Prove this statement is invalid by backwards and develop a model which it's false
$$[\forall b~[\alpha(b) \to\forall a~\beta(a,b)]] \to [\alpha(1) \to \forall a\forall b~ \beta(a,b)]$$
I am kinda stuck on this and I hope someone could help. I know that we have to flip the whole thing backward and prove the right side equation is false but without knowing a and b relationship, I got stuck
Let domain $D=\{x \in \mathbb R: x \leq 1\}, \alpha=(b \geq 1)$, $\beta(a, b)=(a \leq b)$, then clearly under this model the antecedent of the whole implication is true, but the conclusion if false since $\alpha(1)=T$ while $\forall a \forall b~ \beta(a,b)=F$.