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I am new to propositional logic, and I am trying to figure out whether the below statement is correct:

[a |= c] or [ b |= c] → [a^b |= c]

First of all, it is obvious that

a^b ⊆ a and a^b ⊆ b.

Moreover, because either a |= c or b|= c are correct(or both), it seems that a^b must also entail c. In other words:

Ma ^ b ⊆ Ma . Also, when we say that a |= c,

we have M a ⊆ Mc.

But what if the intersection of a and b is empty? Does this argument still hold?

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  • $\begingroup$ " because either a |= c or b|= c are correct(or both), it seems that a^b must also entail c" Why does that seem to be the case? $\endgroup$ Commented Feb 15, 2021 at 14:54
  • $\begingroup$ Because M <sub>a ^ b</sub> ⊆ M <sub>a </sub> . Also, when we say that a |= c, we have M <sub>a </sub> ⊆ M <sub>c</sub> . $\endgroup$
    – Winston
    Commented Feb 15, 2021 at 15:03
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    $\begingroup$ Then "^" cannot possibly be intersection, and statements like "a^b$\subseteq$a" are meaningless. I think "^" is supposed to be "$\wedge$" ("and"). $\endgroup$ Commented Feb 15, 2021 at 15:17
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    $\begingroup$ You are mixing symbols for statements: $a \land b$ is the conjunction of statements $a$ and $b$ (reads: "$a$ and $b$") and symbols for sets: $M_{a \land b}$ is the set of models of statement $a \land b$ and in this case we have intersection and inclusion: $M_{a \land b} \subseteq M_a$. $\endgroup$ Commented Feb 15, 2021 at 15:43
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    $\begingroup$ The two are linked but different: $M_{a \land b}=M_a \cap M_b$ $\endgroup$ Commented Feb 15, 2021 at 15:46

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$a\land b\vDash c$ means that there are no interpretations(aka models) which value $a\land b$ as true but $c$ as false.

If there are no interpretations which value $a\land b$ as true, then the above holds. (See also vacuous truths)

The intersection of the models $\mathbf M_a$ and $\mathbf M_b$ being empty means there are no models which simultaneously value $a$ and $b$ as true.   Thus $\mathbf M_{a\land b}$ will be empty.   Therefore...

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