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This is the first time I see this kind of question.

Ok, I have:

$\{\neg A \vee B, B \to C, A \vee C \} \models B \vee C$

I have to determine whether an inference exist or not.

How do I do so? please help.

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1  
Do you mean the double turnstyle, as written, meaning modl-theoretic argument, or single turnstyle, syntactic argument? –  André Nicolas Jul 18 '13 at 1:19
    
@AndréNicolas Wow .. I don't know half of the words you said .. Amm , I guess it's single turnstyle .. –  Billie Jul 18 '13 at 1:21
    
Did you mean to write the symbol $\models$ or the symbol $\vdash$? –  amWhy Jul 18 '13 at 1:28
    
Then I cannot do it, since the details depend on what axioms and rules of inference you are using for logic. And there are lots of (equivalent) variants. –  André Nicolas Jul 18 '13 at 1:28
    
@amWhy I meant |= –  Billie Jul 18 '13 at 1:30

2 Answers 2

up vote 4 down vote accepted

HINTs:

Note that on the left of the double turnstile $\models$, you have $\lnot A\lor B$, and you have $A\lor C$.

$$\lnot A \lor B \equiv A\rightarrow B\tag{1}$$

$$A \lor C \equiv \lnot A\rightarrow C\tag{2}$$

Now, we know that for any proposition, $A$, we have that $A \lor \lnot A$ is a tautology (always true).

So, given that we have $A\lor \lnot A$, $(1)$, and $(2)$, what can we say about $B\lor C$?

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I got it. thanks! –  Billie Jul 18 '13 at 2:08
    
You're welcome! –  amWhy Jul 18 '13 at 2:09
    
@amWhy: Best kind of feedback! +1 –  Amzoti Jul 18 '13 at 2:56

You used $\models$, which often, but not always, means semantic entailment.

If that is what is meant, then all we need to do is to show that if the sentences on the left are true (under some assignment of truth values to atomic sentences if you are doing propositional logic), then the sentence on the right is true under the same assignment of truth values to atomic sentences.

Suppose that $\lnot A \lor B$ is true If $\lnot A$ is true, then $A$ is false. But then since $A\lor C$ is true, we get that $C$ is true. It follows that the sentence $B\lor C$ on the right is true.

If $\lnot A$ is false, then since $\lnot A\lor B$ is true, it follows that $B$ is true. But then $B\lor C$ on the right is true.

Note that we did not need or us $B\rightarrow C$.

Remark: If the double turnstyle denotes derivability, then the above, in the absence of general theorems, is inadequate. Derivability is always with respect to a specific set of axioms and/or rules of inference. The axioms and rules of inference used vary quite widely. All the common ones one sees are equivalent. However, if one has to cite specific axioms and rules of inference, I cannot do it without knowing the fine details of the presentation of logic in your course.

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