Physicsman567
When P -> Q, why is it true when P is false and Q is true, but why is it false when P is true and Q is false?

If I suppose P mean "Jon is a guy" and Q mean "Mary is a girl". When both P and Q are true it does make sense that this proposition is true because Jon is a guy and Mary is a girl, but when it come to the second and third agreement, where P is false but Q is true still make the proposition true, but the Proposition is a false when P is true and Q is false. I tried replacing P and Q with "Jon is a guy" and "Mary is a girl". And it doesn't make sense why do one of them is right and the another one is wrong?

I believe this is not a smart question, but if there's anyone forgo their time to explain this I would be really appreciate.
Thank you very much.

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Homework Helper
P -> Q means: If P is true then Q is true. So if P is true and Q is false, the statement is not correct (ie. P being true does not imply that Q is also true).

P->Q does not mean: If Q is true then P is true. It does not say what P is if Q is true.

AM

Physicsman567
In propositional calculus, $P\Rightarrow Q$ is defined as $Q \vee (\neg P) = \neg (\neg Q \wedge P)$ (∧ = logical AND, ∨ = logical OR).

Physicsman567
Physicsman567
Thank you very much Andrew Mason and Svein.

It's possible to understand the why the truth table for ##P \implies Q## is defined the way it is by considering a more advanced situation where quantifiers are involved.

For example, consider the statement: For each x, if x is a boy then x has a mother. Which we can abbreviate by:

##\forall x## For each x
## B(x)## x is a boy
## M(x)## x has a mother
## \forall x ( B(x) \implies M(x) ) ## For each x, if x is a boy then x has a mother.

Statements such as ##\forall x ( B(c) \implies M(x)) ## are generalities. In mathematics, for a generality to be considered True, we demand that it is True without exceptions. (- not that it is True only 9 out of 10 times etc.)

Suppose someone wishes to disprove ##\forall x ( B(x) \implies M(x)) ## by presenting an exception. Suppose he says "Let ##x =## my coffee cup. My coffee cup does not have a mother." We do not consider ##x = ## my coffee cup as a valid exception to the rule because a coffee cup is not a boy. The simplest way to prevent ##x = ## my coffee cup from disproving the rule is to declare that ##B(x) \implies M(x)## is a true when ##B(x)## is false. This prevents irrelevant examples from being used to claim exceptions to a generality.

Let ##S(x)## denote "x wears suspenders". Consider the generality ##\forall x ( B(x) \implies S(x))##. To show an exception to this generality we need to provide an example of a boy who does not wear suspenders. This is in accordance with the truth table for ##B(x) \implies S(x)## , which says the implication is False when ##B(x)## is True and ##S(x)## is False.

Physicsman567, QuantumQuest, Logical Dog and 1 other person
Mentor
When P -> Q, why is it true when P is false and Q is true, but why is it false when P is true and Q is false?
In addition to the other fine responses here, the truth value of the implication ##P \Rightarrow Q## is determined by the truth values of the statements P and Q, each of which can be either true or false.

The only pair of truth values for P and Q that makes the implication false is when P is true and Q is false. All other combinations of truth values for P and Q result in a value of true for the implication. Here is the truth table for the implication ##P \Rightarrow Q##:
Code:
P    Q    P => Q
T    T        T
T    F        F
F    T        T
F    F        T

Physicsman567 and QuantumQuest
In propositional calculus, $P\Rightarrow Q$ is defined as $Q \vee (\neg P) = \neg (\neg Q \wedge P)$ (∧ = logical AND, ∨ = logical OR).
Using the fact that a double negation does not change the value of a logic variable, we can derive the following identity: $\neg Q\Rightarrow \neg P = \neg (\neg (\neg P) \wedge \neg Q) = \neg (P \wedge \neg Q)= \neg (\neg Q \wedge P)=P\Rightarrow Q$. This equivalence is called contraposition and is often used in mathematical proofs.