semantically equivalent

Question

question about: $((\forall x A)\lor B)$ is semantically equivalent to $(\forall x(A\lor B))$ with the condition that $x$ is not free in $B$.

I have thought about structure,

$U = \{ k , h \}$ ... where $k$ and $h$ are names of humans

$A:= \{ x$ eats sandwiches $\}$ .. $A =\{ (h,{\rm TRUE}) , (k,{\rm FALSE}) \}$

$B:= \{x$ eats healthy$\}$.. $B= \{ (k,{\rm TRUE}) , (h,{\rm FALSE}) \}$

now we have two options to consider (I think):

1-$( (\forall x~A) \lor (\forall x~B) )$ is semantically equivalent to $( \forall x~(A \lor B) )$

2-$( (\forall x~A) \lor (\exists x~B) )$ is semantically equivalent to $(\forall x~(A \lor B) )$

(I think $(\exists x~B)$ is $x$ is not free in $B$.) (I think $(\forall x~B)$ is $x$ is not free in $B$. )

but if we consider rule number 1,

1-$( (\forall x~A) \lor (\forall x~B) )$ is semantically equivalent to $(\forall x~(A\lor B) )$

then we find the following:

• $\forall x~A$ is false

• $\forall x~B$ is false

• $\forall x~(A \lor B)$ is true ( (forAll x A) or (forAll x B) ) is false so they are not equivalent $( (\forall x~A) \lor (\forall x~B) )$ is not equivalent to $\forall x (A \lor B)$

so what is wrong in what I'm doing?

Answer 1

"$x$ is not free in $B$", means that the variable $x$ does not occur as a free variable in the predicate $B$.

Your example predicate $\{x$ eats healthy$\}$ contains $x$ as a free variable.   Hence it does not fit the require criteria for "$x$ is not free in $B$".

Try $A:\{x$ eats sandwiches$\}$ and $B:\{$sandwiches are tasty$\}$.

Answer 2

In general, $(∀x A)∨(∀x B)$ is not equivalent to $∀x(A∨B)$.

We have $(∀x A)∨(∀x B) \vDash ∀x(A∨B)$, but note vice versa.

To show it, consider a box full of balls: some Black and some White.

It is true that: "all balls in the box are (either Black or White)", but it is not true that "either (all balls in the box are Black) or (all balls in the box are White)".

---

We have that $B$ is equivalent to $∀xB$, if $x$ is not free in $B$; thus, due to the above result, we have that: $(∀x A)∨ B \vDash ∀x(A∨B)$.

So, we have to prove that:

$∀x(A∨B) \vDash (∀x A)∨ B$, with $x$ not free in $B$.

A "semantical" argument by contradiction runs as follow: consider an interpretation where $∀x(A∨B)$ is true and $(∀x A)∨ B$ is false. This means that both $(∀x A)$ and $B$ are false.

For $(∀x A)$ to be false, we have an object $a$ in the domain of the interpretation such that $A$ does not hold for $a$. But this means that for $a$, also $A \lor B$ does not hold (because $B$ is false): contradiction !