Why $ M\models \forall x ( \alpha \to \beta)$ Is False?

Question

if M be a model and $\alpha$ and $ \beta$ be two formula the following is False:

$ M \models \forall x ( \alpha \to \beta)$ if and only if $ M \models \forall x \alpha$ has conclusion $ M \models \forall x \beta $

My question is why this is a false statement !?

Answer 1

It might be helpful to verbalise as follows:

Left hand side: for every $x\in M$, if $\alpha(x)$ is true then so is $\beta(x)$. On the right hand side: if ($\forall x \in M \alpha(x)$) then ($\forall x \in M \beta(x)$)

Then, it should be clear that we can make a following counterexample: $\alpha(0)$ is false, $\alpha(1)$ is true, $\beta(0)$ and $\beta(1)$ are both true. Then $\alpha(0)\rightarrow \beta(0)$ is false, so the left hand side is false, but the right hand side is true. Hope this is helpful!

Answer 2

Take the model $\mathfrak A\,$ whose domain is $\{a,b \}$, and let $\alpha:=F(x)$ and $\beta:=G(x)$, and finally let $F=\{a \}$ and $G=\{b \}$.

Recall that for finite domains $(\forall x)\varphi$ is equivalent to the conjunction $\varphi a_1 \land \varphi a_2 \,...\land \,\varphi a_n \; $ for an $n$ sized domain.

I'll show that $\mathfrak A\,$ is a countermodel to the following statement: $$(1)\;\;\;\;\; For\;all\;models\;M \;\;\;\;\; M\vDash \forall x(\alpha \to \beta) \;\;iff\;\; M\vDash \forall x(\alpha) \Rightarrow \;M\vDash \forall x(\alpha)$$ $(1)$ fails from right to left.

For suppose $\mathfrak A\vDash (Fa\land Fb)\Rightarrow \mathfrak A\vDash(Ga\land Gb)\, $, which is true vacuously, since the antecedent is false. That is, $b\notin F$, so $(Fa\land Fb)$ is false, which makes $\mathfrak A\vDash (Fa\land Fb)$ false.

But $\mathfrak A\vDash (Fa\to Ga)\land (Fa\to Ga)\,$ is false, since in the first conjunct the antecedent is true, but the consequent is false. That is, $a\in F$ but $a\notin G$, so the conditional $Fa\to Ga\,$ is false, which makes the conjunction $(Fa\to Ga)\land (Fa\to Ga)\,$ false.