Is P -> ~(Q ^ ~P) a Tautology or Contradiction?

[joemama69]

Homework Statement

Wrie out the truth table for the statement form P -> ~(Q ^ ~P). Is it a tautology or a contradiction?

Homework Equations

The Attempt at a Solution

First off is it true to say that P -> ~(Q ^ ~P) and P -> (~Q v P) are equal.

P | Q | ~P | ~Q | P -> (~Q v P)
T T F F F
T F F T F
F T T F F
F F T T F

I believe it is a contradiction. Is this correct

 

[Mark44]

Technicality: ~(Q ^ ~P) is equivalent to (not equal to) (~Q v P), so P ==> ~(Q ^ ~P) and P ==> (~Q v P) are also equivalent.

If I were doing the truth table I would have columns for P and Q (which you do), plus one column for ~Q v P.

 

[joemama69]

P | Q | (~Q v P) | P -> (~Q v P)
T__T_____F_________F
T__F_____T_________T
F__T_____F__________F
F__F_____F__________F

is the second row correct. If P, then not Q or P. It sounds like a contradiction but P is true ~QvP.

 

[annoymage]

how come F->F is false?

 

[annoymage]

and there's something wrong in the (~Q v P) column, in the first row.

FvT is false?

 

[joemama69]

not followings u

 

[annoymage]

hmmm

P--->(~QvP)
T_()___T
T_()___T
F_()___F
F_()___F

fill in the blank ()

;P

 

[joemama69]

for ~QvP, it is only true when it Q is false and P is true.

 

[Mark44]

for ~QvP, it is only true when it Q is false and P is true.

No, that's not right. ~QvP is true in all cases other than when P is false and Q is true.

 

[annoymage]

"v" is or. it's only true when either ~Q or P is true

your statement here "for ~QvP, it is only true when it Q is false and P is true."

suppose to be "for ~Q\wedgeP, it is only true when it Q is false and P is true."

 

[joemama69]

P | Q | (~Q v P)
T__T___T
T__F___F
F__T___T
F__F___T

ok i got this part. but when u pu P->(~Q v P) you get True only when (~Q v P) is True and when P is True.

P | Q | (~Q v P) | P->(~Q v P)
T__T___T___________T
T__F___F___________F
F__T___T___________F
F__F___T___________F

 

[annoymage]

P | Q | (~Q v P)
T__T___T
T__F___F
F__T___T
F__F___T

in second row is wrong

ok i got this part. but when u pu P->(~Q v P) you get True only when (~Q v P) is True and when P is True.

and also whenever P is false, it must be true too

because if the premises is already false, either the consequence is false or true, it doesn't matter. The statement must be true

 

[joemama69]

P | Q | (~Q v P) | P->(~Q v P)
T__T___T___________T
T__F___T___________T
F__T___F___________T
F__F___T___________T

 

[annoymage]

yea, that's correct, i hope you understand each of them. and btw, you can also proof it algebraically. using all those law, assiosiative, identity, commut, distributive and so on,

 

[joemama69]

im having a hard time grasping the abstractness of this. do you know of any good sites i can use for added material.

 

[annoymage]

hmm opencourseware MIT, there's a topic on logic, or there's many free books in http://www.e-booksdirectory.com/listing.php?category=3