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I am currently using Rosen's "Discrete Mathematics and Its Applications" (7th ed.) for my discrete mathematics course. We recently talked about quantifiers, more specifically the universal and existential quantifier. What confuses me is when we started talking about restricting the domain (i.e. instead of all real numbers, only all real positive numbers), where does the conditional statement come from when restricting the domain of a universal quantifier and where does the conjunction come from when restricting the domain of a existential quantifier?

This link did help: Universal and Existential quantifier in Propositional logic

However, the part that confuses here is how are the domains restricted in the answer given in the link above?

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A. L.
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2 Answers2

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The restricted domain statement $~\forall x{\in} A ~ P(x)~$ reads, "For all $x$ in $A$, the predicate $P(x)$ holds".

This is equivalent to saying, "For all $x$, if it is in $A$, then the predicate $P(x)$ holds."

Which is $~\forall x~ \big(x\in A~\to~ P(x)\big)~$.

It is a conditional rather than a conjunction for a universal statement because, by restricting the claim to those $x$ that are in $A$, we are making no claim about all the $x$ that are not in $A$.   Hence the restriction is not saying that all $x$ are in $A$, only that if they are there, then the predicate is true for them.

The restricted domain statement $~\exists x{\in} A ~ P(x)$ reads, "There exists some $x$ in $A$, where the predicate $P(x)$ holds".

This is equivalent to saying, "There exists some $x$, that is in $A$ and the predicate $P(x)$ holds."

Which is $~\exists x~\big(x\in A~\wedge~ P(x)\big)~$.

This is a conjunction, because by restricting the domain to $A$ we are not making any claim about any $x$ that exists elsewhere, only that there is some $x$, which make the predicate true, that are in $A$.

Graham Kemp
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  • In case of ∀x∈A P(x), are we allowed to put a value of x which is not in the set A? Because in ∀x (x∈A → P(x)), we are allowed to do so. – Preetom Saha Arko Mar 08 '19 at 07:57
  • The statements $\forall x{\in} A~P(x)$ and $\forall x~(x{\in} A\to P(x))$ are the exact same claim. The former is just a shortened notation for the latter. Thus Universal Elimination on either statement to a term $c$ produces $c{\in} A\to P(c)$. @PreetomSahaArko – Graham Kemp Mar 09 '19 at 23:49
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Technically, the domain that the quantifiers quantify over is always the same. Or, to be exact, you don't change what the quantifiers quantify over by using conditionals or any other logical operator. Thus, if within some context I say:

$\forall x (P(x) \rightarrow Q(x))$

then the $\forall$ still quantifies over the exact same set of objects as in:

$\forall x Q(x)$

Of course, effectively, the latter claims says that all of the objects have property $Q$, while the former says that all of the $P$'s from that domain have property $Q$, and so it looks like the quantification is happening over a more restricted domain (we go from "all objects" to "all P's").

However, the former really says "For all of the objects from the domain: if you are a P then you are a Q", so the universal still quantifies over the whole domain ... as they always do and always will!

Bram28
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