Exactly.

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You don't need a weatherman to know which way the wind blows - Robert A. Zimmerman

In class, I think, we were told that once you prove it one way, it proves up the ladder backwards.

This thread is the reason why, when the test required me to "show my work" I drew a picture of me looking at my neighbors desk . . .

well now im going back in my hw looking at an section that had 4 problems along those lines and seeing if I can make sense of it all. I appreciate the help


well how about this one

[~p ^ (p v q)] -> q show it is a tautology without truth tables

If that ~ wasn't there it would easily go from p ^ (p v q) to just ... p->q (but then it wouldnt be a tautology)

so would it go to just ~p->q and still not be a tautology

There are many different views. That was just mine. Do what you were told in class.

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You don't need a weatherman to know which way the wind blows - Robert A. Zimmerman

I'll have to double check. I only remember the prof doing just one side and making them equivalent and it can go backwards. I'll have to double check.

Yep so just show that q follows from the antecedent.

Start with

1) ~p ^ (p v q)
2) ~p (1, conjunct elimination)
3) p v q (1, conjunct elimination)
4) q (2, 3 disjunction elimination)

Edited for more appropriate notation

OR

4) ~p -> q (3, def. of conditional backwards)
5) q (2, 4, modus ponens)

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You don't need a weatherman to know which way the wind blows - Robert A. Zimmerman

4) ~p -> q (3, def. of conditional backwards)
5) q (2, 4, modus ponens)

now.. I'm not sure thats exactly a tautology..

I got stuck at the ~p->q and wasn't too sure (modus ponens doesnt come into effect until the next section so I probably cant use it


I saw that ~p ^ (p v q) might be able to be equiv to ~p through the absorbtion law(the ~) mucks it up for me.

but now having ~p->q can be changed to ( p v q) by backwards definition of conditional...

P V Q
t t
t f
f t
f f

still not a tautology

~p -> q
f t
f f
t t
t f

not a tautology

The following statement

[~p ^ (p v q)] -> q

is a tautology. All the values for each symbol are written underneath. So first column is the value for ~, second column is value for p. Third column is value for ^. So just like arithmetic you enter values for p and q first. Then enter the values for the negations of p or q if there are any. Then solve inside parentheses. In this case enter values for p v q, then the ^, then the ->. I'm using 1's and 0's. Sorry habit. This is basically the disjunctive elimination theorem.

__________________
You don't need a weatherman to know which way the wind blows - Robert A. Zimmerman

Teacher I had for discrete was an absolute fool. She had a Ph.D in computer science and is an absolute genius with A.I systems. She was pretty hot too for a woman in her 50s.

At least she knew she sucked at teaching and gave the entire class As. That was the only math class I got an A in too

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-Typos courtesy of my mobile phone.

lol i know its a tautology but i had to show equivalence

So you getting it?

__________________
You don't need a weatherman to know which way the wind blows - Robert A. Zimmerman

somewhat. I've got more of this crap to work on tonight and read more. on top of other homework.

My brain just bluescreened.

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1985 500SL Euro w/ AMG bits 130k
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2007 S600 Biturbo 149k Overheated Project, IT'S ALIVE!!!

well if your brain was a mac it would be quacking right now...

I've got a new one thats stumping me.

~p -> (q->r) is equivalent to q -> (p v r)

step 1) ~q v( p v r) definition of implication

step 2) (~q ^ p) v ( ~q ^ r) Distributive Law


then I'm stumped.