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