logical propositions $\left ( \left ( p\Rightarrow q \right )\Leftrightarrow p \right )\iff p \wedge q$

Question

I have been trying to do this exercise but I can t come to the solution so I need a hint pls, i think that there is a step that i don t see (i can't use truth tables to demonstrate this). $$\left ( \left ( p\Rightarrow q \right )\Leftrightarrow p \right )\Leftrightarrow p \wedge q$$

is not so sort but this is all i could did

Answer 1

Note that $a \leftrightarrow b$ is the same as $(a \wedge b) \vee (\neg a \wedge \neg b)$. So you want to show that the following:

$$[((p \to q) \wedge p) \vee (\neg (p \to q) \wedge \neg p)] \leftrightarrow p \wedge q$$

Expanding out the implication in terms of "not" and "or":

$$[((q \vee (\neg p)) \wedge p) \vee (\neg (q \vee (\neg p)) \wedge \neg p)] \leftrightarrow p \wedge q$$

The very first chunk, $(q \vee (\neg p)) \wedge p$, can be multiplied out into $(p \wedge q) \vee (p \wedge \neg p)$, which is equivalent to $(p \wedge q)$.

The second chunk, $\neg (q \vee (\neg p)) \wedge \neg p$, is de Morgan'ed into $\neg (p \vee q \vee \neg p)$, which is equivalent to the constant false.

---

The trick is to spot that $(p \to q) \wedge p$ is already equivalent to $p \wedge q$, so you can strongly suspect that the second component is going to cancel out.