Prove that for every formula in the first-order logic language $l$ there is a quantifier free formula($B$) that have the following property.

Question

T = {$\forall x \forall y\ x = y$}.For every formula $A$ in the language I'm going to find a quantifier free formula $B$ such that :$T\vdash (A\rightarrow B) \land (B\rightarrow A)$. I think if we write the formula A in the form $Q1\ x1 ...Qn\ xn\ B$ ,$Q\in\{\forall , \exists\}$,then B that is quantifier free is the desired formula. if $m$ is a l-structural model such that $m \vDash T$ and $m\vDash A$ it is easy to prove that $m \vDash B$,note that $m\vDash T$ means that the world of the $m$ has at most one element.but the problem is when $m \vDash T$ nd $m \vDash B$ and the world of $m$ has no element and there is a existential quantifier in A,we cannot deduce that $m \vDash A$,where am I wrong? any help would be appreciated.

Answer 1

Since $T$ implies that there is exactly one object in the domain, you can take any formula $A$ and find formula $B$ by dropping all quantifiers, and by replacing all variables with a constant $a$.

For example, if

$$A = \forall x \exists y R(x,y)$$

then $$B = R(a,a)$$

This also works if the quantifiers are not all in front, e.g. if

$$A = \forall x (P(x) \rightarrow \exists y (Q(y) \land R(x,y)))$$

then

$$B = P(a) \rightarrow (Q(a) \land R(a,a))$$