Which law of logical equivalence says $P\Leftrightarrow Q ≡ (P\lor Q) \Rightarrow(P\land Q)$ I'm going through the exercises in the book Discrete Mathematics with Applications. I'm asked to show that two circuits are equivalent by converting them to boolean expressions and using the laws in this table.
$$\begin{array}{lcc}
\hphantom{1}\mathsf{1.\; Commutative\; laws:}       &  p\land q \equiv q\land p                          &  p\lor q \equiv q\lor p\\
\hphantom{1}\mathsf{2.\; Associative\; laws:}       &  (p\land q)\land r \equiv p\land (q\land r)        &  (p\lor q)\lor r \equiv p\lor (q\lor r)\\
\hphantom{1}\mathsf{3.\; Distributive\; laws:}      &  p\land (q\lor r) \equiv (p\land q)\lor (p\land r) &  p\lor (q\land r) \equiv (p\lor q)\land (p\land r)\\
\hphantom{1}\mathsf{4.\; Identity\; laws:}          &  p\land t \equiv p                                 &  p\lor c \equiv p\\
\hphantom{1}\mathsf{5.\; Negation\; laws:}          &  p\lor \lnot p \equiv t                           &  p\land \lnot p  \equiv c\\
\hphantom{1}\mathsf{6.\; Double\; negative\; law:}  &  \lnot(\lnot p) \equiv p                           &\\
\hphantom{1}\mathsf{7.\; Idempotent\; laws:}        &  p\land p \equiv p                                 &  p\lor p \equiv p\\
\hphantom{1}\mathsf{8.\; Universal\; bound\; laws:} &  p\lor t \equiv t                                  &  p\land c\equiv c\\
\hphantom{1}\mathsf{9.\; De\; Morgan}\text{'}\mathsf{s\; laws:}     &  \lnot(p\land q) \equiv \lnot p\lor \lnot q        &  \lnot(p\lor q) \equiv \lnot p\land\lnot q\\
\mathsf{10.\; Absorption\; laws:}                   &  p\lor (p\land q) \equiv p                         &  p\land (p\lor q) \equiv p\\
\mathsf{11.\; Negations\; of\; t\; and\; c:}        &  \lnot t \equiv c                                  &  \lnot c \equiv t\\
\end{array}$$
so as which law/s of logical equivalence says $P\Leftrightarrow Q ≡ (P\lor Q) \Rightarrow(P\land Q)$
I can see their equivalence clearly with a truth table. But the book is asking me to show it using the equivalence laws in the above table, and I can't see how any of them apply here. So, do any of those laws apply here in a way I'm not understanding? Or is there some other known law that does apply here?
 A: The truth table involves "t" and "c" throughout.  So, you can rewrite the truth table using the negation laws.  That is, instead of using "t" and "c" in your truth table, use (x∨¬x), and (x $\land$ $\lnot$x) respectively.  Then, since you'll have (x∨¬x) throughout the last column of the truth table, you use the equivalence law (x∨¬x)=t in an additional column, and thus you've showed the formula true using the equivalence laws...  specifically your last two columns will look like this:
(x∨¬x)  t (by the equivalence law (x∨¬x)==t)   
(x∨¬x)  t (by the equivalence law (x∨¬x)==t)
(x∨¬x)  t (by the equivalence law (x∨¬x)==t)
(x∨¬x)  t (by the equivalence law (x∨¬x)==t)

A: Let's assume the following definitions:

$\begin{array}{lc}
\mathsf{12.\; Definition \; of \; Implication:}       &  p \Rightarrow q \equiv \neg p \lor q \\
\mathsf{13.\; Definition \; of \; Biconditional:}     &  p \Leftrightarrow q \equiv (p \Rightarrow q) \land (q \Rightarrow p)
\end{array}$

Then we have:
$$\begin{array}{rll}
P \Leftrightarrow Q &\equiv (P \Rightarrow Q) \land (Q \Rightarrow P) &\text{by (13)} \\
&\equiv  (\neg P \lor Q) \land (\neg Q \lor P) &\text{by (12)} \\
&\equiv  (\neg P \land (\neg Q \lor P)) \lor (Q \land (\neg Q \lor P)) &\text{by (3)} \\
&\equiv  ((\neg P \land \neg Q) \lor (\neg P \land P)) \lor ((Q \land \neg Q) \lor (Q \land P)) &\text{by (3)} \\
&\equiv  ((\neg P \land \neg Q) \lor (P \land \neg P)) \lor ((Q \land \neg Q) \lor (P \land Q)) &\text{by (1)} \\
&\equiv  ((\neg P \land \neg Q) \lor c) \lor (c \lor (P \land Q)) &\text{by (5)} \\
&\equiv  ((\neg P \land \neg Q) \lor c) \lor ((P \land Q) \lor c) &\text{by (1)} \\
&\equiv  (\neg P \land \neg Q) \lor (P \land Q) &\text{by (4)} \\
&\equiv  \neg (P \lor Q) \lor (P \land Q) &\text{by (9)} \\
&\equiv  (P \lor Q) \Rightarrow (P \land Q) &\text{by (12)} \\
\end{array}$$
