I want to find a model and a countermodel for the following formula: $¬∀a¬∃b((P(a)∧P(f(b)))→Q(f(f(b))))$
I tried:
- Model 1: $A = \{x, y\}, P^M = \{x,y\}, Q^M = \{x\}, f(b) = b$ which satisfies the formula.
- Model 2: $A = \{x, y\}, P^M = \{x,y\}, Q^M = \{\}, f(b) = b$ which does not satisfy the formula.
I am very unsure about the fact that the formula says $\lnot \forall a$. Is $\lnot \forall a$ false if $\forall a$ holds? Because this would be the case here, but I thought $\lnot \forall = \exists$ and so the formula could be rewritten to $∃a∃b((P(a)∧P(f(b)))→Q(f(f(b))))$.
Can I choose any function for $f$? I mean, can I also say $f(b) = \lnot b$?