# Fitch-style natural deduction

How to prove the following questions?

(a) p from assumption ¬(p → q)

(b) ¬¬p → p from no assumptions.

Hint

For (a) you need (¬E) rule.

First, assume both p and ¬p and use the contradiction to get \$q\$.

Then, use (→I) to get p → q. Now you have a new contradiction.

For more details, you have to specify the set of rules you are allowed to use. See e.g. Natural Deduction.

• Yes, I was trying to use RAA to get not q, but then how to get p from not q? Any suggestion? – Kenny Jones Mar 16 '19 at 0:36
• I suggest not trying to do that. As you appear to have Reduction to Absurdity (RAA) available, then as Mauro suggests: Assume ¬p for an indirect proof of p. Inside this proof you derive the needed contradiction by assuming p for a conditional proof of p → q, which contradicts the premise. For part b, do similarly : assume ¬¬p for a conditional proof of ¬¬p → p. Inside this proof, assume ¬p for indirect proof of p. – Graham Kemp Mar 19 '19 at 3:30

For the second one here is how you would use a proof checker to enter the problem and then attempt a solution: TFL (Truth Functional Logic) is checked. There are no assumptions so nothing is placed in the premises box. The conclusion is entered.

Then press "CREATE PROBLEM". The problem is set up for you.

I start the first line with an assumption "~~P". On the second line I use double negative elimination (DNE). Between lines 1 and 2 I have the desired conditional and write that on line 3 eliminating the assumption on line 1.

The rules you need to use may be different.