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# Using logical equivalences, show that (¬Q n (P => Q)) => ¬P is a tautology watch

1. this is what i have so far:

(¬Q n (P => Q)) => ¬P

(¬Q n (P v ¬Q)) => ¬P (using implication law)

(¬Q n P) v (¬Q n ¬Q) => ¬P (using distributive law)

not sure where to go from here or if i've even done the question right. any ideas?

all help is much appreciated thanks in advance
2. (Original post by baws.)
this is what i have so far:

(¬Q n (P => Q)) => ¬P

(¬Q n (P v ¬Q)) => ¬P (using implication law)

(¬Q n P) v (¬Q n ¬Q) => ¬P (using distributive law)

not sure where to go from here or if i've even done the question right. any ideas?

all help is much appreciated thanks in advance
Your first step is incorrrect. Check the implication law.
3. There are two variables thus 4 possible valuations. Just check 'em all.
4. ghostwalker: i corrected it but i end up with (¬Q n ¬P) v (¬Q n Q) ?

mark85: the question tells me i have to use logical equivalences
5. (Original post by baws.)
ghostwalker: i corrected it but i end up with (¬Q n ¬P) v (¬Q n Q) ?
I presume you mean all that => ¬P

You can carry on from there.

What's (¬Q n Q)?
6. ah.. think i'm being stupid

(¬Q n Q) is false, so i'm left with ¬Q n ¬P => ¬P which is correct, since ¬P is true on both sides. right?

keep treating => as =
7. (Original post by baws.)
ah.. think i'm being stupid

(¬Q n Q) is false, so i'm left with ¬Q n ¬P => ¬P which is correct, since ¬P is true on both sides. right?

keep treating => as =
Brackets are important. I know what you mean, but as it's written (in bold) it's ambiguous.

Well, I'd go all the way and reduce it to "T" using logical equivalences.
I'd apply the implication law again, etc.

But if you've covered that particular formula already, it's probably fine to leave it as is.
8. thank you for all your help man
9. (Original post by baws.)
thank you for all your help man
np.

Corrected last post. Meant implication law, not distribution law.
10. (Original post by Mark85)
There are two variables thus 4 possible valuations. Just check 'em all.
Doing it that way implicitly assumes the completeness theorem for propositional logic, which is non-trivial.

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