$[(\sim p \vee q) \wedge p ] \Rightarrow q $

What should be done next in order to apply direct proof to the example above?

The following process has been already done but seemingly it's incorrect:

1st. Apply distributive property

2nd. After applying complementation it results into $[ F \vee (q \wedge p) ]$

3rd. Applying distributive property a second time.

Eventualy the result is $q \wedge P$ , which cannot conclude anything... what shall I do?

  • 1
    $\begingroup$ Hint: $[\neg p \lor q ]\equiv [p\implies q]$ $\endgroup$ – Dan Christensen Aug 19 '14 at 15:38

Since you have the antecedent $[(\sim p \vee q) \wedge p ]$ reduced to $q \land p$ (which is correct), you simplify (or you might call it $\land$-elimination) to get $q$, as desired.

That is, $$q\land p$$

$$\therefore q$$

Hence, we can claim that $[(\sim p \vee q) \wedge p ] \Rightarrow q$.

(Therefore $p$ follows as well, but you are asked to show that the antecedent implies the consequent, so $p$ isn't relevant here.)


Use conjunction elimination, or take the long route.

$\left.\begin{align} (p\wedge q) \to q & & \text{premise} \\ (\neg p \vee \neg q) \vee q & &\text{conditional} \\ \neg p \vee (\neg q\vee q) & &\text{disjunctive associativity} \\ \neg p \vee \top & &\text{disjunctive negation} \\ \top & &\text{universal bound} \end{align}\right\}\quad(p\wedge q)\implies q \quad \text{conjunctive elimination}$


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