Sr Examen
Expresión ¬(¬((A↔B)∧(¬A→(¬B∨C))→C))
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
¬(¬(((a⇔b)∧((¬a)⇒(c∨(¬b))))⇒c))
$$\neg \left(\left(\left(a ⇔ b\right) \wedge \left(\neg a \Rightarrow \left(c \vee \neg b\right)\right)\right) \not\Rightarrow c\right)$$
Solución detallada
$$a ⇔ b = \left(a \wedge b\right) \vee \left(\neg a \wedge \neg b\right)$$
$$\neg a \Rightarrow \left(c \vee \neg b\right) = a \vee c \vee \neg b$$
$$\left(a ⇔ b\right) \wedge \left(\neg a \Rightarrow \left(c \vee \neg b\right)\right) = \left(a \wedge b\right) \vee \left(\neg a \wedge \neg b\right)$$
$$\left(\left(a ⇔ b\right) \wedge \left(\neg a \Rightarrow \left(c \vee \neg b\right)\right)\right) \Rightarrow c = c \vee \left(a \wedge \neg b\right) \vee \left(b \wedge \neg a\right)$$
$$\left(\left(a ⇔ b\right) \wedge \left(\neg a \Rightarrow \left(c \vee \neg b\right)\right)\right) \not\Rightarrow c = \neg c \wedge \left(a \vee \neg b\right) \wedge \left(b \vee \neg a\right)$$
$$\neg \left(\left(\left(a ⇔ b\right) \wedge \left(\neg a \Rightarrow \left(c \vee \neg b\right)\right)\right) \not\Rightarrow c\right) = c \vee \left(a \wedge \neg b\right) \vee \left(b \wedge \neg a\right)$$
$$\neg a \Rightarrow \left(c \vee \neg b\right) = a \vee c \vee \neg b$$
$$\left(a ⇔ b\right) \wedge \left(\neg a \Rightarrow \left(c \vee \neg b\right)\right) = \left(a \wedge b\right) \vee \left(\neg a \wedge \neg b\right)$$
$$\left(\left(a ⇔ b\right) \wedge \left(\neg a \Rightarrow \left(c \vee \neg b\right)\right)\right) \Rightarrow c = c \vee \left(a \wedge \neg b\right) \vee \left(b \wedge \neg a\right)$$
$$\left(\left(a ⇔ b\right) \wedge \left(\neg a \Rightarrow \left(c \vee \neg b\right)\right)\right) \not\Rightarrow c = \neg c \wedge \left(a \vee \neg b\right) \wedge \left(b \vee \neg a\right)$$
$$\neg \left(\left(\left(a ⇔ b\right) \wedge \left(\neg a \Rightarrow \left(c \vee \neg b\right)\right)\right) \not\Rightarrow c\right) = c \vee \left(a \wedge \neg b\right) \vee \left(b \wedge \neg a\right)$$
Simplificación
[src]
$$c \vee \left(a \wedge \neg b\right) \vee \left(b \wedge \neg a\right)$$
c∨(a∧(¬b))∨(b∧(¬a))
Tabla de verdad
+---+---+---+--------+ | a | b | c | result | +===+===+===+========+ | 0 | 0 | 0 | 0 | +---+---+---+--------+ | 0 | 0 | 1 | 1 | +---+---+---+--------+ | 0 | 1 | 0 | 1 | +---+---+---+--------+ | 0 | 1 | 1 | 1 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 0 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+
FNC
[src]
$$\left(a \vee b \vee c\right) \wedge \left(a \vee c \vee \neg a\right) \wedge \left(b \vee c \vee \neg b\right) \wedge \left(c \vee \neg a \vee \neg b\right)$$
(a∨b∨c)∧(a∨c∨(¬a))∧(b∨c∨(¬b))∧(c∨(¬a)∨(¬b))
FND
[src]
Ya está reducido a FND
$$c \vee \left(a \wedge \neg b\right) \vee \left(b \wedge \neg a\right)$$
c∨(a∧(¬b))∨(b∧(¬a))
FNDP
[src]
$$c \vee \left(a \wedge \neg b\right) \vee \left(b \wedge \neg a\right)$$
c∨(a∧(¬b))∨(b∧(¬a))
FNCD
[src]
$$\left(a \vee b \vee c\right) \wedge \left(c \vee \neg a \vee \neg b\right)$$
(a∨b∨c)∧(c∨(¬a)∨(¬b))