# translating to english the logic statements

im having a difficulty translating it to english. SO basically, i will type the corresponding statements in words. (Universal x ( x is not 0) -> existential y(xy = 1)

then we need to evaluate it as true or false.

heres my approach. i translated it to english as :

" For some x OR some y, xy = 1 is valid. " and my evaluation is "true" (T)

The reason why there is an or because i converted the "p->q" into simpler operations (negate p, or q). Then negate q means negating the universal x so i made it to some x.

I hope things get sorted out if Im wrong. Thanks stackexchange

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It seems to me a bit unclear what does "for some $x$ OR some $y$" mean? Can you have $xy$ without choosing an $x$, for example?

In this case, translating the implication to conjunction does not change the quantifiers: "$\forall$x.$\exists$y.$\neg[(x\neq0)\wedge(xy\neq1)]$". So now, using De Morgan: $\forall$x.$\exists$y.$[(x=0)\vee(xy=1)]$.

The law of translation is $(p \implies q) \equiv \neg(p \wedge\neg q)$.

However, why not try translating it directly? I think you will find it much less confusing.

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for all x, if x is not 0, then there exists y such that xy=1

for all x, either x is 0, or there exists y s.t. xy=1.

if I get the scope of universal quantifier right, and it is the case that I'm right this statement is true in all models of division rings, Fields (such as fiels of Reals R).

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Fully English translation (rather than just a literal symbol-by-symbol translation) proceeds as follows:

For all x(x is not 0) there exists y (xy = 1)

For all x besides 0, there exists a y such that xy = 1

For all numbers besides 0, a number exists that will multiply with the first number to produce 1.

All numbers besides 0 have a multiplicative inverse.

Or, if you like:

If x is not 0, x has a multiplicative inverse.

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