Why is this counter-example valid?

I don't understand why the counter example of the following argument is valid:

$\forall x\exists y(Ax\iff By)$

$\exists xBx \land \exists x\sim Bx$

$\forall x(Ax \to \sim Cx)$

Therefore: $\sim \forall x Cx$

Let there be two objects in the universe, 0, and 1.

A: $\{ \}$

B: $\{0\}$

C: $\{0,1\}$

The first premise is true because everything is such that something is such that 'A' is true of the first if and only if 'B' is true of the second. In fact, 'A' is true of nothing at all. And no matter what there is, there is something that 'B' is not true of, namely 1. So there is always something that makes the biconditional true. The second premise is clearly true since 'B' is true of something, namely 0, and 'B' is also false of something, namely 1. The third premise is true because 'A' is true of nothing, so that every instance is a conditional with a false antecedent. The conclusion is false because 'C' is indeed true of everything.

What I don't understand is how premise 1 is true. Left to right of the IFF is true because A is true of nothing: vacuous truth. But right to left doesn't look true to me: $\forall x\exists y(By\to Ax)$ Clearly there is something B is true of (i.e. 0), so the antecedent is true. But the consequent is false - so how is this IFF true?

Also, the explanation mentioned that 'no matter what there is, there is something that 'B' is not true of, namely 1'; I don't get the relevance of this either. The concerned proposition is about something which is true of B, not something which is not true of B. Also $\exists x By$ and $\exists x \sim By$ can both be true!

Thank you so much!

• You know that in a given finite universe, the quantifiers can be converted into $\land$ and $\lor$ expressions, and everything is just a propositional logic expression? – DanielV Jun 12 '16 at 12:13
• Yes but I am not sure how that fits in here...? – Daniel Mak Jun 12 '16 at 14:39

$P\Leftrightarrow Q$ if $P$ and $Q$ are both false.

$\forall x \sim \!\!Ax$. So to make the "iff" in the first premiss true, it is enough to find $y$ where $By$ is false. And $y=1$ will serve, because $y\notin B$.

To prove $\exists y\;By\rightarrow Ax$, it is enough to find a $y$ where either $By$ is false or $Ax$ is true. It doesn't matter if there is also a different $y$ where $By$ is true.

• You beat me to it. – Maik Pickl Jun 12 '16 at 11:47
• I don't think this is quite you were talking about; but the way you put it reminds me of contraposition. Since $\forall x\exists y(By \to Ax) \iff \forall x\exists y(\sim Ax \to \sim By)$, so when By is false and Ax is true, that would prove the conditional, which I suppose is kinda the same as you mean...? – Daniel Mak Jun 12 '16 at 14:38
• Indeed I wasn't thinking of what you say, but what you say is true. $By\rightarrow Ax$, $\sim Ax\rightarrow \sim By$ and $Ax \vee \sim By$ are equivalent. – Rosie F Jun 12 '16 at 14:54

I will assume that $Ax$ means $x \in A$ just to make it easier to interpret. Similarly for $B$ and $C$. Now what you want to prove is: For all $x$ there is a $y$ such that $x \in A$ iff. $y \in B$.

First case: Let $x=0$ and $y=1$ then $0 \in A$ is false and $1\in B$ is false and therefore $0\in A \Leftrightarrow 1 \in B$ is true.

Second case: Let $x=1$ and $y=1$ then $1 \in A$ is false and $1\in B$ is false and therefore $1\in A \Leftrightarrow 1 \in B$ is true.

Overall $\forall x \exists y Ax\Leftrightarrow By$ is true.