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Little question (math logic) - Manu - 2011-04-08
Didn't really wanna make a thread for this but my other logic post was necro.
how do I deny (p<=>q)?
I'm thinking something like morgan but not sure:
(~p _v_ ~q) where _v_ is exclusive 'or' so it's one true but not both for it to work.
Little question (math logic) - OB3LISK - 2011-04-08
Make one true and one false.
True cannot imply false.
EAT THAT 21
Little question (math logic) - Manu - 2011-04-08
OB3LISK Wrote:Make one true and one false.
True cannot imply false.
EAT THAT 21
That's not imply, that's a <=> double arrow, not sure how you call it in english but it means both true or both false for it to be true.
Little question (math logic) - OB3LISK - 2011-04-08
Exactly it's a double implication. So if p is true and q is false then the whole thing is false.
Unless you're asking a different question.
Little question (math logic) - modular - 2011-04-08
Taking "not p and q or p and not q" would falsify the statement... aka "p xor q"
English equivalent of <=> is iff / if and only if / xnor
(I think I don't know the proper notation...)
Little question (math logic) - Manu - 2011-04-08
oh so it's the way I said it, wasn't sure I was making something up.
Thanks for the help
Little question (math logic) - octopusprime - 2011-04-09
Darkmaniak Wrote:oh so it's the way I said it, wasn't sure I was making something up.
Thanks for the help
it's called equivalence unless i am misunderstanding and what obelisk said is correct.
p<=>q
Little question (math logic) - Kaasoljoyyx - 2011-04-09
Just to add-on, if you are talking about the double arrow in mathematics
<==>
Like someone said, it means "if and only if"
In order to prove it, you have to prove it in one direction, and then prove it in the other
i.e. Prove that p --> q, and q --> p
If you want to deny it, you only have to disprove one of them i.e. Make a contradiction happen in either direction