Proof Rules in Set Theory

Question

This is probably a very dumb question, but I can't see where am I going wrong.

So Let F be formula, where $F \equiv xy = 5.$

Prove that $ \forall x F \vDash y = 5 $. Using the proof Rules, If $ S ∪ {[F]x→t}\vDash G $ Then $ S ∪ {∀x F} \vDash G $, where t is a term

$ [F]x→1 \equiv 1 * y = 5$, * is multiplication.

Assuming we are in a field, it is easy to show $ S ∪ {[F]x→1} \vDash y = 5 $. , where S is a premise set that contains the field axioms

Then $ S ∪ {∀x F} \vDash G $. But it is easy to see that if x = 2, then y is 5/2. What did I do wrong. I know it is probably very dumb, but I cant see it.

Answer 1

Ignoring various notational trouble, you have correctly proved that if you assume the field axioms plus $\forall x(xy=5)$ then you can conclude $y=5$.

You have also almost correctly concluded from the same assumptions that $y=5/2$ (the reason that this is only almost correct is that you need $2\ne 0$ in order to be able to divide by $2$, and your assumptions do not guarantee that).

So the assumption $\forall x(xy=5)$ leads to the two apparently conflicting results $y=5$ and $y=5/2$. But so what? A priori there are at least two ways this can make sense:

• Either $\forall x(xy=5)$ simply cannot be true in the first place. In that case it should not surprise you that you can prove nonsense from it.
• Or your conclusions are not actually nonsense. It could be that you find yourself in a field where $5=5/2$, and then there's no contradiction after all.

The second of these is what's actually happening. In $\mathbb F_5$ or another field of characteristic $5$ we do indeed have $5=5/2$ because both sides are actually zero! And $\forall x(xy=5)$ is true exactly if we're in a field of characteristic $5$ and $y$ happens to be $0$.