6.7: Prueba por contraejemplo
- Page ID
- 118220
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Recordar. La equivalencia
sostiene para cualquier conjunto de casos\(C_1, C_2, \ldots, C_m\) tal que\(C_1 \lor \cdots \lor C_m\) sea una tautología. (Ver Sección 6.4.)
Entonces si no\(P \land C_i \rightarrow Q\) es una tautología para al menos una\(i\text{,}\) entonces\(P \rightarrow Q\) tampoco puede ser una tautología. De nuevo, esto también funciona en el caso universal ya que\(\forall\) distribuye sobre\(\land\) (Proposición 4.2.1).
relativo a la implicación lógica\(P \Rightarrow Q\text{,}\) una declaración\(C\) tal que\(P \land C \rightarrow Q\) es falsa
En el Ejercicio 6.12.8, se le solicita acreditar la siguiente declaración demostrando el contrapositivo.
Si\(2^n - 1\) prime, entonces\(n\) es primo.
Demostrar que lo contrario de esta afirmación es falso.
Solución
La declaración converse es “Si\(n\) es primo, entonces\(2^n - 1\) es primo”. Pero el caso\(n=11\) es un contraejemplo:
no es primo aunque\(n = 11\) sea primo.
Comprueba tu comprensión. Intento Ejercicio 6.12.9.