Study if the $\forall x \exists y R(x,y)\land \forall y \neg R(c,y)$ is logically valid, a contingency or a contradiction.

Question

Study if the $\forall x \exists y R(x,y)\land \forall y \neg R(c,y)$ is logically valid, a contingency or a contradiction.

Take $A = \{ a, c \}$. Since for $\forall x \exists y R(x,y)$ we can set our $y$ to be some constant $a$ and $x$ to be $c$, which leads us to $R(c,a)$. But since $\forall y \neg R(c,y)$, we can take our $y$ to be $a$ again, that is, we get $\neg R(c,a)$. So any model on $A$ ends up with $R(c,a) \land \neg R(c,a)$, and therefore, the formula is a contradiction.

Is the argument above true? Thank you in advance!

Answer 1

Take $x=c$. Then $$\exists yR(c, y)\wedge \forall y\neg R(c, y).$$ But $\forall y\neg R(c, y)$ equivalent to $\neg\exists y R(c, y)$ so the formula is $$\exists yR(c, y)\wedge \neg\exists y R(c, y),$$ a contradiction.

Answer 2

No, you cannot argue that the formula is a contradiction by considering only models with the particular universe $\{a,c\}$. For that you need to either argue that every model, no matter what the universe is, will fail to satisfy the sentence, or actually infer an explicit contradiction from the formula.

The latter seems to be easiest. You have $\forall x\exists y\,R(x,y)$ and $\forall y\,\neg R(c,y)$ separately. Instantiate $x$ to $c$ to get $\exists y\,R(c,y)$ -- but that is (equivalent to) the negation of $\forall y\,\neg R(c,y)$, so with a minimum of additional footwork (the details of which depend on your precise proof system) you have an explicit contradiction right there.