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I have asked a concrete example here.

The accepted answer said: $$\forall x\in A: P(x)\iff \forall x~(x\in A\to P(x))\\\exists y\in B: Q(y)\iff \exists y~(y\in B\land Q(y))$$

which means universal statements are restricted by conditional and existential statements are restricted by a conjunction.

But I still cannot convince myself why this is the case.

Imagine if C is a subset of D, is it equally right to say that $\forall x~(x\in C \land x \in D)$ because all C is in D?

JOHN
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  • If $\exists x\in A.P(x)$ is supposed to be equivalent to $\neg\forall x\in A.\neg P(x)$, then what does that mean for the translation? – Derek Elkins left SE Feb 05 '19 at 03:38
  • "For every $x$, $x$ is in $C$ and $x$ is in $D$" definitely doesn't mean $C\subseteq D$; take e.g. $C$ to be the set of multiples of $4$ and $D$ to be the set of multiples of $2$. $C\subseteq D$ is clearly true, but "Every $x$ is in $C$ and in $D$" is clearly false. – Noah Schweber Feb 05 '19 at 13:38

2 Answers2

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Imagine if C is a subset of D, is it equally right to say that $\forall x~(x\in C \land x \in D)$ because all C is in D?

No.   That is saying that everything is in both C and D.   This asserts that nothing is outside their intersection.

Where as you want to declare that: if anything is in C, then it will be in D.

$$C\subseteq D\quad\iff\quad\forall x~(x\in C \to x\in D)$$

The negation of this is: the claim that something is in C but not in D.$$C\nsubseteq D\quad\iff\quad\exists x~(x\in C\land x\notin D)$$

Graham Kemp
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Realize that your hypothetical scenario started with "If C is a subset of D" this is equivalent to the logic connective $\rightarrow$, which restricts the domain for the universal quantifier.

So your proposition should be $ \forall x(C \subset D) \rightarrow (x \in C \land x \in D)$

Above you showed that this makes sense.

Cedric Martens
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