Question

I'm suppose to proof a problem with the fitch method, the problem is ¬(¬P ∨ Q) proofs P ∧ ¬Q I'm stuck and have no idea how to prove this plz help :)

if you don't know fitch method check this out http://en.wikipedia.org/wiki/Fitch-style_calculus

Answer 1

Disclaimer: Just a 2-cent worth of suggestion, the proof could be made shorter I guess.

1. ¬(¬P ∨ Q) (Premise)
2. ¬P (Assumption)
2.1. ¬P ∨ Q (Disjunction Introduction)
2.2. (¬P ∨ Q) ∧ (¬(¬P ∨ Q)) (Conjunction Intro, 1, 1.1)
2.3. ¬P → (¬P ∨ Q) ∧ (¬(¬P ∨ Q)) (....)
3. ¬¬P (....)
4. P (Negation Elimination)
5. Q (Assumption)
5.1. ¬P ∨ Q (Disjunction Introduction)
5.2. (¬P ∨ Q) ∧ (¬(¬P ∨ Q)) (Conjunction Intro, 1, 2.1)
5.3. Q → (¬P ∨ Q) ∧ (¬(¬P ∨ Q))
6. ¬Q
7. P ∧ ¬Q
Q.E.D.

Sorry I learned it so long ago I can't remember the names of some rules. You have to fill in some of the blanks yourself.

Answer 2

What I don't understand is what you did on step 3, how are you allowed to do a negation there?

Answer 3

because (¬P ∨ Q) ∧ (¬(¬P ∨ Q)) is a contradiction

Answer 4

so I'm suppose to show that ¬Cube(b) ∨ Cube(c)) ∧ ¬( ¬Cube(b) ∨ Cube(c)) leads to ⊥?

Answer 5

i think it's one of the rules of inference i.e. you don't need to show this, A and (not A) is always a contradiction