3
$\begingroup$

I have a question like this

Let $P(x,y,z)$ denote $xy=z$, $E(x,y)$ denote $x=y$, $G(x,y)$ denote $x>y$. Transcribe the following into logical notation. Assume that the universe of discourse is the set of integers.

(1) If $y=1$, then $xy=x$

The solution given is: $\forall y [E(y,1)\rightarrow \forall xP(x,y,x)] $

Why is it not $\exists y [E(y,1)\rightarrow \forall xP(x,y,x)] $ ?

I though the $y$ only have the value of $1$, why using $\forall$??

$\endgroup$

3 Answers 3

3
$\begingroup$

In fact the answer given as correct is wrong: there are no quantifiers in the statement ‘if $y=1$, then $xy=x$’, so a correct transcription is $E(y,1)\to P(x,y,x)$.

Apparently you’re supposed to interpret ‘if $y=1$, then $xy=x$’ as having implicit quantifiers giving it the meaning

(1) there is an integer $y$ such that $y=1$ and $xy=x$ for all integers $x$

or

(2) if $y$ is an integer equal to $1$, then $xy=x$ for all integers $x$.

I would transcribe (1) as $\exists y\big(E(y,1)\land\forall x P(x,y,x)\big)$, and I consider this a perfectly reasonable answer to the question if it’s been made clear that you’re to supply implicit quantifiers in some reasonable way.

On the face of it (2) should be transcribed $E(y,1)\to\forall xP(x,y,x)$, but this leaves $y$ unquantified $-$ a free variable, if you’ve learned that term yet. Apparently you’re expected to quantify the expression completely, leaving no free variables and turning it into what is technically known as a sentence, so we need to decide whether $\forall y$ or $\exists y$ is more appropropriate. (2) is implicitly a statement about an arbitrary integer $y$: if that integer is $1$, then something happens. Thus, it really does say something about all integers $y$: for any integer $y$, either $y\ne 1$, or $xy=x$ for all integers $x$. This meaning is captured by the universal quantifier: $\forall y\big(E(y,1)\to\forall x P(x,y,x)\big)$.

It also says something about the integer $1$, namely, that $\forall x P(x,1,x)$. The statement $$\exists y\big(E(y,1)\to\forall x P(x,y,x)\big)$$ does not; it says only that there is some integer $y$ that either (a) is not equal to $1$, or (b) has the property that $xy=x$ for all integers $x$. This is a much weaker statement: it does not say that $1x=x$ for all integers $x$. Exhibiting the integer $0$ and pointing out that $\lnot E(0,1)$ is enough to show that $\exists y\big(E(y,1)\to\forall x P(x,y,x)\big)$ is a true statement, because $E(0,1)\to\forall x P(x,0,x)$ is a true statment, simply because $E(0,1)$ is false.

But the problem is very badly posed.

$\endgroup$
1
  • $\begingroup$ I think I was confused about the usage of for all and there exist. Now I know (after 3 hours). When using "There exist", All the cases in the statement can be truth. but using "For all", only the case(s) interest can be truth. Thanks Brain. ^^ $\endgroup$
    – Samuel
    Commented Oct 13, 2012 at 10:00
1
$\begingroup$

The original solution is correct for the reasons explained by Brian Scott. It is, as he says, the correct interpretation of his sentence 2. And sentence 2 is exactly like the original, except that sentence two makes the domain of integers explicit.

$\endgroup$
0
$\begingroup$

Statement $1$ sais "If $y=1$, then $xy=x$".

So, for every integer $y$ - if it is $1$ then $xy=x$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .