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Explain why $(p\lor \lnot q) \land (q \lor ¬r) \land (r \lor ¬p)$ is True when p,q,and r have the same truth value and it is false otherwise. (Without using a truth table )

Please help me solve this

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  • $\begingroup$ You have been given P AND Q AND R, OR, not Q and not P and not R. Any attempts at solving this? It should be completely intuitive with a little mathematical maturity. $\endgroup$ – Display Name Oct 20 '13 at 12:17
  • $\begingroup$ If $p,q$ and $r$ are all true, then... if $p,q$ and $r$ are all false, then... $\endgroup$ – Git Gud Oct 20 '13 at 12:19
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When you're completely lost, sometimes the best thing to do is to draw a truth-table, even if you can't hand that in for your assignment. Doing so can be illuminating, and may help you grasp what's going on, intuitively, so that you can confirm its truth and better understand why it's true.

Note that you can also rewrite your expression:

$$(p\lor \lnot q) \equiv q \rightarrow p$$

$$(q \lor \lnot r) \equiv r \rightarrow q$$

$$(r \lor \lnot p) \equiv p \rightarrow r$$

"Anding" them in reverse order gives $$(p\rightarrow r) \land (r \rightarrow q) \land (q\rightarrow p)$$
Maybe seeing this equivalent expression will help you understand intuitively why the statement is true if and only if $p, q, r$ all share the exact same truth-value.

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  • $\begingroup$ Thanks @amWhy it helped me as a beginner $\endgroup$ – MathDisease Oct 20 '13 at 12:30
  • $\begingroup$ You're welcome, @user2715467 $\endgroup$ – Namaste Oct 20 '13 at 12:31
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The first thing you must observe -according with the answer of amWhy- is that $$(p→r)∧(r→q)∧(q→p)$$ is a tripartite conjunction, so not an only one conjunct is allowed to be negative if these expression is wanted to be true. Let's say that the variables are bounded by the formal or internal reason that relates, or better, constrains -by means of the conditional operator-, the veritative circumstances that allows the positive verification of this expression. So you will observe that the 1st expression is negative only if $p=1$ and $r=0$. You are constrained from the very beginning to keep in mind these forbidden values, so the following constraints to simultaneously avoid are $(r=1, q=0)$ and $(q=1, p=0)$ for the second and third expressions.

So let's convert your expression in other terms to its shorter, equivalent forms: $(p→q)∧(q→p)$ by hypothetical syllogism into the 1st and the 2nd original conjuncts, so it can be asserted $(p≡q)$ that verificates only if both $p$ and $q$ equates. It follows by the same means upon the 3rd and the 1st original conjuncts $(q→r)∧(r→q)$ that is $(q≡r)$ which verifies too only if both $q$ and $r$ equates. From these two expressions it can be declared by addition $(p≡q)∧(q≡r)$ or even by similar conversions $(p≡q)∧(p≡r)$ or even more $(q≡r)∧(r≡p)$, all of them verifing iff the three variables have the same veritative value.

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  • $\begingroup$ Heard of full-stops? $\endgroup$ – BlackAdder Dec 16 '13 at 3:33

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