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