SOLUTION: Use an ordinary proof (not conditional or indirect proof): 1. K∨(S • N) 2. ∼(K •∼Q) 3. ∼(N •∼Q) / Q

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Question 1155133: Use an ordinary proof (not conditional or indirect proof):

1. K∨(S • N)
2. ∼(K •∼Q)
3. ∼(N •∼Q) / Q

Found 2 solutions by Edwin McCravy, jim_thompson5910:
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
 1. K∨(S • N) 
 2. ∼(K •∼Q) 
 3. ∼(N •∼Q) / Q

 4. ~K∨~~Q        2, DeMorgan's law
 5. ~K∨Q          4, Double negation
 6. ~N∨~~Q        3, DeMorgan's law
 7. ~N∨Q          6, Double negation
 8. Q∨~K          5, Commutation
 9. Q∨~N          7, Commutation
10. (Q∨~K)•(Q∨~N) 8,9, Conjunction
11. Q∨(~K•~N)     10, Distribution
12. Q∨~(K∨N)      11, DeMorgan's law
13. (K∨S)•(K∨N)   1, Distribution
14. (K∨N)•(K∨S)   13, Commutation
15. K∨N           14, Simplification
16. ~~(K∨N)       15, Double negation
17. ~(K∨N)∨Q      12, Commutation
18. Q             17,16, Disjunctive syllogism

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

I'm going to use an ampersand symbol in place of a dot.
NumberStatementLine(s) UsedReason
1K v (S & N)
2~(K & ~Q)
3~(N & ~Q)
ConclusionQ
4(K v S) & (K v N)1Distribution
5(K v N) & (K v S)4Commutation
6K v N5Simplification
7~K v ~~Q2De Morgan’s Law
8~K v Q7Double Negation
9~N v ~~Q3De Morgan’s Law
10~N v Q9Double Negation
11K -> Q8Material Implication
12N -> Q10Material Implication
13(K -> Q) & (N -> Q)11,12Conjunction
14Q v Q13,6Constructive Dilemma
15Q14Tautology
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