1. Jan 5, 2017

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.

Last edited by a moderator: Jan 6, 2017
2. Jan 5, 2017

Andrew Mason

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

3. Jan 6, 2017

Svein

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

4. Jan 6, 2017

Physicsman567

Thank you very much Andrew Mason and Svein.

5. Jan 12, 2017

Stephen Tashi

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.

6. Jan 12, 2017

Staff: Mentor

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 (Text):

P    Q    P => Q
T    T        T
T    F        F
F    T        T
F    F        T

7. Jan 13, 2017

Svein

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.

8. Jan 18, 2017

Physicsman567

Thank you all so much!