Sr Examen
Expresión ∼(∼p→r)∧∼(q∧∼p)
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
(¬(q∧(¬p)))∧(¬(¬(p⇒r)))
$$\neg \left(q \wedge \neg p\right) \wedge \neg \left(p \not\Rightarrow r\right)$$
Solución detallada
$$\neg \left(q \wedge \neg p\right) = p \vee \neg q$$
$$p \Rightarrow r = r \vee \neg p$$
$$p \not\Rightarrow r = p \wedge \neg r$$
$$\neg \left(p \not\Rightarrow r\right) = r \vee \neg p$$
$$\neg \left(q \wedge \neg p\right) \wedge \neg \left(p \not\Rightarrow r\right) = \left(p \wedge r\right) \vee \left(\neg p \wedge \neg q\right)$$
$$p \Rightarrow r = r \vee \neg p$$
$$p \not\Rightarrow r = p \wedge \neg r$$
$$\neg \left(p \not\Rightarrow r\right) = r \vee \neg p$$
$$\neg \left(q \wedge \neg p\right) \wedge \neg \left(p \not\Rightarrow r\right) = \left(p \wedge r\right) \vee \left(\neg p \wedge \neg q\right)$$
Simplificación
[src]
$$\left(p \wedge r\right) \vee \left(\neg p \wedge \neg q\right)$$
(p∧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 | 0 | +---+---+---+--------+ | 1 | 0 | 0 | 0 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 0 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+
FND
[src]
Ya está reducido a FND
$$\left(p \wedge r\right) \vee \left(\neg p \wedge \neg q\right)$$
(p∧r)∨((¬p)∧(¬q))
FNC
[src]
$$\left(p \vee \neg p\right) \wedge \left(p \vee \neg q\right) \wedge \left(r \vee \neg p\right) \wedge \left(r \vee \neg q\right)$$
(p∨(¬p))∧(p∨(¬q))∧(r∨(¬p))∧(r∨(¬q))