# Given that Γ ∪ {¬P} is unsatisfiable, show that Γ ⊨ Q → P?

I honestly have no clue as to where to start with this question. My TA told us to think of it parts:

``````1) Consider what having a set unified with 'not P' means.
2) What this union being unsatisfiable would mean.
3) What this would mean for the relationship of 'if Q then P' with Gamma.
``````

We barely touched this module yet so I don't know what to start off with. Any help would be greatly appreciated!

The fact that Γ ∪ {¬P} is unsatisfiable, implies that Γ ⊨ P.

This is proved using the definition of the relation of logical consequence ( ) :

every truth assignement (or model) that satisfy the formulae in Γ will also satisfy P.

To say that Γ ∪ {¬P} is unsatisfiable means that there is no truth assignment that satisfy it; thus, every truth assignment that satisfy Γ will not satisfy ¬P, i.e. will necessarily satisfy P.

And this in turn means : Γ ⊨ P.

Now we apply the tuth table for the conditional () :

Q → P is true either when Q is false or P is true.

Thus :

every truth assignment that satisfy P will also satisfy Q → P,

and in conclusion :

every truth assignment that satisfy Γ will also satisfy Q → P,

i.e. :

Γ ⊨ Q → P.

• So to show Γ ⊨ Q → P, provide a truth table for Q → P when P = false? – K.Wong Oct 17 '16 at 17:09
• But to show a set unsatisfiable, the consequent would have to be false wouldn't it? – K.Wong Oct 17 '16 at 21:01