# propositional logic entailment proof

I have a question, I am doing this exercise but I am a bit lost, it is the following:

Let $\Sigma = \{\phi_1,...,\phi_n\}$ and $\varphi$ a proposition. Show that $\Sigma \vDash\varphi \iff \left(\phi_1,...,\phi_n\right)\rightarrow\phi$ is a tautology.

I am trying to prove that by contradiction, first proving it from left to right, this is what I have:

Let suppose that $\Sigma \vDash \varphi \land \left(\phi_1,...,\phi_n\right)\rightarrow\phi$ is not a tautology, then we have:

$(\forall$ valuation $V, (V(\Sigma)=1 \land V(\varphi) = 1)) \land \exists V_0, (V_0(\phi)=0\land V_0(\Sigma) = 1)$

I am stuck on that part. I don't know if the problem statement is wrong and it is really $\left(\phi_1,...,\phi_n\right)\rightarrow\varphi$, because is that were the case and I were good in my reasoning, it would be easy to reach a contradiction. Could someone give me a hand?

• What kind of thing is $\Sigma$? What are the $\phi_i$s? Aug 21, 2015 at 2:25
• Yes, it does help to define the nature of the symbols. Aug 21, 2015 at 2:28
• MY BAD I have corrected it Aug 21, 2015 at 2:42
• You have to rewrite your condition as : $(∀V(V(Σ)=1 → V(\varphi)=1)) \quad ∧ \quad ∃V_0(V_0(Σ)=1∧V_0(\varphi)=0)$; in this way, the contradiction is clear, because the right conjunct is the negation of the left one : $\lnot (p \to q) \equiv (p \land \lnot q)$. Aug 23, 2015 at 16:39

Note: $$A\Leftrightarrow B \qquad\equiv\qquad (A\wedge B) \vee (\neg A\wedge\neg B)$$
$$A\nLeftrightarrow B \qquad\equiv\qquad (A\wedge \neg B) \vee (\neg A\wedge B)$$
What you are to prove is a tautology is: $$\Sigma \vDash\varphi \iff (\phi_1, \ldots,\phi_n)\to\varphi$$
This is not (necessarily): $\Sigma \vDash\varphi \wedge (\phi_1, \ldots,\phi_n)\to\varphi$