Translation of : The disjunction of two contingencies can be a tautology.

Question

The statement is: "The disjunction of two contingencies can be a tautology." The predicates are: $C(x)$: "$x$ is a contradiction." $T(x)$: "$x$ is a tautology."

The book says the answer is $$\exists{x}\exists{y}(\lnot T(x) \wedge \lnot C(x) \wedge \lnot T(y) \wedge \lnot C(y) \wedge T(x\lor y)) $$ However, I was thinking something more along the lines of $$\exists{x}\exists{y}(\lnot T(x) \wedge \lnot C(x) \wedge \lnot T(y) \wedge \lnot C(y) \rightarrow T(x\lor y)) $$

What's wrong with my line of thinking?

Answer 1

The book's answer says "There exist $x,y$ such that both are neither tautologies nor contradictions, and their disjunction is a tautology", which is exactly what the sentence says.

Your answer says (assuming you meant to put parentheses around the entire part before '$\to$') "There exist $x,y$ such that the fact that both are neither tautologies nor contradictions implies that their disjunction is a tautology".

This is problematic, since $F\to T$ and $F\to T$ both hold, if, for example, $x$ is a tautology, the part on the left of '$\to$' is false and thus $x,y$ satisfy the predicate and the entire predicate may be true even though $x$ is not a contingency, which means that, for example, in a model containing only one contingency and some tautologies, the truth value of the predicate would change.

Answer 2

Which book is in question??? On the face of it, the suggested answer is simply ill-formed. Propositional connectives like $\lor$ connect propositions not terms, so in standard first-order syntax $T(x \lor y)$ is not a well-formed formula.