Why is this logical formula invalid for expressing the statement "anything bought it not human"?

Question

In my first order logic notes, one example that comes up is to express that "anything bought is not human". Is says this can be expressed as:

$\forall x(\exists y \, bought(y,x) \rightarrow \neg human(x))$

where $bought(y,x)$ means y bought x. It explicitly says not to express it as:

$\forall x \exists y( \, bought(y,x) \rightarrow \neg human(x))$

I read the second one as for all x, there exists y, such that if y bought x, then x is not human. I read the first one as for all x, if there exists a y such that y bought x, then x is not human.

So how does the second formula not express the fact that anything bought is not human? They both seem equivalent in this regard.

Answer 1

The formulas are not equivalent since for example the first one is unfortunately not true in the real world – there are human beings that were bought – but the second one is true in the real world. I have never bought a human being, so I am the witnessing $y$ for any $x$ – for every $x$, if I bought it, then it is not human. By the way this shows that even the stronger statement $∃y ∀x (bought(y, x) → ¬human(x))$ is true in the real world.

If you don't like the real world model, you can build a toy formal model. For example our model may consist of two entities $A$, $B$ such that $A$ is a human that bought itself and $B$ bought nothing. $A$ witnesses that the first statement is false. $B$ witnesses that the second statement is true.

You may also analyze the statements syntactically: the second formula is equivalent to $∀x∃y(¬bought(y, x) ∨ ¬human(x))$, which is equivalent to $∀x((∃y¬bought(y, x)) ∨ (∃y¬human(x)))$, which is equivalent to $∀x((∃y¬bought(y, x)) ∨ ¬human(x))$. On the other hand, the first formula is equivalent to $∀x(¬(∃y(bought(y, x))) ∨ ¬human(x))$. So the key difference is the contravariance of the first argument of the implication.

Answer 2

Since we don't care about the real world meaning, but the underpinning logic, we'll simply use the relation $B$ and univariate $C$ to shorten the math statements.

The first statement is of the form: $\forall x~((\exists y~B(x,y))\to\neg C(x))$.   Anything will not satisfy $C$ if it is $B$-related to something.  

The second statement is of the form: $\forall x~\exists y~(B(x,y))\to\neg C(x))$.   Everything has something such that, the former does not satisfy $C$ if it is $B$-related to the second.

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In the universe $\{a,b,c\}$ where $B(a,c), C(a), C(c)$ are the only facts for $B,C$, then $\forall x~\exists y~(B(x,y)\to\neg C(x))$ is true but $\forall x~((\exists y~B(x,y))\to\neg C(x))$ is false.

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Using equivalences:

$${\quad\forall x~((\exists y~B(x,y))\to C(x))\\ \equiv \forall x~((\neg \exists y~B(x,y))\vee C(x))\qquad\text{implication equivalence}\\\equiv \forall x~((\forall y~\neg B(x,y))\vee C(x))\qquad\text{quantifier duality}\\\equiv \forall x~\forall y~(\neg B(x,y)\vee C(x))\qquad~~~\text{quantifier distribution}\\\equiv \forall x~\forall y~(B(x,y)\to C(x))\qquad\quad\text{implication equivalence}}$$

Which is certainly not equivalent with $\forall x~\exists y~(B(x,y)\to C(x))$, the form of the second statement.