Logical entailment in propositional logic.

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Suppose θ, ψ ⊨ ¬φ. Which of the following does NOT hold:

(A) θ,φ,ψ ⊨ ⊥

(B) ⊨ ψ → (θ → ¬φ)

(C) θ,φ ⊨ ¬ψ

(D) θ ⊨ φ → ¬ψ

(E) φ ⊨ ¬θ ∧ ¬ψ

Having difficulty answering this questions,

What I understand of the questions are,

  1. Suppose (θ is True, ψ is True) Conclude That (φ is not True)
  2. Which of the following does NOT hold: there are 4 in the following functions that are Correct (A..E) and we need to pick which one is not true as per condition 1.
  3. However I seems to get more than one that is suppose to be a wrong answer and instead can only see one that is correct.

a. For example for A, ‘all 3 are true conclude that its false’ (ie none of them can be true at the same time). I would say in A we cant say or not hold, because ¬φ only happen if (θ& ψ) happen and they cant happen at the same time?

b. For B,. ψ → (θ → ¬φ) is always True, but by pitching if ψ True and (θ → ¬φ) is False and they are not True Hence not always True

c. And etc.

  1. I seems to be going it in the wrong direction am I? any hints or tips will be perfect
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Your assumption is telling you that $\neg\phi$ is a logical consequence of $\theta$ and $\psi$. This means that if you assume $\theta$ and $\psi$ to evaluate to $\top$ (true), then $\neg\phi$ is true as a consequence, i.e. $\phi$ is false. Formally, this condition says that each interpretation $I$ that is a model of both $\theta$ and $\psi$ is not a model of $\phi$.

Given this condition, you can now proceed by (dis)proving the statements. Ideally, you would start by considering an interpretation $I$ that is a model for the formulas that are on the l.h.s. of the entailment symbol $\models$, to (dis)prove that $I$ is a model of the formulas on the r.h.s. In point (B), you should (dis)prove that your formula is a tautology, i.e. each interpretation is a model of it.

As an example, let's try to solve (C).

Assume that $I$ is a model for $\theta$ and $\phi$. Using an equivalent version of our condition, i.e. $\phi \models \neg \theta \lor \neg \psi$, we can already see that each model of $\phi$ makes at least one between $\theta$ or $\psi$ false. Being $I$ also a model of $\theta$, it follows that $\psi$ must be false. In the end, $\theta, \phi \models \neg \psi$.

A good strategy involves thinking how to reformulate your condition and how models of formula interact in the different cases.