Proving a logic theorem I am trying to prove this theorem : $((p ⇒ ¬ r) ∧ (s ∧ r ≡ ¬ q) ≢ ¬ ((r ⇒ ¬ s) ≡ q) ⇒ r ∧ p) ≡ p ∧ (p ⇒ r)$, but I am not sure where to start.
$$((p ⇒ ¬ r) ∧ (s ∧ r ≡ ¬ q) ≢ ¬ ((r ⇒ ¬ s) ≡ q) ⇒ r ∧ p)$$
$$=    ⟨ \text{Definition of ≢ with } p, q ≔ (p ⇒ ¬ r) ∧ (s ∧ r ≡ ¬ q), ¬ (r ⇒ ¬ s ≡ q) ⇒ r ∧ p) ⟩$$
$$¬ ((p ⇒ ¬ r) ∧ (s ∧ r ≡ ¬ q) ≡ ¬ ((r ⇒ ¬ s) ≡ q) ⇒ r ∧ p)$$
Can you give me a hint how to continue? I am interested in a proof using logical laws instead a truth table.
 A: Using material implication we can replace $(p \Rightarrow \lnot r)$ by $(\lnot p \lor \lnot r)$. Using the Morgan's law we can then replace it by $\lnot (p\land r)$.
Using material implication we can replace $(r \Rightarrow \lnot s)$ by $(\lnot r \lor \lnot s)$. Using the Morgan's law we can then replace it by $\lnot (r\land s)$. Using commutativity of conjuction, we can then replace it by $\lnot (s\land r)$
We have an expression $(\lnot (s\land r) \equiv q)$. It can be replaced by $((s\land r) \equiv \lnot q)$.
We have an expression $\lnot((s\land r) \equiv \lnot q) \Rightarrow (r \land p)$. Using material implication  it can be replaced by $\lnot\lnot((s\land r) \equiv \lnot q) \lor (r \land p)$. Using double negation it can be replaced by $((s\land r) \equiv \lnot q) \lor (r \land p)$. Using commutativity of conjuction, we can then replace it by $((s\land r) \equiv \lnot q) \lor (p \land r)$.
We have an expression
$$ (\lnot (p \land r) \land ((s\land r) \equiv \lnot q)) ≢ ((s\land r) \equiv \lnot q) \lor (p \land r) $$
which is of the form $ (\lnot x \land y) ≢ (y \lor x)$ where $x = p\land r$, $y = (s\land r) \equiv \lnot q$. By the definition of $≢$, it can be replaced by
$$ \big((\lnot x \land y) \land \lnot(y \lor x)\big) \lor \big(\lnot(\lnot x \land y) \land (y \lor x)\big)$$
Using de Morgan's laws it can be replaced by
$$ \big((\lnot x \land y) \land (\lnot y \land \lnot x)\big) \lor \big((\lnot\lnot x \lor \lnot y) \land (y \lor x)\big)$$
Using assosciativity of conjugation, $(\lnot x \land y) \land (\lnot y \land \lnot x)$ can be replaced by $\lnot x \land (y \land \lnot y) \land \lnot x$, which is a conjuction that includes a alwayys false expression $y \land \lnot y$. Therefore, this whole expression can be replaced by ${\bf 0}$. Using double negation, $\lnot\lnot x$ can be replaced by $x$ so we get
$$ {\bf 0} \lor \big((x \lor \lnot y) \land (y \lor x)\big)$$
which can be replaced by
$$ (x \lor \lnot y) \land (y \lor x)$$
$$ (x \lor \lnot y) \land (x \lor y)$$
$$ x \lor (\lnot y \land y)$$
$$ x \lor {\bf 0}$$
$$ x$$
In this way, the whole expression is now
$$ x \equiv p \land (p \Rightarrow r)$$
that is
$$ p \land r \equiv p \land (p \Rightarrow r)$$
I'll leave proving that to you, it should be simple enough.
