you have to do is assume $p$ and $(q \Rightarrow p) \Rightarrow (p \Rightarrow q)$ and from that prove $q$. But that's easy: from the first assumption $p$, infer $q \Rightarrow p$, whence by the second assumption you can infer $p \Rightarrow q$ and then using the assumption $p$ again, you can infer $q$.