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I need to check whether the following second-order formula is a tautology according to the standard second-order (fully impredicative) semantics. I think the empty set should play some role here, but I don't see exactly how. Any hints? Thanks in advance.

$\forall R \neg \forall X \exists x \forall y (Rxy \leftrightarrow Xy)$

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    Your question is not clear. You have to specify the arity of the relation-symbols somewhere... – 2017-01-28

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$\forall R\ \neg \forall X\ \exists x\ \forall y\ (R(x,y) \leftrightarrow X(y))$

Assuming that $R$ is a $2$-input predicate and $X$ is a $1$-input predicate, the given statement says that something holds for every such $R$. Well, what holds? Some $X$ exists such that... Specifically we can push the negations in, obtaining the equivalent:

$\forall R\ \exists X\ \forall x\ \exists y\ (\neg R(x,y) \leftrightarrow X(y))$

To help the intuition, let's restate $R$ as a 2d-array comprising rows, where $R(x,y)$ is the item in row $x$ and column $y$. Then "$\exists y\ (\neg R(x,y) \leftrightarrow X(y))$" compares a row of $R$ against $X$, so we can imagine $X$ to be a 1d-array arranged parallel to the rows of $R$. The statement then says:

Given any 2d-array $R$ of truth-values, there is a 1d-row $X$ of truth-values that, for every row in $R$, at some position gives the opposite truth-value.

Does that sound possible or impossible to you? It should also tell you how to prove or disprove it.

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    Helpful illustration thanks. However, according to the duality of the quantifiers the material equivalence should be in the scope of the negation operator, right? – 2017-01-28
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    @sequitur: Yes, but they are the same thing in this special case. A not equivalent to B is the same as the negation of A being equivalent to B (in classical logic). =) – 2017-01-28
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    OK, thanks a lot! – 2017-01-28