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Consider the following:

If a |= c or b |= c, then a ∨ b |= c. Prove whether this statement is true or false.

My gut instinct is to compare truth tables, but I don't think a truth table is possible with double turnstiles. Can someone offer me a better way to tackle this problem. I'm more interested in advice and hints.

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The question deals with i.e. with Logical consequence:

A formula A is a logical consequence within some formal system of a set of statements S if and only if there is no model in which all members of S are true and A is false.

We can apply the definition above: by assumption, we have that every truth valuation that satisfy a will also satisfy c and every valuation that satisfy b will also satisfy c.

But a valuation that satisfy a ∨ b must satisfy either a or b.

  • Using what you said in the second paragraph. Since a ∨ b is true if either a or b is true or both. Then would this prove that a ∨ b entails c? – Amous Oct 18 '17 at 14:40
  • @Amous - Correct. – Mauro ALLEGRANZA Oct 18 '17 at 14:47
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This is false in general, as in the following example.

Let a = c = some false statement, and let b be some true statement. Then:

  • (a |= c) is true, therefore ( a|=c or b|=c ) is also true.
  • b is true, therefore "a or b" is true.
  • c is false.
  • By the previous two points, ((a or b) |= c ) is false.

You can get the same result by analysing the formula "(( a -> c ) v ( b -> c )) -> ( a v b -> c )".

But if you assume that both a|=c AND b|=c hold, then you can conclude that (a or b)|=c also holds.

  • Did you look at Mauro's answer? I think you're misreading the symbols. – virmaior Oct 24 '17 at 2:04

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