SOLUTION: 1. ~P∨Q 2. Q→R ∴ P→R 1. R∨~Q 2. P→Q 3. ~R ∴ ~P 1. P 2. ~Q 3. ~R 4. (P&~Q)→(R∨S) ∴ S 1. (P&Q)&#8

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Question 935397: 1. ~P∨Q
2. Q→R ∴ P→R
1. R∨~Q
2. P→Q
3. ~R ∴ ~P
1. P
2. ~Q
3. ~R
4. (P&~Q)→(R∨S) ∴ S
1. (P&Q)→R
2. (P&R)→Q
3. P ∴ R↔Q
1. ~P&~Q
2. (P∨Q)∨(R∨S) ∴ (R∨S)
1. ~P ∴ ~(P&Q)
1. (P→Q)→S
2. S→~(T∨W)
3. (~T&~W)→R ∴ (P→Q)→R
1. (P&~Q)&(~R∨S)
2. ~((S&P)&~Q) ∴ ~R

1. P→Q
2. P→~(R∨Q) ∴ ~P

10. ∴ (P→~Q)↔(Q→~P)


Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
That's too many to post.  I'll just do 2:

1.	P
2.	~Q
3.	~R
4.	(P&~Q)→(R∨S) ∴ S 

5.      P&~Q         1,2,conjunction of premises
6.      R∨S         4,5,hypothetical syllogism
7.      S            6,3,disjunctive syllogism

--------------------

1.	(P→Q)→S
2.	S→~(T∨W)
3.	(~T&~W)→R ∴ (P→Q)→R

4.      (P→Q)→~(T∨W)    1,2,hypothetical syllogism
5.      (P→Q)→(~T&~W)     4,DeMorgan's theorem
6.      (P→Q)→R          5,3,hypothetical syllogism    

Edwin
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