Let $S$ be a nonempty set of natural numbers. Is the following formula $$ \exists p\ \bigl(\text{$p$ is prime } \rightarrow \forall x \text{ ($x$ is prime)}\bigr) $$ true or false on $S$? I know the answer to this question, but what would be the shortest way to arrive to the conclusion using some deduction system?
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Hint: $A \to B$ is true if $A$ is false. – Robert Israel Feb 15 '19 at 15:25
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2See Drinker's Paradox : the formula is an instance of an universally valid formula. – Mauro ALLEGRANZA Feb 15 '19 at 15:27
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The "shortest way to arrive at the conclusion" will depend on which deduction system you mean when you say "some deduction system". If I can choose one myself, I'l define a deductive system where I've declared the drinker formula to be a logical axiom; then the proof is very short! – hmakholm left over Monica Feb 15 '19 at 15:29
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@ArnaudMortier - in the linked post there are many proofs of it: it is enough to replace $P(x)$ with "$x \text { is prime }$" in one of them. – Mauro ALLEGRANZA Feb 15 '19 at 15:31