SOLUTION: (R->B)&(B->~C),D <-> (P&~K),R&C |- M I recently started taking an intensive Logic class and I am falling behind. Can you please help me solve this exercise so i can get an idea

Algebra ->  Proofs -> SOLUTION: (R->B)&(B->~C),D <-> (P&~K),R&C |- M I recently started taking an intensive Logic class and I am falling behind. Can you please help me solve this exercise so i can get an idea      Log On


   



Question 255427: (R->B)&(B->~C),D <-> (P&~K),R&C |- M
I recently started taking an intensive Logic class and I am falling behind. Can you please help me solve this exercise so i can get an idea how to do these type of exercises? thanks so much
JT

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
1.  (R -> B) & (B -> ~C)
2.  D <-> (P & ~K)
3.  R & C                 /:.  M
-----------------------------------
4.  (B -> ~C) & (R -> B)            1        Commutation
5.  R -> B                          1        Simplification
6.  B -> ~C                         4        Simplification
7.  C & R                           3        Commutation
8.  R                               3        Simplification
9.  C                               7        Simplification
10. B                               5,8      Modus Ponens
11. ~C                              6,10     Modus Ponens
12. C & ~C                          9,11     Conjunction
13. (C & ~C) v M                    12       Addition
14. F v M                           13       Contradiction
15. M                               14       See note below


Note: Recall that P v Q is true when either P or Q (or both) are true. If P is known to be false, then the truth value of P v Q is entirely dependent on Q. In other words, if P is known to be false, then P v Q is true if Q is true or P v Q is false if Q is false. If you aren't convinced of this observation, make a truth table.


Also, remember that P & Q is only true when both P and Q are true. Since P can only be true or false (but not both), then ~P will be the opposite making P & ~P automatically false for both truth values of P. Again, make a truth table if you aren't convinced.