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        riddles >> medium >> Is p->q->q a tautology?
        
(Message started by: techno on Sep 26^th, 2012, 4:36pm)

Title: Is p->q->q a tautology?
Post by techno on Sep 26^th, 2012, 4:36pm

note that there are no parenthesis. Thanks for the help ;D

Title: Re: Is p->q->q a tautology?
Post by Noke Lieu on Sep 26^th, 2012, 7:07pm

Depends on what you mean...

Not in this case.

Title: Re: Is p->q->q a tautology?
Post by techno on Sep 26^th, 2012, 9:03pm

I think that's exactly what I meant. Thanks. But how would you solve it? What I did was this.

p->q->q

-p || (q->q)

-p || (-q||q)

-p || (T)

T.

Can you please explain me where I did something wrong? thanks for the help.

Or is it like this? I think this makes more sense but still.

p->q->q

(-p || q)->q

-(-p || q) || q

(p && q) || q

(p || q) && (q || q)

q && (p || q)

Thanks for replying so quickly.

Title: Re: Is p->q->q a tautology?
Post by techno on Sep 26^th, 2012, 9:06pm

Actually I was wrong on the last thing I did.

It's

(p && -q) || q

(q || p) && (-q || q)

( q || p ) && T

( q || p )

Title: Re: Is p->q->q a tautology?
Post by peoplepower on Sep 27^th, 2012, 5:18am

A self-implication is a tautology, so implying a self-implication is still a tautology.
I have personally seen implication as a right-associative operation exclusively, where we have in general
p->(q->r) <=> ~pV(~qVr) <=> ~(p&q)Vr <=> (p&q)->r.

Title: Re: Is p->q->q a tautology?
Post by techno on Sep 27^th, 2012, 8:56am

That was my main concern. So it is a right associative operation. Thanks for the help appreciate it :)