SOLUTION: hey can you help me with these proofs? 1. ~(J & K) prem 2. ~(L & M) prem 3. J v L / ~(K & M) prem/conc 4. 5. 6. 7. 8. 1.

Question 531393: hey can you help me with these proofs?
1. ~(J & K) prem
2. ~(L & M) prem
3. J v L / ~(K & M) prem/conc
4.
5.
6.
7.
8.

1.~ P v Q prem
2.~ R > ~Q prem
3.~(R & ~S) / P > S prem / conc
4.
5.
6.
7.
8.
Found 2 solutions by solver91311, jim_thompson5910:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

You'll have to come up with the reasons but it should go something like this:

~(J & K) -> ~J v ~K

~(L & M) -> ~L v ~M

Since J v L, ~(~J & ~L)

~(~J & ~L) -> ~K v ~M -> ~(K & M)

That last step, in English: Since J and L cannot both be false, one or the other or both of K and M must be false, so both K and M cannot be true.

John

My calculator said it, I believe it, that settles it

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

1. ~(J & K) prem 2. ~(L & M) prem 3. J v L / ~(K & M) prem/conc 4. ~J v ~K 1 DM 5. ~L v ~M 2 DM 6. J -> ~K 4 MI 7. L -> ~M 5 MI 8. (J -> ~K) & (L -> ~M) 6,7 Conj 9. ~K v ~M 8,3 CD 10. ~(K & M) 9 DM

Abbreviations/Acronyms Used:

CD = Constructive Dilemma
Conj = Conjunction
MI = Material Implication
DM = De Morgan's Law

==============================================================================================

1. ~P v Q prem 2. ~R > ~Q prem 3. ~(R & ~S) / P > S prem / conc 4. ~~Q > ~~R 2 Trans 5. Q > R 4 DN 6. P > Q 1 MI 7. P > R 6,5 HS 8. ~R v ~~S 3 DM 9. ~R v S 8 DN 10. ~~R > S 9 MI 11. R > S 10 DN 12. P > S 7,11 HS

Abbreviations/Acronyms Used:

Trans = Transposition
DN = Double Negation
MI = Material Implication
HS = Hypothetical Syllogism
DM = De Morgan's Law
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