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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!!
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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?
(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. @jhaldy10325 what do you think?
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.