# Homework Help: Tautology or contradiction

1. Aug 30, 2010

### joemama69

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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

2. Aug 30, 2010

### Staff: Mentor

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.

3. Aug 30, 2010

### 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.

4. Aug 30, 2010

### annoymage

how come F->F is false?

5. Aug 30, 2010

### annoymage

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

FvT is false?

6. Aug 30, 2010

### joemama69

not followings u

7. Aug 30, 2010

### annoymage

hmmm

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

fill in the blank ()

;P

8. Aug 30, 2010

### joemama69

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

9. Aug 30, 2010

### Staff: Mentor

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

10. Aug 30, 2010

### 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$$\wedge$$P, it is only true when it Q is false and P is true."

11. Aug 30, 2010

### 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

12. Aug 30, 2010

### annoymage

in second row is wrong

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

13. Aug 30, 2010

### 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

14. Aug 30, 2010

### 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,

15. Aug 30, 2010

### joemama69

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

16. Aug 30, 2010