How do you prove that $p → q$ is equivalent to $p \lor q ↔ q$? I gotta draw $p \lor q ↔ q$ from $p → q$, logically. not by a truth table.
While it seems obvious, I cannot find a formal proof.
This is how far I came up to:
$\quad p \lor q$
$\equiv (p \land T) \lor q$
$\equiv q \lor (p \land T)$
$\equiv (q \lor p) \land (q \lor T)$
$\equiv (q \lor p) \land T$
$\equiv (q \lor p) \land (\neg p \lor q)$
I know that by drawing a venn-diagram here i can intuitively know that it is equivalent to q, but how do I draw such conclusion logically?
 A: $(p\lor q)↔q$
$=((p\lor q)→q)\land((p\lor q)←q)$
$=(\overline{(p\lor q)}\lor q)\land((p\lor q)\lor \overline {q})$
$=((\overline{p}\land \overline{q})\lor q)\land(p\lor (q\lor \overline {q}))$
$=((\overline{p}\land \overline{q})\lor q)\land(p\lor T)$
$=((\overline{p}\land \overline{q})\lor q)\land T$
$=((\overline{p}\land \overline{q})\lor q)$
$=(\overline{p}\lor q)\land (\overline{q}\lor q)$
$=(\overline{p}\lor q)\land T$
$=(\overline{p}\lor q)$
$=(p→q)$ 
A: You got this far:
$$p \vee q \equiv (\sim p \vee q) \wedge (p \vee q)$$
Use distributivity:
$$p \vee q \equiv (\sim p \wedge p) \vee q$$
By Law of Contradiction, we know that the former part of this disjunction is always false, so using disjunctive syllogism we can conclude:
$$p \vee q \equiv q$$
A: so we must prove: $(p\Rightarrow q)\Leftrightarrow [(p\vee
q)\Leftrightarrow q)]$.
suppose $(p\Rightarrow q)$ and prove the equivalence $(p\vee
q)\Leftrightarrow q$
$\Rightarrow$): $ (p\Rightarrow q)\Rightarrow[ (p\Rightarrow q)
\wedge (q\Rightarrow q)] \Rightarrow [(p\vee q)\Rightarrow q]$.
$\Leftarrow$): $q\Rightarrow(p\vee q)$ obvious
suppose $(p\vee q)\Leftrightarrow q$ and prove that $p\Rightarrow
q$
so $[(p\vee q)\Leftrightarrow q)]\Rightarrow [(p\vee q)\Rightarrow
q)]\Rightarrow [(p\Rightarrow q)\wedge (q\Rightarrow
q)]\Rightarrow (p\Rightarrow q)$.
A: One way to prove that the statements $p \rightarrow q$ and $p \vee q \iff q$ are logically equivalent is as follows.  It suffices to show that their truth tables (which have $4$ rows each) are the same.  Recall that the only assignment of truth values for $p$ and $q$ for which $p \rightarrow q$ evaluates to $F$ is when $p$ is $T$ and $q$ is $F$.  For which truth values does $p \vee q \iff q$ evaluate to $F$?  Precisely when $p \vee q$ and $q$ have different truth values.  If $q$ is $T$, so is $p \vee q$.  If $q$ is $F$, the only way for $p \vee q$ to be $T$ is if $p$ is $T$.  So, the only truth values for which $p \vee q \iff q$ is $F$ is when $p$ is $T$ and $q$ is $F$. This truth table is the same as that of $p \rightarrow q$. 
