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I wounder if I have done right. You can find the question at Question 1b

I know that $(1)$ $P_1∧(P_2∨P_3)$ ≈ $(P_1∧P_2)∨(P_1∧P_3)$ which means that these two formulas have the same truth value in every interpretation. Furthermore, I know that $$( P_1∧(P_2∨P_3) ⊧ (P_1∧P_2)∨(P_1∧P_3)$$ means that the formula on the right side of ⊧ is true in every interpretation in which the formula on the left side are true. Since these two formulas, on the right and left side, have the same value in every interpretation, the logical consequences are true, or?

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    $\begingroup$ If two formulas are equivalent, then one certainly entails the other. $\endgroup$ – Jared Sep 17 '14 at 20:25
  • $\begingroup$ @Jared Yes. But are my "argument" right? Have I understand the definition of $"⊧"$ right? The expression ask me to show that the formula on the right side of $"⊧"$ is right in every interpretation in which the formula on the right side are? And by $(1)$ we can see that it is true $\endgroup$ – anna Sep 17 '14 at 20:27
  • $\begingroup$ I generally find that entailment is very similar to implication. The difference is that implication ($\rightarrow$) deals with specific values while entailment does not. It seems to me that to prove entailment is to prove that the implication is a tautology: $x \models y\ \equiv\ \models (x \rightarrow y)$. Simply using Rule of Replacement, you can prove that $P_1 \wedge (P_2 \vee P_3) = (P_1 \wedge P_2) \vee (P_1 \wedge P_3)$. $\endgroup$ – Jared Sep 17 '14 at 20:39
  • $\begingroup$ When you have an entailment, you can suppose anything on the left hand side of the |= true. So, what happens if assume the left hand side of the |= meta-statement true? $\endgroup$ – Doug Spoonwood Sep 17 '14 at 22:02

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