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If I encounter an statement like the following: $$\exists x P(x)\land\exists y Q(y)$$

Should this be interpreted as if x and y refer necessarily to different objects or it is to be interpreted as they may or may not be the same object?

Is it therefore equivalent to: $$\exists x P(x)\land\exists x Q(x)$$ ?

Martx
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    It is equivalent. There are no further conditions on satisfying objects so being equal is allowed. – drhab Jan 07 '22 at 13:48
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    $\exists x P(x)\land\exists x Q(x)$ is not the same as $\exists x P(x)\land Q(x)$. In the last instance, it has to be the same $x$. I prefer using two different letters ($x$ and $y$) to stress the difference, especially for persons starting in logic. – MasB Jan 07 '22 at 13:54
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    The variable $x$ is a bound variable, and its "scope" is the part of the sentence surrounded by the existential quantifier - it does not "exist" outside. The same for $y$. In other words, even if you write $(\exists x)P(x)\land (\exists x)Q(x)$, the "$x$" in the first part is different from the "$x$" in the second part. Thus, why not remove ambiguity, use different letters for different things, and write $(\exists x)P(x)\land (\exists y)Q(y)$ ? Both are correct but the latter is clearer. –  Jan 07 '22 at 13:59
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    Both versions say "P is non-empty and Q is non-empty." – Ned Jan 07 '22 at 16:47
  • @StinkingBishop Thank You! So, just to be clear, in this kind of statements the scope of the first existential quantifier finishes at the $\land$ symbol? – Martx Jan 07 '22 at 23:18

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just to be clear, in this kind of statements, the scope of the first existential quantifier finishes at the $∧$ symbol?

Yes, the convention is that quantifiers apply to as little as construeable; so, parentheses are required when the quantification is meant to apply beyond this. For example, \begin{align}\big(\forall x\, A(x)\big)\to B(x)\quad\equiv\quad\forall x \, A(x)\to B(x)\quad\not\equiv\quad\forall x\;\big(A(x)\to B(x)\big).\end{align}

(For unambiguity and good practice, the first, second, fourth, and fifth occurrences of $x$ in the above ought to be replaced with $y.$)

Is $$\exists x P(x)\land\exists y Q(y)\tag1$$ equivalent to $$\exists x P(x)\land\exists x Q(x)\;?\tag2$$

Yes, sentences $(1)$ and $(2)$ are logically equivalent to each other.

Should this be interpreted as if $x$ and $y$ refer necessarily to different objects, or it is to be interpreted as they may or may not be the same object?

The latter. For example, if predicates $P$ and $Q$ symbolise “is Taiwanese” and “likes to read”, respectively, and Brigitte is a Taiwanese bookworm, then both variables $x$ and $y$ in sentence $(1)$ can refer to her.

ryang
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  • Thank you! So, $\forall x , A(x)\to B(x)$ should be written as $\forall y , A(y)\to B(x)$ and be interpreted as: whenever every element y makes A(y) true then another element x makes B(x) true? – Martx Jan 08 '22 at 11:39
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    @Martx 1. $∀yA(y)→B(x)$ is not actually a sentence, due to the variable $x$ hanging free. 2. On the other hand, $∃x\Big(∀yA(y)→B(x)\Big)$ and $∀yA(y)→∃xB(x)$ (they are equivalent sentences) are understood as "whenever every $y$ makes $A(y)$ true, some $x$ makes $B(x)$ true". Notice that, unlike your translation, mine avoids saying "another $x$" or "some other $x$"? This is because, as my Brigitte example illustrates, both variables $x$ and $y$ *can* still point to the same object. – ryang Jan 08 '22 at 12:38
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They are equivalent statements in theory . because $ x$ and $y$ concern only the statement they are in (they are called bound variables as mentioned in the comments).

So in theory you can name both of them $ x$

However in practice , you should note them differently, it is necessary to do so actually .

Otherwise during the reasoning for example you can't differentiate the $ x$ that satisfies $ P$ from the one that satisfies $ Q$ which will lead to an unnecessary ambiguity .

Oussama Akar
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  • I don't think this is right. The question is tagged "logic" so presumably is about the topic "logic" as a branch of mathematics. So given a formula $\phi$ it may very well use $x$ at varies places tied to different quantifiers. If we replace some of these with other symbols and get a formula $\hat{\phi}$ then this is not equal to $\phi$. Depending on what we are doing we will have to check out that if we can prove one we can prove the other etc etc; or that structures which model one model the other etc etc. So without knowing the OP's precise logic we don't know if we can juggle the letters. – ancient mathematician Jan 07 '22 at 15:46