Find a counter-model for a system for $\forall x (S(x) \lor T(x)) \vdash \forall x S(x) \lor \forall x T(x)$ or a proof of natural deduction

Question

I think it should be possible to prove, but since I can not for the life of me find a natural deduction proof that prove it, I have concluded that it should be possible to find a model that prove it is not valid.

The problem is that if I for example have a model $M$ where $A = \{1,2\}$, $S^M = \{1\}$ and $T^M = \{2\}$ so that $M \vDash \forall x (S(x) \lor T(x))$ is valid, then also $M \vDash \forall x S(x) \lor \forall x T(x)$ is valid. Can not get around it.

Am I correct that it is valid? If so how should I approach it in order to construct a proof in natural deduction?

Answer 1

For example, let $S(x)$ iff $x$ is sad. and $T(x)$ iff $x$ is tall.

And we know that

Everyone is sad or tall

$$∀x(S(x)∨T(x))$$

Does this implies

Everyone is sad or everyone is tall ? i.e.

$$∀xS(x)∨∀xT(x)$$

What about in a world with two people, one of them is tall but not sad, another is sad but not tall? In this case $∀x(S(x)∨T(x))$ hold, but $∀xS(x)∨∀xT(x)$ does not hold, this give a counter example.

This example is actually equivalent to your model, $$A=\{1,2\} , S^M=\{1\}, T^M=\{2\}$$ There are two people in world $A$, i.e. '$1$' is sad but not tall, '$2$' is tall but not sad. So it's not the case that Everyone is sad or everyone is tall, that $∀xS(x)∨∀xT(x)$ does not hold.

Answer 2

Here is an intuitive counterexample:

Domain: Natural numbers

$S(x)$ : $x$ is even

$T(x)$ : $x$ is odd