# Can you help me get started on this proof?

• Sep 15th 2010, 11:02 PM
Lprdgecko
Can you help me get started on this proof?
I'm having trouble on where to begin with this proof. I'm pretty sure it is a valid argument, since I tried to come up with a counter-example but couldn't. I will use the symbol ~ to stand for "not", since I don't know how to make the "not" symbol online.

Directions: Derive ~Z -> ~Y

(1) ~X -> Y
(2) ~X -> Z
-------------------
(3) ~Z -> ~(~X).......... (1) Contrapositive

I think I should start with the contrapositive, but I have no clue where to go from here. Any help would be appreciated :)
• Sep 15th 2010, 11:49 PM
MathoMan
if ~Z->~(~X) and ~X->Y then ~Z->~Y. Errr... right? Guys?
• Sep 16th 2010, 01:34 AM
PiperAlpha167
Quote:

Originally Posted by Lprdgecko
I'm having trouble on where to begin with this proof. I'm pretty sure it is a valid argument, since I tried to come up with a counter-example but couldn't. I will use the symbol ~ to stand for "not", since I don't know how to make the "not" symbol online.

Directions: Derive ~Z -> ~Y

(1) ~X -> Y
(2) ~X -> Z
-------------------
(3) ~Z -> ~(~X).......... (1) Contrapositive

I think I should start with the contrapositive, but I have no clue where to go from here. Any help would be appreciated :)

It's not a valid argument.

Consider the valuation v, where v(Z) = false, and v(Y) = v(X) = true.

Under v, both premisses are true, and the conclusion is false.
• Sep 16th 2010, 06:13 AM
Lprdgecko
Quote:

Originally Posted by PiperAlpha167
It's not a valid argument.

Consider the valuation v, where v(Z) = false, and v(Y) = v(X) = true.

Under v, both premisses are true, and the conclusion is false.

Thanks! I could've sworn I tried that counter-example, but I guess not.