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Let $U = \{s_1, s_2, s_\cdots, s_n\}$ be the universal set and $A = \{a_1, a_2, a_3,\cdots, a_m\}$ is the set under consideration.

Now, I am proving $\forall_{x \in \{\}} p(x) = T$ as follows

$$\forall_{x \in A} p(x) = \forall_{x \in u} (x \in A \implies p(x)) $$ $$= (s_1 \in A \implies p(s_1)) \land (s_2 \in A \implies p(s_2)) \land \cdots (s_n \in A \implies p(s_n))$$

Hence if $A = \{\}$, then premise in every conditional statement becomes false and the result will be true.

But it is not working to prove $\exists_{x \in \{\}} p(x) = T$;

How to prove it?

hanugm
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    $\exists_{x \in {}} p(x)$ is $\exists x (x \in { } \land p(x))$. – Mauro ALLEGRANZA Jun 06 '18 at 17:08
  • What will be the expansion for $\exists_{x \in A} p(x)$ – hanugm Jun 06 '18 at 17:10
  • Like this

    $$\forall_{x \in A} p(x) = \forall_{x \in u} (x \in A \implies p(x)) $$ $$= (s_1 \in A \implies p(s_1)) \land (s_2 \in A \implies p(s_2)) \land \cdots (s_n \in A \implies p(s_n))$$

    Just to replace $\land$ by $\lor$?

     – hanugm Jun 06 '18 at 17:11
  • $\forall$ is like a conjunction while $\exists$ is like a disjunction. Thus $\exists x P(x)$ will be : $P(s_1) \lor P(s_2) \lor \ldots \lor P(s_n)$. – Mauro ALLEGRANZA Jun 06 '18 at 17:16
  • So, is it true that

    $$\exists{x \in A} p(x) = \exists{x \in u} (x \in A \implies p(x)) $$ $$= (s_1 \in A \implies p(s_1)) \lor (s_2 \in A \implies p(s_2)) \lor \cdots (s_n \in A \implies p(s_n))$$

     – hanugm Jun 06 '18 at 17:18
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    NO; as said above $\exists_{x \in A}P(x)$ is $\exists x (x \in A \land P(x))$. – Mauro ALLEGRANZA Jun 06 '18 at 17:20

1 Answers1

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This won't work because $\exists x\in\{\},p(x)$ is false.

Like Mauro says in the first comment, $ \exists x\in\{\},p(x) $ is syntactic sugar for $\exists x , (x\in\{\} \wedge p(x))$. But $x\in\{\}$ can never be true, so $x\in\{\} \wedge p(x)$ is never true, so $\exists x , (x\in\{\} \wedge p(x))$ is false.

You seem hung up on expanding quantification to the individual elements of a set. That's not necessary, and seems to limit you logically to a finite universal set. But these statements are true irrespective of the cardinality of $U$.

Matthew Leingang
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