Material_implication_(rule_of_inference) Knowpia

Material implication
Type	Rule of replacement
Field	Propositional calculus
Statement	P implies Q is logically equivalent to not- or . Either form can replace the other in logical proofs.
Symbolic statement

In propositional logic, material implication^[1]^[2] is a valid rule of replacement that allows for a conditional statement to be replaced by a disjunction in which the antecedent is negated. The rule states that P implies Q is logically equivalent to not- $P$ or $Q$ and that either form can replace the other in logical proofs. In other words, if $P$ is true, then $Q$ must also be true, while if $Q$ is not true, then $P$ cannot be true either; additionally, when $P$ is not true, $Q$ may be either true or false.

$P\to Q\Leftrightarrow \neg P\lor Q$

Where " $\Leftrightarrow$ " is a metalogical symbol representing "can be replaced in a proof with," P and Q are any given logical statements, and $\neg P\lor Q$ can be read as "(not P) or Q". To illustrate this, consider the following statements:

$P$ : Sam ate an orange for lunch
$Q$ : Sam ate a fruit for lunch

Then, to say, "Sam ate an orange for lunch" implies "Sam ate a fruit for lunch" ( $P\to Q$ ). Logically, if Sam did not eat a fruit for lunch, then Sam also cannot have eaten an orange for lunch (by contraposition). However, merely saying that Sam did not eat an orange for lunch provides no information on whether or not Sam ate a fruit (of any kind) for lunch.

Partial proof edit

Suppose we are given that $P\to Q$ . Then, we have $\neg P\lor P$ by the law of excluded middle^{[clarification needed]} (i.e. either $P$ must be true, or $P$ must not be true).

Subsequently, since $P\to Q$ , $P$ can be replaced by $Q$ in the statement, and thus it follows that $\neg P\lor Q$ (i.e. either $Q$ must be true, or $P$ must not be true).

Suppose, conversely, we are given $\neg P\lor Q$ . Then if $P$ is true that rules out the first disjunct, so we have $Q$ . In short, $P\to Q$ .^[3] However if $P$ is false, then this entailment fails, because the first disjunct $\neg P$ is true which puts no constraint on the second disjunct $Q$ . Hence, nothing can be said about $P\to Q$ . In sum, the equivalence in the case of false $P$ is only conventional, and hence the formal proof of equivalence is only partial.

This can also be expressed with a truth table:

P	Q	¬P	P→Q	¬P ∨ Q
T	T	F	T	T
T	F	F	F	F
F	T	T	T	T
F	F	T	T	T

Example edit

An example: we are given the conditional fact that if it is a bear, then it can swim. Then, all 4 possibilities in the truth table are compared to that fact.

If it is a bear, then it can swim — T
If it is a bear, then it can not swim — F
If it is not a bear, then it can swim — T because it doesn’t contradict our initial fact.
If it is not a bear, then it can not swim — T (as above)

Thus, the conditional fact can be converted to $\neg P\vee Q$ , which is "it is not a bear" or "it can swim", where $P$ is the statement "it is a bear" and $Q$ is the statement "it can swim".

References edit

^ Patrick J. Hurley (1 January 2011). A Concise Introduction to Logic. Cengage Learning. ISBN 978-0-8400-3417-5.
^ Copi, Irving M.; Cohen, Carl (2005). Introduction to Logic. Prentice Hall. p. 371.
^ Math StackExchange: Equivalence of a→b and ¬ a ∨ b

[Hurley2011-1] Patrick J. Hurley (1 January 2011). A Concise Introduction to Logic. Cengage Learning. ISBN 978-0-8400-3417-5.

[2] Copi, Irving M.; Cohen, Carl (2005). Introduction to Logic. Prentice Hall. p. 371.

[3] Math StackExchange: Equivalence of a→b and ¬ a ∨ b

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Summary

Partial proof edit

Example edit

References edit