3 ms·

Is this related to mistaking an "if" statement in math as "if and only if" ?

by j7ake 5y ago

gota 5y ago
Well, sort of - but we use "if" in day-to-day as a different thing to the material implication in logic. I've yet to run into a person using 'if and only if' in a way that doesn't map to the material equivalence, though, but I'm sure it must happen Actually, this wikipedia article for has got a specific section about that https://en.wikipedia.org/wiki/Material_conditional https://en.wikipedia.org/wiki/Material_conditional Edit: I also just learned abou the 'Wason selection task' from that link. Cool stuff. https://en.wikipedia.org/wiki/Wason_selection_task https://en.wikipedia.org/wiki/Wason_selection_task
kozd 5y ago
I think so. The loosey-goosey language of "if" is usually (would need to actually check this) interpreted as an implication relationship "B if A" or "If A, then B" interpreted as A -> B The fallacy would then be to do the following A -> B B ------------ A On the other hand, iff is interpreted as A <=> B i.e. A -> B and B -> A So you can get A or B if you have B or A respectively.
Banana699 5y ago
yes, "if" statments in proofs are the "p->q" forms from logic, if you have p you also have q (therefore if no q then no p). If you imagine the set of all possible logical propositions as nodes in a graph, p->q tells you that q is reachable from p. If you prove that q is not reachable from whatever axioms you have (i.e. false), then p is false too by necessity, because otherwise q would be reachable and we know that's not the case. But if all you have is that q is reachable, that's not enough to assert that p is also reachable. Because you have no reason to believe that the road from p to q is the only one, it might very well be that p is false but q is reachable from another statment s or something. "if and only if" is "p->q and q->p" (also equivalent to "p->q and not(p)->not(q), the more intuitive sense of the phrase), it basically establishes equivalence: any proof that requires p also requires q and vice versa, any proof that guarantees p also guarantees q and vice versa.

Is this related to mistaking an "if" statement in math as "if and only if" ?

by j7ake 5y ago

Is this related to mistaking an "if" statement in math as "if and only if" ?

gota 5y ago

Well, sort of - but we use "if" in day-to-day as a different thing to the material implication in logic. I've yet to run into a person using 'if and only if' in a way that doesn't map to the material equivalence, though, but I'm sure it must happen Actually, this wikipedia article for has got a specific section about that https://en.wikipedia.org/wiki/Material_conditional https://en.wikipedia.org/wiki/Material_conditional Edit: I also just learned abou the 'Wason selection task' from that link. Cool stuff. https://en.wikipedia.org/wiki/Wason_selection_task https://en.wikipedia.org/wiki/Wason_selection_task

kozd 5y ago

I think so. The loosey-goosey language of "if" is usually (would need to actually check this) interpreted as an implication relationship "B if A" or "If A, then B" interpreted as A -> B The fallacy would then be to do the following A -> B B ------------ A On the other hand, iff is interpreted as A <=> B i.e. A -> B and B -> A So you can get A or B if you have B or A respectively.

Banana699 5y ago

yes, "if" statments in proofs are the "p->q" forms from logic, if you have p you also have q (therefore if no q then no p). If you imagine the set of all possible logical propositions as nodes in a graph, p->q tells you that q is reachable from p. If you prove that q is not reachable from whatever axioms you have (i.e. false), then p is false too by necessity, because otherwise q would be reachable and we know that's not the case. But if all you have is that q is reachable, that's not enough to assert that p is also reachable. Because you have no reason to believe that the road from p to q is the only one, it might very well be that p is false but q is reachable from another statment s or something. "if and only if" is "p->q and q->p" (also equivalent to "p->q and not(p)->not(q), the more intuitive sense of the phrase), it basically establishes equivalence: any proof that requires p also requires q and vice versa, any proof that guarantees p also guarantees q and vice versa.