# Tautology, truth table

• Feb 23rd 2009, 09:32 AM
premierplayer
Tautology, truth table
Alright so I have to make a truth table of the following to prove its a Tautology, meaning the whole value of the table is true. I have an issue of sometimes thinking things are correct when I am really messing up on the table somewhere. Let me show you what the problem is, and my solution. If someone could double check this for me that would be very awesome. Forgive me if my coding isnt right... please let me know.
-> = implies but I think u all know this lol.

(~q ^ (p -> q )) -> ~p

Code:

p  q  p->q  ~q      ~q ^ (p->q)  ~p

T  T    T      F          F        F            T
T  F    F      T          F        F            T
F  T    T      F          F        T            T
F  F    T      T          T        T            T

• Feb 23rd 2009, 10:22 AM
Plato
The table is correct. You may want to all a bit to the headings.
• Feb 23rd 2009, 10:28 AM
premierplayer
Quote:

Originally Posted by Plato
The table is correct. You may want to all a bit to the headings.

yeah I just got sick of typing out. I have it all written out but the table itself did look good to me. I just wanted to double check. Thanks!(Hi)
• Feb 23rd 2009, 05:37 PM
benes
Quote:

Originally Posted by premierplayer
Alright so I have to make a truth table of the following to prove its a Tautology, meaning the whole value of the table is true. I have an issue of sometimes thinking things are correct when I am really messing up on the table somewhere. Let me show you what the problem is, and my solution. If someone could double check this for me that would be very awesome. Forgive me if my coding isnt right... please let me know.
-> = implies but I think u all know this lol.

(~q ^ (p -> q )) -> ~p

Code:

p  q  p->q  ~q      ~q ^ (p->q)  ~p

T  T    T      F          F        F            T
T  F    F      T          F        F            T
F  T    T      F          F        T            T
F  F    T      T          T        T            T

Premierplayer.some times you do not have to write the whole truth table to show that the proposition is a tautology as the following proof of the above shows:

Assume that : (~q^(p---->q))---->~p is false,then according to the definition of the conditional statement,

(~q^(p--->q)) must be true and ~p false ,or p true.

But if (~q^(p---->q)) is true ,then according now to the definition of conjunctive statement , ~q must be true and also (p--->q) must be true

If however ~q is true ,then q is false.

So far we have proved p true and q false,thus P---->q ,false.A CONTRADICTION .Hence the above statement is a tautology
• Feb 23rd 2009, 07:12 PM
premierplayer
yeah I was asked to do a truth table for this one. I know that you can use proofs as well. Thank you for your help.