Why does P -> Q hold true when P is false and Q is true in mathematical logic?

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Discussion Overview

The discussion centers on the logical implication P -> Q, specifically exploring why the statement is considered true when P is false and Q is true, while it is false when P is true and Q is false. Participants examine the definitions and truth values associated with implications in mathematical logic.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the truth of the implication P -> Q when P is false and Q is true, seeking clarification on the logical reasoning behind this.
  • Another participant explains that P -> Q means if P is true then Q must also be true, and clarifies that it does not imply that if Q is true, then P must be true.
  • A participant provides a formal definition of the implication in propositional calculus, stating it can be expressed as Q ∨ (¬P) or ¬(¬Q ∧ P).
  • One participant introduces the concept of quantifiers and generalities, illustrating the truth of implications through examples involving boys and mothers, emphasizing that a false antecedent does not invalidate the implication.
  • Another participant reiterates the truth table for the implication, noting that the only scenario where P -> Q is false is when P is true and Q is false.
  • A later post discusses the concept of contraposition, explaining its equivalence to the original implication and its utility in mathematical proofs.

Areas of Agreement / Disagreement

Participants generally agree on the definitions and truth values associated with implications, but there is no consensus on the deeper philosophical implications or interpretations of these logical constructs.

Contextual Notes

The discussion involves various interpretations of logical implications and their applications, with some participants providing formal definitions while others explore conceptual understandings. There are unresolved nuances regarding the implications of false antecedents in different contexts.

Physicsman567
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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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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
 
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In propositional calculus, P\Rightarrow Q is defined as Q \vee (\neg P) = \neg (\neg Q \wedge P) (∧ = logical AND, ∨ = logical OR).
 
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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.
 
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Physicsman567 said:
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
 
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Svein said:
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.
 
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Thank you all so much!
 

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