$\left(\forall x \cdot p(X) \Rightarrow q(X)\right) \wedge p(Y) \Rightarrow q(Y)$
At first glance this seems like a tautlogy and that's what my notes say. But an interpretation where $p$ is always true and $q$ is always false seems to be a counterexample.
Can someone confirm this or show me where I've gone wrong?
Thanks.
On
It is valid; after restoring the parentheses for better readibility, we have :
$(∀x[p(x) \to q(x)] ∧ p(y)) \to q(y)$.
Proof
1) $∀x[p(x) \to q(x)] ∧ p(y)$ --- assumed
2) $∀x[p(x) \to q(x)]$ --- from 1) by b$\land$-elimination
3) $p(y)$ --- from 1) by b$\land$-elimination
4) $p(y) \to q(y)$ --- from 2) by $\forall$-elimination
5) $q(y)$ --- from 3) and 4) by $\to$-elimination
6) $(∀x[p(x) \to q(x)] ∧ p(y)) \to q(y)$ --- from 1) and 5) by $\to$-introduction.
It is not a counterexample. In the case where $p$ is always true and $q$ is always false you have $$(\forall X ,p(X) \Rightarrow q(X)) \qquad \mbox{ false}$$ $$(\forall X ,p(X) \Rightarrow q(X)) \wedge p(Y)\qquad \mbox{ false}$$ $$(\forall X ,p(X) \Rightarrow q(X)) \wedge p(Y) \Rightarrow q(Y)\qquad \mbox{ true}$$ since $\mbox{ false} \Rightarrow \mbox{ true}$ is true.