SOLUTION: please help me solve this problem: Construct a proof of the following theorem: [(P → Q)&(R → ¬Q)] → ¬(P&R)

Question 1191577: please help me solve this problem: Construct a proof of the following theorem: [(P → Q)&(R → ¬Q)] →
¬(P&R)
Answer by Edwin McCravy(20054) (Show Source): You can put this solution on YOUR website!


[(P → Q)&(R → ¬Q)] → ¬(P&R)

Put TTTTFFFF under each P.
Put TTFFTTFF under each Q.
Put TFTFTFTF under each R.

[(P → Q)&(R → ¬Q)] → ¬(P&R)
  T   T   T    T       T T      
  T   T   F    T       T F
  T   F   T    F       T T
  T   F   F    F       T F
  F   T   T    T       F T
  F   T   F    T       F F
  F   F   T    F       F T
  F   F   F    F       F F

Under the ¬ of ¬Q write the opposite of what's under
the Q.  Then erase what's under the Q

[(P → Q)&(R → ¬Q)] → ¬(P&R)
  T   T   T   F        T T      
  T   T   F   F        T F
  T   F   T   T        T T
  T   F   F   T        T F
  F   T   T   F        F T
  F   T   F   F        F F
  F   F   T   T        F T
  F   F   F   T        F F

Under the → of P → Q write F for T → F
and T for everything else. Then erase what's 
under P and Q:

[(P → Q)&(R → ¬Q)] → ¬(P&R)
    T     T   F        T T      
    T     F   F        T F
    F     T   T        T T
    F     F   T        T F
    T     T   F        F T
    T     F   F        F F
    T     T   T        F T
    T     F   T        F F

Under the → of R → ¬Q write F for T → F
and T for everything else. Then erase what's 
under R and ¬:

[(P → Q)&(R → ¬Q)] → ¬(P&R)
    T       F          T T      
    T       T          T F
    F       T          T T
    F       T          T F
    T       F          F T
    T       T          F F
    T       T          F T
    T       T          F F

Under the & of P&R write T for T&T
and F for everything else. Then erase what's 
under P and R:

[(P → Q)&(R → ¬Q)] → ¬(P&R)
    T       F           T      
    T       T           F
    F       T           F
    F       T           F
    T       F           F
    T       T           F
    T       T           F
    T       T           F

Under the & of (P → Q)&(R → ¬Q) write T for T&T
and F for everything else. Then erase what's 
under (P → Q) and (R → ¬Q):

[(P → Q)&(R → ¬Q)] → ¬(P&R)
        F               T      
        T               F
        F               F
        F               F
        F               F
        T               F
        T               F
        T               F

Under the ¬ of ¬(P&R) write the opposite of what's under
the &.  Then erase what's under the &



[(P → Q)&(R → ¬Q)] → ¬(P&R)
        F            F        
        T            T   
        F            T    
        F            T   
        F            T    
        T            T    
        T            T   
        T            T   


Under the → of [(P → Q)&(R → ¬Q)] → ¬(P&R) write 
F for T → F and T for everything else. Then erase what's 
under [(P → Q)&(R → ¬Q)] and ¬(P&R):

[(P → Q)&(R → ¬Q)] → ¬(P&R)
                   T          
                   T     
                   T      
                   T     
                   T      
                   T      
                   T     
                   T    

Since we end up with all T's, the theorem is proved 
true in all 8 possible cases.

Edwin

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