Formal Logic

LSATHopeful-2

Hey, quick question. Can one go from

(X ---> Y) ---> Z
~Z ---> ~(X ---> Y)

To

Z & X ---> Y?

Intuitively I feel I can, but I don't recall that exact translation being in the CC. I recall that I could go from (X ---> Y) ---> Z to X & Y ---> Z, but not the one above. If i can make this inference, how and why?


Thanks!!

Cant Get Right

That's an interesting question. I'm leaning towards no, but not with as much confidence as I'd like!

So with the (X ---> Y) as a sufficient condition, we aren't making any determinations about whether or not that relationship holds up. X may ---> Y or it may not. All we know is that IF it does, then Z. So, in Z & X ---> Y, it seems to me that we're having to satisfy this condition first. Z is a necessary condition, so we could theoretically deny the sufficient condition in which case Z is free to go either way. But by denying the sufficient, we're denying the relationship between X and Y. So they have nothing to do with each other. So in that scenario, Y is not strictly necessary from Z & X.

Sami

(X ---> Y) ---> Z is lawgically equivalent to:
(/X or Y) --->Z this is equivalent to:
/(/X or Y) or Z this can be simplified to:
(X and /Y) or Z
I don't think this can be simplified any further.

This above statement {(X and /Y) or Z} gives us two possibilities:
/Z-----> X and /Y
/(X and /Y)--> Z

This means whenever Z doesn't happen we will have both X and /Y occur.
And if we don't have (X and /Y), then we have to have Z occur.


This was tough for me and I am not sure about this at all. @"Cant Get Right" what do you think?

Jonathan Wang

If all apples are bad, then candy is awesome. (A->B) -> C

If candy is not awesome, then not all apples are bad. ~C -> ~(A->B)

So now, your proposed inference:

If candy is awesome and I have apples, then I have something bad. C and A -> B

What do you think?