Table of Contents
What is the meaning of implication in logic?
implication, in logic, a relationship between two propositions in which the second is a logical consequence of the first. In most systems of formal logic, a broader relationship called material implication is employed, which is read “If A, then B,” and is denoted by A ⊃ B or A → B.
Why does false imply true?
So the reason for the convention ‘false implies true is true’ is that it makes statements like x<10→x<100 true for all values of x, as one would expect. You want “real life”, eh? If the policeman sees you speeding, then you will have to pay a fine. This is true.
What does this mean P -> Q?
Implication œ Conditional Statement. p → q (p implies q) (if p then q) is the proposition that is false when p is true and q is false and true otherwise. Equivalent to —not p or q“ Ex.
What is the negation of P -> Q?
The negation of compound statements works as follows: The negation of “P and Q” is “not-P or not-Q”. The negation of “P or Q” is “not-P and not-Q”.
Why is implication true when p is false?
An implication is the compound statement of the form “if p, then q.” It is denoted p⇒q, which is read as “p implies q.” It is false only when p is true and q is false, and is true in all other situations.
How do you read logical implications?
What happens if p and q are both false?
But if P and Q are both false, then P ∧ Q is false, and we are led to the conclusion that if both antecedent and consequent are false, a conditional is true. From Herbert Enderton’s “A Mathematical Introduction to Logic” page 21:
Why is a promise unbroken when p → q is true?
If the condition p turns out not to be met, then the promise stands unbroken, regardless of q. That’s why it’s said to be “vacuously true”. That ( p → q) is True when both p, q are False is different from saying the conclusion q is True (which would be a contradiction).
Why is $P \o Q)$ said to be ‘vacuously true’?
That’s why it’s said to be “vacuously true”. That $(p \o q)$ is True when both $p$, $q$ are False is different from saying the conclusion $q$ is True (which would be a contradiction). Rather, this is more like saying “we cannot show $p \o q)$ to be false here” and Not False is True.
Is $\\begingroup$ an implication of $P$ and $Q$?
$\\begingroup$In $p ightarrow q$, three of the four possibilities are true. In $ eg (p ightarrow q)$ only one of the four possibilities is true. So it is not an implication in any simple way. Similarly, when you negate $p$ AND $q$, the result is not an AND statement.