# Question #0970c

Jun 7, 2017

See explanation...

#### Explanation:

From:

$\exists x \neg P \left(x\right)$

we have some ${x}_{1}$ such that $\neg P \left({x}_{1}\right)$

Then:

$\forall x \left(P \left(x\right) \vee Q \left(x\right)\right) \vdash P \left({x}_{1}\right) \vee Q \left({x}_{1}\right)$

Then:

$\left(P \left({x}_{1}\right) \vee Q \left({x}_{1}\right)\right) \wedge \neg P \left({x}_{1}\right) \vdash Q \left({x}_{1}\right)$

We have

$\forall x \left(\neg Q \left(x\right) \vee S \left(x\right)\right) \vdash \neg Q \left({x}_{1}\right) \vee S \left({x}_{1}\right)$

Then:

$\left(\neg Q \left({x}_{1}\right) \vee S \left({x}_{1}\right)\right) \wedge Q \left({x}_{1}\right) \vdash S \left({x}_{1}\right)$

We have:

$\forall x \left(R \left(x\right) \to \neg S \left(x\right)\right) | - R \left({x}_{1}\right) \to \neg S \left({x}_{1}\right)$

Then:

$\left(R \left({x}_{1}\right) \to \neg S \left({x}_{1}\right)\right) \wedge S \left({x}_{1}\right) \to \neg R \left({x}_{1}\right)$

Finally:

$\neg R \left({x}_{1}\right) \vdash \exists x \neg R \left(x\right)$

$\textcolor{w h i t e}{}$
Rules used

The rules we have used are:

• $\exists x P \left(x\right) \vdash P \left({x}_{1}\right) \text{ }$ for some ${x}_{1}$

• $\forall x P \left(x\right) \vdash P \left({x}_{1}\right) \text{ }$ for any ${x}_{1}$

• $\left(P \vee Q\right) \wedge \neg P \vdash Q$

• $\left(\neg P \vee Q\right) \wedge P \vdash Q$

• $\left(P \to \neg Q\right) \wedge Q \vdash \neg P$