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For instance, a -> (b -> c) simplifies to a + -b -> -c. What about (-a -> b) -> c. How can I simplify it?
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For instance, a -> (b -> c) simplifies to a + -b -> -c. What about (-a -> b) -> c. How can I simplify it?
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6 comments
@leeginnyy488, the first two are spot on, but for the third conditional statement A -> (B->C) = A and not B -> C, the contrapositive should be (B and not C) > not A since your embedded necessary condition is (B > C).
If A exists then the relation B > C kicks in. If the relation B > C cannot kick in, then A cannot exist.
So,
(A -> B) -> C = not C -> A and not B
(not A -> B) -> C = not C -> not A and not B
A -> (B->C) = A and not B -> C
?
@shinesther715 another way to think of not(A>B) would be "you can have A and not B." So not(-A>B) would be "you can have not A and not B." (-A>B) is just an "or relation" in which "one must always be in." Saying nope to this would be "you don't need to always have one in," which is "you can have both out (not A and not B). I hope this helped.
Would someone help me with this one? (I'm reviewing JY's explanation, but just making sure I'm understanding this concept...)
this I get:
not (A -> B) would be A some -B
this I'm not so sure:
not (-A -> B)
Any help is much appreciated!
To simplify: (-a -> b) -> c, think of the expression (-a -> b) as y. That is, y=(-a -> b). Then, we get
y -> c,
or, -c -> -y (taking contrapositive)
or, -c -> (-a -> b)
or, -c AND -a -> b
I believe the second solution is the proper approach.
(-a->b)->c
because you have a conditional within a conditional. View the lessons on Demorgans laws & mastery to see a similar issue that JY solves.
But unless=negate sufficient, so negate a and sufficient, then only if= necessary
so -a->b is its only conditional, but it is a conditional that has a conditional that applies to it as well, so (-a->b) servers as the entire sufficient, and then only if ties that it together.
Hope that makes sense, these are difficult to explain, if your having issues like I said watch the videos on Demorgans law.