Sr Examen
Expresión (P→Q)∧(P↔Q)
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Solución detallada
$$p ⇔ q = \left(p \wedge q\right) \vee \left(\neg p \wedge \neg q\right)$$
$$p \Rightarrow q = q \vee \neg p$$
$$\left(p ⇔ q\right) \wedge \left(p \Rightarrow q\right) = \left(p \wedge q\right) \vee \left(\neg p \wedge \neg q\right)$$
$$p \Rightarrow q = q \vee \neg p$$
$$\left(p ⇔ q\right) \wedge \left(p \Rightarrow q\right) = \left(p \wedge q\right) \vee \left(\neg p \wedge \neg q\right)$$
Simplificación
[src]
$$\left(p \wedge q\right) \vee \left(\neg p \wedge \neg q\right)$$
(p∧q)∨((¬p)∧(¬q))
Tabla de verdad
+---+---+--------+ | p | q | result | +===+===+========+ | 0 | 0 | 1 | +---+---+--------+ | 0 | 1 | 0 | +---+---+--------+ | 1 | 0 | 0 | +---+---+--------+ | 1 | 1 | 1 | +---+---+--------+
FND
[src]
Ya está reducido a FND
$$\left(p \wedge q\right) \vee \left(\neg p \wedge \neg q\right)$$
(p∧q)∨((¬p)∧(¬q))
FNC
[src]
$$\left(p \vee \neg p\right) \wedge \left(p \vee \neg q\right) \wedge \left(q \vee \neg p\right) \wedge \left(q \vee \neg q\right)$$
(p∨(¬p))∧(p∨(¬q))∧(q∨(¬p))∧(q∨(¬q))