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The rule states that P implies Q is logically equivalent to not-or and that either form can replace the other in logical proofs. In other words, if P {\displaystyle P} is true, then Q {\displaystyle Q} must also be true, while if Q {\displaystyle Q} is not true, then P {\displaystyle P} cannot be true either; additionally, when P {\displaystyle ...
Modus tollens is a mixed hypothetical syllogism that takes the form of "If P, then Q. Not Q. Therefore, not P." It is an application of the general truth that if a statement is true, then so is its contrapositive. The form shows that inference from P implies Q to the negation of Q implies the negation of P is a valid argument.
The proposition to be proved is P. We assume P to be false, i.e., we assume ¬P. It is then shown that ¬P implies falsehood. This is typically accomplished by deriving two mutually contradictory assertions, Q and ¬Q, and appealing to the law of noncontradiction. Since assuming P to be false leads to a contradiction, it is concluded that P is ...
The sun being above the horizon is a necessary condition for direct sunlight; but it is not a sufficient condition, as something else may be casting a shadow, e.g., the moon in the case of an eclipse. The assertion that Q is necessary for P is colloquially equivalent to "P cannot be true unless Q is true" or "if Q is false, then P is false".
For example, even though material conditionals with false antecedents are vacuously true, the natural language statement "If 8 is odd, then 3 is prime" is typically judged false. Similarly, any material conditional with a true consequent is itself true, but speakers typically reject sentences such as "If I have a penny in my pocket, then Paris ...
In propositional logic, modus ponens (/ ˈ m oʊ d ə s ˈ p oʊ n ɛ n z /; MP), also known as modus ponendo ponens (from Latin 'method of putting by placing'), [1] implication elimination, or affirming the antecedent, [2] is a deductive argument form and rule of inference. [3] It can be summarized as "P implies Q. P is true. Therefore, Q must ...
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But p asserts the negation of Bew(G(p)). Thus the system would be inconsistent, proving both a statement and its negation. This contradiction shows that p cannot be provable. If the negation of p were provable, then Bew(G(p)) would be provable (because p was constructed to be equivalent to the negation of Bew(G(p))).