Q.Show that $p \to q$ and $q \to p$ are not equivalent.

Logical equivalence. Two compound statements $A$ and $B$ are logically equivalent (written $A\equiv B$ or $A\Leftrightarrow B$) if their truth-table columns are identical in every row. Equivalently -- and this is the practical test -- $A\equiv B$ exactly when $A\leftrightarrow B$ is a tautology.

The standard laws of equivalence (each provable by truth table, and usable afterwards to prove new equivalences symbolically, without redrawing a table):

1. Idempotent: $p\vee p\equiv p$, $\quad p\wedge p\equiv p$.
2. Commutative: $p\vee q\equiv q\vee p$, $\quad p\wedge q\equiv q\wedge p$.
3. Associative: $p\vee(q\vee r)\equiv(p\vee q)\vee r$, $\quad p\wedge(q\wedge r)\equiv(p\wedge q)\wedge r$.
4. Distributive: $p\vee(q\wedge r)\equiv(p\vee q)\wedge(p\vee r)$, $\quad p\wedge(q\vee r)\equiv(p\wedge q)\vee(p\wedge r)$.
5. Identity: $p\vee\mathbb T\equiv\mathbb T$, $\ p\vee\mathbb F\equiv p$; $\quad p\wedge\mathbb T\equiv p$, $\ p\wedge\mathbb F\equiv\mathbb F$.
6. Complement: $p\vee\neg p\equiv\mathbb T$, $\ p\wedge\neg p\equiv\mathbb F$; $\quad \neg\mathbb T\equiv\mathbb F$, $\ \neg\mathbb F\equiv\mathbb T$.
7. Involution (Double Negation): $\neg(\neg p)\equiv p$.
8. De Morgan's Laws: $\neg(p\wedge q)\equiv\neg p\vee\neg q$, $\quad \neg(p\vee q)\equiv\neg p\wedge\neg q$.
9. Absorption: $p\vee(p\wedge q)\equiv p$, $\quad p\wedge(p\vee q)\equiv p$.

Two named equivalences worth memorising on their own:

• $p\to q\equiv\neg p\vee q$ -- a conditional is "not-$p$ or $q$", which is exactly why a false hypothesis makes $p\to q$ automatically true.
• $p\leftrightarrow q\equiv(p\to q)\wedge(q\to p)\equiv(p\wedge q)\vee(\neg p\wedge\neg q)$ -- "if and only if" really does mean "both directions hold", and can also be read as "$p,q$ agree". …