(P->Q) v (R & S,), ~ (P -> Q): S?

Question

(Q & S) -> P, ~~ (Q & S) :P

P & Q : P v S

(~~P v S) -> R, -R, -S :P

(Q v P) & S, S -> (T v Q) -T : ~~Q

And can you provide the reason and proofs.

Answer 1

I use Kalish-Montague-Mar style notation because it's my favorite and I think the best and easiest to read. You can convert it to any style of proof or derivation you want, plus converting will help you learn.





1. Show P ............... Direct, 5

2. .. ~~ (Q ^ S) ........ P1

3. .. (Q ^ S) ............. 2, Double Negation

4. .. (Q ^ S) → P ..... P2

5. .. P .................... 3, 4, Modus Ponens



1. Show P v S ......... Direct, 4

2. .. P ^ Q ............... P1

3. .. P ..................... Simplification

4. .. P v S ............... Addition (+S)



This next one does not work. As the first couple were very simple, I assume you are supposed to just Modus Tollens and Disjunction. I think the first Premise was meant to be more like ~(P v S) → R or ~(~~P v S) → R.



1. Show P .................. Direct, 7

2. .. ~(~~P v S) → R ... P1

3. .. ~R ...................... P2

4. .. ~~P v S ............... 2, 3, Modus Tollens, Double Negation

5. .. ~S ...................... P3

6. .. ~~P .................... 4, 5, Disjunction

7. .. P ........................ 6, Double Negation



1. Show ~~Q ................ Direct, 8

2. .. (Q v P) ^ S ............. P1

3. .. S .......................... 2, Simplification

4. .. S → (T v Q) ........... P2

5. .. T v Q ..................... 3, 4, Modus Ponens

6. .. ~T ......................... P3

7. .. Q .......................... 5, 6, Disjunction

8. .. ~~Q ...................... 7, Double Negation

Answer 2

a million. P => Q V R, premise 2. ~P V (Q V R), by way of definition of conditional assertion 3. (~P V Q) V (~P V R), by way of distributive sources 4. (P => Q) V (P => R), by way of definition of conditional assertion

Answer 3

maths?