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P => Q: why does this evaluate to True, where Q is True and P is False?

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Suppose you impose yourself a rule: if it rains then you bring your umbrella.

If it doesn't rain, does that mean *this rule* is broken?

That's the intuition behind making such "implication" true when the precondition is false.
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One thing important to point out. The => is *not* the same as the "if...then..." construction in usual language; but it's closest translation of "if...then..." in the context of propositional logic. That's why the => is called the "material implication" rather than just implication, the word adjective "material" tells you that it only care about truth values. The natural language's "if...then..." is a hypothetical, it sets up a situation for you to consider. That's why people find it weird when you think of => as "if...then..." but the left side is false, because with "if...then..." you're not supposed to think about the situation when the left side is false. But that is the best we can do in the context of propositional logic, which is not expressive enough to handle a hypothetical.

There are many other possible translation of "if...then..." in the context of logic. This particular translation deal with the situation where both sides always return a truth value True or False, and the operator can only produces an output based on these truth value and nothing else.
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Because intuitively you want not(P => Q) being : P *and* not(Q). That's how you use counter examples.
Are all continuous (P) functions derivable (Q) ? No, the Weierstrass function is a continuous function (P) which is not derivable (not Q), therefore we have not(continuous => derivable).

Now you can use the truth table of P *and* not(Q) to get P => Q and you'll see how it unravels.

As a bonus, we have an expression of the => operator in terms of the *or* operator, that is P => Q is logically equivalent to not(P) *or* Q.
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Suppose you have the statement “if P, then Q.”

This can also be read as “P is a sufficient condition for Q.”

That is to say, P is a sufficient, but not necessary, condition for Q. The not necessary part is important, because it means that even if P were false, we may still be able to satisfy Q.

Let’s say P is the statement “I go to school,” and Q is “I have books.” So “if P, then Q,” is “if I go to school, then I have books.” While it may be true that me going to school means I have books, it could also be that I have books, but I don’t go to school (in which case, the statement is still true, even if P is false).
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-If you study enough, you'll pass the exam.

-That's not true, sometimes I don't study at all and I pass the exam.

-It's the exam about propositional logic?

-Yes.

-I would keep studying a little bit more if I were you.
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Something that is even weirder for me it’s that false -> false is true
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Let's say P means "I have children" and Q means "I have toys in my house" and so P=> Q means "if I have children then I have toys in my house". It doesn't say anything about whether I have toys anyway.

Now's let's say that I have some toys. That could happen because I have children (true => true is true). But it might also happen because I have toys of my own (false => true is true).
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A false assumption implies anything.
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There are already lots of great answers but I'll try to help to: basically its the idea that if you start with a premise that's false, and come up with some conclusion based on that premise, you can really come up with any crazy false thing you want. Ex. if I could sprout wings and fly at the speed of sound, I can get to my parents house in minutes. IF it were true that I could sprout wings, there's a lot of things that could be true based on that false premise
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I'd like to add my own intuitive example.

Let's have a statement such as "If i fall, then i would have an injury"

P is "If I fall", and Q is "I would have an injury".

So if I fall (P is true), then I must have an injury (Q is true).

However, if I have an injury (Q is true), it doesnt necessarily have to be because I fell (P).

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