SOLUTION: Can someone help me construct this truth table? P->(P ->(Q ^ P)). Now, does this mean that first, you work the Q^P? or do you work the p-> which is inside the ( ) first. Then d

Algebra ->  Proofs -> SOLUTION: Can someone help me construct this truth table? P->(P ->(Q ^ P)). Now, does this mean that first, you work the Q^P? or do you work the p-> which is inside the ( ) first. Then d      Log On


   

Question 147603: Can someone help me construct this truth table? P->(P ->(Q ^ P)). Now, does this mean that first, you work the Q^P? or do you work the p-> which is inside the ( ) first. Then do you take the result of whichever is used and connect it with p-> which is oustide the ( ). Do that make sense. I'm confused too.
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
Can someone help me construct this truth table? P->(P ->(Q ^ P)). Now, does this mean that first, you work the Q^P? or do you work the p-> which is inside the ( ) first. Then do you take the result of whichever is used and connect it with p-> which is oustide the ( ). Do that make sense. I'm confused too.

Start with this:

P|Q| P->(P ->(Q ^ P))
T|T|   
T|F|    
F|T|  
F|F| 

Now transfer TTFF under every P, and TFTF under every Q

P|Q| P->(P ->(Q ^ P))
T|T| T   T    T   T  
T|F| T   T    F   T   
F|T| F   F    T   F  
F|F| F   F    F   F

Now take care of the ^, because that's the innermost
operation.  If ^ is between two T's it's T, otherwise it's F
So we use this rule to fill in under the ^, but everytime
we do we scratch through the letters on each side of it,
like this:

P|Q| P->(P ->(Q ^ P))
T|T| T   T    T T T  
T|F| T   T    F F T   
F|T| F   F    T F F  
F|F| F   F    F F F

Now I'll just erase them, but you can just leave them
scratched:

P|Q| P->(P ->(Q ^ P))
T|T| T   T      T    
T|F| T   T      F     
F|T| F   F      F    
F|F| F   F      F

Now take care of the second ->, because that's now the innermost
operation.  If -> has a T on the left and an F on the right,
then it's F, otherwise it's T. So we use this rule to fill 
in under the ->, but as before, every time we do we scratch 
through the letters on each side of it, like this:  

P|Q| P->(P ->(Q ^ P))
T|T| T   T  T   T    
T|F| T   T  F   F     
F|T| F   F  T   F    
F|F| F   F  T   F

Again I'll erase the ones scratched, but you can just leave them
scratched:

P|Q| P->(P ->(Q ^ P))
T|T| T      T       
T|F| T      F        
F|T| F      T       
F|F| F      T   

Finally we take care of the first ->, because that's now 
the innermost operation.  We use the same rule again to fill 
in under the first ->,

P|Q| P->(P ->(Q ^ P))
T|T| T T    T       
T|F| T F    F        
F|T| F T    T       
F|F| F T    T   

And erasing those marked thru:

P|Q| P->(P ->(Q ^ P))
T|T|   T           
T|F|   F            
F|T|   T           
F|F|   T     

Edwin