# To prove (KB ⊭ S) and (KB ⊭ ¬S) is satisfiable

In question (e), I have to prove: A ≡ (KB ⊭ S) and (KB ⊭ ¬S) is satisfiable, where KB and S are propositional variables. I am not able to follow the solution given in the image above as to why it is replacing S with another propositional variable P. Even if we do take P, one of (KB ⊭ P) and (KB ⊭ ¬P) becomes true and other false making A as false. So anyway here's what I tried, I find some interpretation to make statement A false, if I cannot find such interpretation then A is a tautology. From what I understand, (KB ⊭ S) is true (i.e. KB does not entail S is true) when premise KB is true and the consequence S is false. So if I take KB ≡ TRUE and S ≡ TRUE making (KB ⊭ S) ≡ FALSE making A ≡ FALSE, hence A is not a tautology, whereas the answer given in above image is Yes.

• So I read the question wrong. It was to prove the statement in question was satisfiable by some interpretation and that's what the selected answer proves. Right? @MohammadRostami. So I am editing the question. Commented Aug 18, 2021 at 3:59
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– D.W.
Commented Aug 18, 2021 at 6:34

When you say $$\alpha\models KB$$, this means that $$models(\alpha)\subseteq models(KB)$$ that $$models$$ of $$\alpha$$ is used for an assignment that evaluates formula $$\phi$$ to true.
Under above interpretation, suppose $$KB\equiv True$$, and $$S$$ be a formula that isn't Tautology and just satisfiable (i.e. $$A\vee B$$), in propositional logic, hence we can conclude that
$$True\nvDash S\wedge True\nvDash \neg S.$$
Because $$models(KB)\nsubseteq models(S)\wedge models(KB)\nsubseteq models(\neg S).$$