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Determine whether or not $((P\land Q)\implies R)$ tautologically implies $((P\implies R)\lor (Q\implies R))$

How do I determine that $((P\land Q)\implies R)$ tautologically implies $((P\implies R)\lor(Q\implies R))$? The problem is not to show equivalence, but to determine whether one formula tautologically implies another, and I'm confused about the difference. Any clarification would be appreciated.

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2 Answers 2

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A formula A either will tautologically imply another formula B, or it will not do so. If A does NOT tautologically imply B, then there exists some truth-value assignment such that A holds true, and B qualifies as false. Suppose ((P→R)∨(Q→R)) false. Then, (P→R)qualifies as a false, and so does (Q→R). Thus, P qualifies as true, Q qualifies as true, and R qualifies as false. If those conditions hold, then ((P∧Q)→R) qualifies as false also. So, it is not the case that A does not tautologically imply B. Thus, because of the content of the initial sentence, ((P∧Q)→R) tautologically implies ((P→R)∨(Q→R)).

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    $\begingroup$ I'm upvoting this answer for obvious reasons, but I'm leaving mine so that (1)I'll blush every time I'll remember this in the future for my lack of attention, and (2) so that others, and in particular the OP, can see the difference. $\endgroup$
    – DonAntonio
    Commented Oct 4, 2013 at 5:25
  • $\begingroup$ Thank you for correcting this, Doug. $\endgroup$ Commented Oct 4, 2013 at 22:25
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You have to check whether any time $\;(P\wedge Q)\rightarrow R\;$ gets a $\;T\;$ value, also $\;(P\rightarrow R)\vee(Q\rightarrow R)\;$ gets$\;T\;$, but:

Check that the second formula gets a value $\;F\;$ iff both its two parts get this value, meaning: only if $\;P=F\;,\;R=T\;$ and also $\;Q=F\;,\;R=T\;$, and this means there's one case where the first formula is true:

$$(\stackrel{F}P\wedge \stackrel{F}Q)\rightarrow \stackrel{T}R$$

But the second formula is false:

$$((\stackrel{F}P\rightarrow (\stackrel{T}R)\vee((\stackrel{F}Q\rightarrow (\stackrel{T}R)$$

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  • $\begingroup$ Would a truth table be sufficient to explain this? $\endgroup$ Commented Oct 2, 2013 at 18:04
  • $\begingroup$ Also, thanks for the clarification. $\endgroup$ Commented Oct 2, 2013 at 18:04
  • $\begingroup$ Of course it would: that's the way these things are done. Yet I'm not sure how to do that Using LaTeX in this site and, besides, it was easier and faster as above. $\endgroup$
    – DonAntonio
    Commented Oct 2, 2013 at 18:06
  • $\begingroup$ If P=F, then the disjunction holds true, since the left conditional holds true. This answer isn't correct. $\endgroup$ Commented Oct 4, 2013 at 4:04
  • $\begingroup$ @DerrekWhistle, Doug's right. Check this. $\endgroup$
    – DonAntonio
    Commented Oct 4, 2013 at 5:16

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