SOLUTION: prove validity using inference rules and rules of replacement and say which rule was used at each step: 1) 1. {X+[Y+(~(A.F)>B)]} 2. (X>C) 3. (Y>C) 4. {~B.[Z.(~C+~C

Algebra ->  Proofs -> SOLUTION: prove validity using inference rules and rules of replacement and say which rule was used at each step: 1) 1. {X+[Y+(~(A.F)>B)]} 2. (X>C) 3. (Y>C) 4. {~B.[Z.(~C+~C      Log On


   

Question 1056542: prove validity using inference rules and rules of replacement and say which rule was used at each step:
1)
1. {X+[Y+(~(A.F)>B)]}
2. (X>C)
3. (Y>C)
4. {~B.[Z.(~C+~C)]}
5. /(F.Z)
2)
1. (A+C)+(A+B)
2. (A+(B+C))>D
3. /D
3)
1. (A>C)+(B>D)
2. /(A.B)>(C+D)
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Unless your teacher tells you otherwise, do not number
the conclusion [what follows the "/"] until the end.
It should be written out to the right of the last given 
premise.  At least, that's the way logic is normally 
taught.

1.	X+[Y+(~(A•F) ⊃B)]
2.	X⊃C
3.	Y⊃C
4.	~B•[Z•(~C+~C)]    /F•Z

6.      Z•(~C+~C)              4, simp.
7.      Z•(~C)                 6, taut.
8.      Z                      7, simp.
9.      ~C•Z                   7, comm.
10.     ~C                     9, simp.
11.     ~X                     2,10, MT  
12.     ~B                     4, simp.
13.     Y+(~(A•F)⊃B)           1,11, DS
14.     Y+{[~B⊃(~~(A•F)]}      13, trans.
15.     Y+[~B⊃(A•F)]           14, DN
16.     ~Y                     3,10, MT        
17.     ~B⊃(A•F)               15,16, DS
18.      A•F                    17,12, MP
19.      F•A                    18, comm.
20.      F                      19, simp.
21.      F•Z                    20,8, conj.

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

1. (A+C)+(A+B)
2. (A+(B+C))⊃D      /D 

3. A+[C+(A+B)]                  1, Assoc.
4. A+[(A+B)+C]                  3, Comm.  
5. [A+(A+B)]+C                  4, Assoc.
6. [(A+A)+B]+C                  5, Assoc.
7. (A+B)+C                      6, taut.
8. A+(B+C)                      7, Assoc.
9. D                            2,8, MP   

Two problems is the limit.

Edwin