I've got a homework question. I am confused with the two variables $x$ and $y$.

In the domain of integers, consider the following predicates: Let $N(x)$ be the statement "$x \ne 0$." Let $P(x, y)$ be the statement "$xy = 1$."

Translate the following statement into the symbols of predicate logic:

For all integers $x$, there is some integer $y$ such that if $x \ne 0$, then $xy = 1$.

This is what I have, not sure if it's right:

$(\forall x)(\exists y)(N(x) \to P(x,y))$

  • $\begingroup$ Looks okay, except the brackets, different books have slightly different notation, modern books have $\forall x \exists y(N(x) \to P(x,y))$ old books have $(x)(\exists y)(N(x) \to P(x,y))$ $\endgroup$ – Willemien Sep 9 '13 at 8:33

Yes, that's perfectly correct, cheers.

Note that if you have the signature of unitary rings (i.e., you can use -- among others -- the ring multiplication, the ring zero $0$ and the unity $1$ in formulae) in predicate logic, then you can also simply write it as:

$$(\forall x)(\exists y)(x \ne 0 \to xy = 1)$$

which is more readable than using predicate letters. (But these predicates may be dictated to you by the question, in which case you're best off using them :).)


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