Proving logical consequence of a set

Question

Prove or disprove the following:

$(P \wedge Q), (\neg Q) \vDash (\neg P)$

I don't see how $\neg P$ could be a logical consequence of the set since it isn't similiar to any of the formulae within the set.

Answer 1

Let $\Gamma = \{P \land Q, \lnot B\}$

Then $\Gamma \vdash P$

And a valid demonstration can be

1. (Hipothesis) $P \land Q$
2. (Logic Axiom) $(P \land Q) \rightarrow P$
3. (Modus Ponens of 1 and 2) P

So, if $M$ is a model of $\Gamma$

$M \not\vDash \lnot P$

(Because $M \vDash P$ by soundness)

Answer 2

In order to show that :

$P∧Q,¬Q \vDash \lnot P$

using the definition of the relation of logical consequence : $\vDash$, we have to show that for every truth assignment $v$ :

if $v(P \land Q)=v(\lnot Q)=T$, then $v(\lnot P)=T$.

But there is no such $v$, becase in order to satisfy $v(P \land Q)=T$ we have that $v(P)=v(Q)=T$, and this is incompatible with $v(\lnot Q)=T$.

Thus, according to the fact that :

if no truth assignment satisfies every member of $\Gamma$, then for any formula $\alpha$, it is vacuously true that $\Gamma \vDash \alpha$

we can conclude with :

$P∧Q,¬Q \vDash \lnot P$.

Answer 3

Right to the point of your question: since you have $P \wedge Q$ in your set of assumptions, and we know that $P \wedge Q \vdash P$ (by $\wedge$E), it follows that

$P \wedge Q, \neg B \nvdash (\neg P)$

i.e. you are right, $\neg P$ not a logical consequence of your premises (note that $\neg B$ is actually irrelevant for our conclusion).