Complete the truth table:
A | B | not B | A => B | A not implies > B | A and not B
T | T | F | T | F | F
T | F | T | F | T | T
I will leave this problem for now and edit it later on. It looks, though, like (A does not imply B) and (A and not B) are the same.
5. I also got: ( A => B) => ~(A ^ ~B) => (~A V B)
ReplyDeleteThis is all new to me so I don't know if it's right or useful.
For 5, I concluded ~A=>B = (A^~B)
DeleteFor 4, I have:
A | B | ~B | A=>B | A=|>B | A^~B
T | T | F | T | F | F
T | F | T | F | T | T
F | T | F | T | F | F
F | F | T | T | F | F
Correction: for 5, I concluded A =|> B = A^~B where
DeleteA =|> B means A not implies B
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ReplyDeleteNeeded to make a correction to my wording
ReplyDeleteMaybe a word description is what is wanted.
I suggest
The truth/falsity of a negative conditional is equivalent to the conjunction of the antecedent with a negative consequent