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**emakarov** Sorry, this depends on the definition of a correct expression. I assume you are talking about syntactically well-formed expressions, regardless of whether they are true or false. This again depends on the definition of an expression, or a formula.

Usually, $\displaystyle x\in T\subset S$ is not considered a well-formed formula, though it makes sense to view it as an abbreviation for $\displaystyle x\in T\land T\subset S$. In fact, I don't think "$\displaystyle {}\subset S$" part is needed in the formula at all.

Concerning semantics, this formula is true according to the description of $\displaystyle R$, $\displaystyle T$ and $\displaystyle S$.

No, if you are talking about the negation of $\displaystyle \forall x\,(x\in T\Rightarrow\sim R(x))$, it is $\displaystyle \exists x\,(x\in T\land R(x))$.

The formula $\displaystyle \forall x\,(x\in W\Rightarrow R(x))$ does not follow from $\displaystyle \forall x\,(x\in T\Rightarrow\sim R(x))$, or its negation, alone. The latter formula says that elements of $\displaystyle T$ make $\displaystyle R$ false, but it does not say anything about elements outside $\displaystyle T$.

On the other hand, $\displaystyle \forall x\,(x\in T\Leftrightarrow\sim R(x))$ does imply $\displaystyle \forall x\,(x\in W\Rightarrow R(x))$ (in fact, the implication to the left is enough).