Rules of Inference: Determine whether a given argument is valid.

Question

So I'm new to Discrete math and I'm having difficulty proving this proof: $$¬ (¬ p∧q)$$ $$p → r$$ $$¬ r$$ therefore, ¬ q So far I've only managed to get this far:

1. $¬ r$ - Premise
2. $p → r$ - Premise
3. $¬ p$ - Modus tollens (1)(2)
4. $¬ (¬ p∧q)$- Premise
5. $(p∧q)$- Double Negation (4)
6. $p$ - by Simplification (5)

At this point I have no idea what to do next. Any help will be appreciated.

Answer 1

$\def\fitch#1#2{~\begin{array}{|l}#1\\\hline#2\end{array}}$

No, you cannot apply double negation elimination to $\neg(\neg p\wedge q)$.   That is not a double negated term; it is the negation of a conjunction where one of the conjuncts is a negation.

Still, the modus tollens is correct (if that is an available rule). Now as you wish to prove $\neg q$, you should have first assumed $q$ and demonstrated that would produce a contradiction.

$$\fitch{ 1.\quad ¬ (¬ p∧q) \\ 2.\quad p → r \\3.\quad ¬ r }{\fitch{4.\quad q\qquad\quad\text{assumption}}{5.\quad\neg p\qquad~~\text{2, 3, modus tollens }\\\vdots\\?.\quad\bot}\\\Box.\quad\neg q\qquad\quad\text{4 - ?, negation introduction}}$$

You should be able to complete this.

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Alternatively, if deMorgan's Rules are accepted, it sets things up nicely.

$$\fitch{ 1.\quad ¬ (¬ p∧q) \\ 2.\quad p → r \\3.\quad ¬ r }{4.\quad\neg p\qquad\qquad~~~\text{2, 3, modus tollens}\\5.\quad \neg\neg p\vee \neg q\qquad\text{1, de Morgan's}\\\Box.\quad\neg q\qquad\qquad~~\text{4,5, some reason}}$$

(PS: Still not double negation)

Answer 2

You made a mistake going from 4 to 5.

If you apply DeMorgan's on 4' you get:

$5. \neg \neg p \lor \neg q$

And now you can do Double Negation to get:

$6. p \lor \neg q$

Finally, use Disjunctive Syllogism on $6$ and $3$ to get:

$7. \neg q$