Sr Examen
Expresión (P⇒(Q∧R))⇔¬(P⇒Q)
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
(¬(p⇒q))⇔(p⇒(q∧r))
$$\left(p \Rightarrow \left(q \wedge r\right)\right) ⇔ p \not\Rightarrow q$$
Solución detallada
$$p \Rightarrow q = q \vee \neg p$$
$$p \not\Rightarrow q = p \wedge \neg q$$
$$p \Rightarrow \left(q \wedge r\right) = \left(q \wedge r\right) \vee \neg p$$
$$\left(p \Rightarrow \left(q \wedge r\right)\right) ⇔ p \not\Rightarrow q = p \wedge q \wedge \neg r$$
$$p \not\Rightarrow q = p \wedge \neg q$$
$$p \Rightarrow \left(q \wedge r\right) = \left(q \wedge r\right) \vee \neg p$$
$$\left(p \Rightarrow \left(q \wedge r\right)\right) ⇔ p \not\Rightarrow q = p \wedge q \wedge \neg r$$
Tabla de verdad
+---+---+---+--------+ | p | q | r | result | +===+===+===+========+ | 0 | 0 | 0 | 0 | +---+---+---+--------+ | 0 | 0 | 1 | 0 | +---+---+---+--------+ | 0 | 1 | 0 | 0 | +---+---+---+--------+ | 0 | 1 | 1 | 0 | +---+---+---+--------+ | 1 | 0 | 0 | 0 | +---+---+---+--------+ | 1 | 0 | 1 | 0 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+