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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?

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    $\begingroup$ @Zev: Wow.${}{}$ $\endgroup$
    – Asaf Karagila ♦
    Commented May 21, 2013 at 0:56
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    $\begingroup$ Haha thanks :) I get carried away sometimes... $\endgroup$
    – Zev Chonoles
    Commented May 21, 2013 at 1:00
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    $\begingroup$ Nice long list, but obviously not enough, we need something that mentions $\longrightarrow$ and $\longleftrightarrow$. $\endgroup$
    – André Nicolas
    Commented May 21, 2013 at 1:05
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    $\begingroup$ @GastónBurrull: Other common mistakes include writing $<a,b>$ instead of $\langle a,b\rangle$ or $\{ x|\varphi(x)\}$ instead of $\{x\mid \varphi(x)\}$ or $|x|$ instead of $\lvert x\rvert$ or $||x||$ instead of $\lVert x\rVert$. On a different note, - instead of – or — (in proper LaTeX you typeset endash by two minuses -- and emdash by three ---). Also, some people write $\to$ instead of $\mapsto$, but I've seen it done on blackboard, too... $\endgroup$
    – tomasz
    Commented May 21, 2013 at 2:17
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    $\begingroup$ @GastónBurrull: you can right-click any math formula and then choose “Show Math As -> TeX Commands”. $\endgroup$
    – tomasz
    Commented May 21, 2013 at 2:30

2 Answers 2

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

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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)
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  • $\begingroup$ I guess maybe I would do better to add here that since truth tables consist of no more than an exhaustive set of truth-value calculations, and since truth-value calculations can get formalized via the prtothetic of Lesniewski (p. 66-67 of Arthur Prior's Formal Logic), the above procedure should qualify as a means to write an informal proof. $\endgroup$
    – Doug Spoonwood
    Commented May 24, 2013 at 17:09

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