Is this a valid interpretation that shows this sentence, (∀x)(Mx ⊃ Kx) ⊃ (∃x)(Mx & Kx), is not quantificationally true?

Question

Here's the interpretation I thought up.

UD: Set of all positive integers

Mx: x is less than 1

Kx: x is greater than 0 (edited)

So the antecedent is vacuously true, because Mx is always false and Kx always true for any x, so (∀x)(Mx ⊃ Kx) is always true. But since Mx is false, the consequent, (∃x)(Mx & Kx), cannot be true. So the whole statement is false, and the sentence is not quantificationally true.

Answer 1

Here, you have that both the quantified antecedent (call it $p$) is false, and the quantified consequent (call it $q$) is false. So the entire statement can easily be represented by the implication: $p\to q$.

Now what do you know about any statement of the form $p\to q$, when $p$, and $q,$ are false?

Put another way, an implication is False ONLY when the antecedent is true, and the consequent is true.

So, $F\to T, T\to T, F\to F$ are all true.

Just as a reminder of the definition of material implication see the truth-table below:

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Now, given your subsequent edit, When $M(x)$ is false, but $K(x)$ is true,you are correct that you have an antecedent that is clearly is true, while the consequent is false, hence the entire statement is false!

Answer 2

Algebraically:

$$\begin{align} \lnot\bigg((\forall x ~ Mx \Rightarrow Kx) \Rightarrow (\exists x ~ Mx \land Kx)\bigg) &= (\forall x ~ Mx \Rightarrow Kx) \land \lnot (\exists x ~ Mx \land Kx) \\ &= (\forall x ~ \lnot Mx \lor Kx) \land (\forall x ~ \lnot Mx \lor \lnot Kx) \\ &= \forall x ~ \bigg((\lnot Mx \lor Kx) \land (\lnot Mx \lor \lnot Kx)\bigg) \\ &= \forall x ~ \bigg(\lnot Mx \land (Kx \lor \lnot Kx)\bigg) \\ &= \forall x ~ \lnot Mx \end{align} $$

So the statement is false in binary logic exactly when $M$ is always false; $K$ is irrelevant.