23. El esquema ~(pq)^~ p es equivalente a: a) (p+q)^~p c)~(~q-p) ^~ p e) ~[~(pq) →p] b) ~[-p~(pq)] d) ~[~(pq) V p] Son ciertas: i) a, b, d ii) b, d, e iii) b, c, e iv) a, c, d v) Todas.
Question
- El esquema ~(pq)^~ p es equivalente a: a) (p+q)^~p c)~(~q-p) ^~ p e) ~[~(pq) →p] b) ~[-p~(pq)] d) ~[~(pq) V p] Son ciertas: i) a, b, d ii) b, d, e iii) b, c, e iv) a, c, d v) Todas.
Solution
Para resolver esta pregunta, necesitamos aplicar las leyes de la lógica proposicional.
Primero, vamos a analizar el esquema original ~(pq)^~ p.
Este esquema se puede reescribir como ~[(p∧q)∧~p] utilizando la notación estándar de la lógica proposicional.
Aplicando la ley de Morgan, esto se convierte en ~(p∧q)∨p.
Ahora, vamos a comparar esto con las opciones dadas:
a) (p+q)^~p se convierte en (p∨q)∧~p, que no es equivalente a ~(p∧q)∨p.
b) ~[-p~(pq)] se convierte en ~[~p∧(p∧q)], que no es equivalente a ~(p∧q)∨p.
c) ~(~q-p) ^~ p se convierte en ~(~q∨p)∧~p, que no es equivalente a ~(p∧q)∨p.
d) ~[~(pq) V p] se convierte en ~[(~p∨~q)∨p], que es equivalente a ~(p∧q)∨p.
e) ~[~(pq) →p] se convierte en ~[(~p∨~q)→p], que no es equivalente a ~(p∧q)∨p.
Por lo tanto, la única opción que es equivalente al esquema original es d).
Entonces, la respuesta correcta es ii) b, d, e.
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