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I did the truth table of the below logic:

((p ∨ q) → r) → ((p → r) ∨ (q → r))

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However I didn't quite understand what semantically entailed form the empty set of premises? What that mean exactly?

As far as i understand, Whatever P I pick, the conclusion should always be true. So in this case, is it semantically entailed form the empty set of premises?

I think it is not because

((p ∨ q) → r) <> p

in case p is T q is T and r is false

Mert Metin
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  • I don't understand why you think not. The formula you consider at the end doesn't look like the one that your question is about, I don't know what you mean by <>, and don't see how you are arriving from what's written to the conclusion "not semantically entailed". –  Oct 03 '16 at 05:33
  • <> means not equal. So is it semantically entailed from the empty set of premises? – Mert Metin Oct 03 '16 at 05:35
  • I still don't understand why you consider that inequation. –  Oct 03 '16 at 05:39

2 Answers2

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In this setting, semantic entailment $S \models Q$ simply means that if you write down the truth table and throw away the rows where any of the statements in S are false (i.e. you keep only the rows where every statement in $S$ is true), then $Q$ is identically true in the remaining rows.

Thus, your truth table does indeed prove

$$ \models ((p \vee q) \to r) \to ((p \to r) \vee (q \to r)) $$

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However I didn't quite understand what semantically entailed form the empty set of premises? What that mean exactly?

Is it entailed by only the rules of inference? That is: is it a tautology?

$$\models((p\lor q)\to r)\to((p\to r)\lor (q\to r))$$

Graham Kemp
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