Differences between free and bounded variables in a statement

Question

If given $y=5x$ where $x,y$ are real, we have an open statement, but if I write $\forall x \exists y: y = 5x$ we have a closed statement that is true, and we can even write $\forall x \exists ! y: y = 5x$ as the mapping $f: x \mapsto 5x$ is one-to-one. Now how do I make sense of the following: $\forall x: y = 5x$ (here $y$ is free) and $y=5x$ (here both $x,y$ are free). Thanks! Also, what is the difference between $\forall x \exists ! y: y = 5x$ and $\exists ! y: y = 5x$

Answer 1

By convention, in mathematical logic “$M$ satisfies $\varphi$” means that $M$ satisfies its universal closure: the sentence derived from $\varphi$ by universally quantifying it. So a model $M$ would satisfy $y=5x$ and $\forall x(y=5x)$ if only if it satisfied $\forall y\forall x(y=5x)$. Likewise, $M$ would satisfy $\forall x\exists!y(y=5x)$ if and only if it satisfied $\exists!y(y=5x)$.

The convention does not say that $y=5x$ and $\forall y\forall x(y=5x)$ mean the same thing, just that we should regard "$M$ satisfies $y=5x$" as meaning the same as "$M$ satisfies $\forall y\forall x(y=5x)$.

As I said, this is a convention. To illustrate further, let's say $M$ is $(\mathbb{R},+,\cdot)$. The standard definition of satisfaction (due to Tarski, and denoted $\models$) in logic says that $$M\not\models\forall y\forall x(y=5x)$$ but $$M\models\forall x\exists!y(y=5x)$$ The Tarski definition does not assign a truth-value to "$M\models y=5x$"; or rather, it assigns one only if you pick values for $x$ and $y$.

But now the convention kicks in. According to it, "$M\models y=5x$" means that $y=5x$ holds in $M$ regardless of what values you pick for $x$ and $y$. So $M\not\models y=5x$. On the other hand, $M\models\exists!y(y=5x)$.

To elaborate: $M\models y=5x$ iff $M\models\forall x\forall y(y=5x)$ because in fact, $M\not\models y=5x$ and $M\not\models\forall x\forall y(y=5x)$. What if we replace $y=5x$ with its negation? Then the convention says that $M\models y\neq 5x$ iff $M\models\forall x\forall y(y\neq 5x)$, and again $M$ satisfies neither. On the other hand, the negation of $M\models\forall x\forall y(y=5x)$ is $M\models\exists x\exists y(y\neq 5x)$, which is of course true.

What's the point of this convention? For one thing, it is useful for writing down some axioms with a minimum of clutter. We can write "$xy=yx$" for the commutative law of multiplication instead of "$\forall x\forall y(xy=yx)$".

It's also useful in setting up inference systems. The rule of universal generalization says that from $\Gamma\vdash\varphi(x)$, you can infer $\Gamma\vdash\forall x\varphi(x)$. (Here $\Gamma$ is a bunch of assumptions.) This enables you to convert to formal logic a common mode of mathematical reasoning. Namely, where you derive a result about (say) $x$ and $y$, and then conclude that since $x$ and $y$ were arbitrary, the result holds for all $x$ and $y$.

Of course, the formulas $y=5x$ and $\forall x\forall y(y=5x)$ are not equivalent for all purposes or in all senses. The first holds for some pairs $(x,y)$, the latter is just false. Negating the first gives us $\neg y=5x$, same at the inequality $y\neq 5x$, while negating $\forall x\forall y(y=5x)$ gives us a true statement equivalent to $\exists x\exists y(y\neq 5x)$.

In general, $M\not\models\varphi$ implies $M\models\neg\varphi$ only when $\varphi$ is a closed formula. So the convention must be handled with care.