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Expresión ¬ab⇔a¬c∨b(a∨¬c⇒ab(¬a∨¬b∨ac))
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
(b∧(¬a))⇔((a∧(¬c))∨(b∧((a∨(¬c))⇒(a∧b∧((¬a)∨(¬b)∨(a∧c))))))
$$\left(b \wedge \neg a\right) ⇔ \left(\left(a \wedge \neg c\right) \vee \left(b \wedge \left(\left(a \vee \neg c\right) \Rightarrow \left(a \wedge b \wedge \left(\left(a \wedge c\right) \vee \neg a \vee \neg b\right)\right)\right)\right)\right)$$
Solución detallada
$$\left(a \wedge c\right) \vee \neg a \vee \neg b = c \vee \neg a \vee \neg b$$
$$a \wedge b \wedge \left(\left(a \wedge c\right) \vee \neg a \vee \neg b\right) = a \wedge b \wedge c$$
$$\left(a \vee \neg c\right) \Rightarrow \left(a \wedge b \wedge \left(\left(a \wedge c\right) \vee \neg a \vee \neg b\right)\right) = c \wedge \left(b \vee \neg a\right)$$
$$b \wedge \left(\left(a \vee \neg c\right) \Rightarrow \left(a \wedge b \wedge \left(\left(a \wedge c\right) \vee \neg a \vee \neg b\right)\right)\right) = b \wedge c$$
$$\left(a \wedge \neg c\right) \vee \left(b \wedge \left(\left(a \vee \neg c\right) \Rightarrow \left(a \wedge b \wedge \left(\left(a \wedge c\right) \vee \neg a \vee \neg b\right)\right)\right)\right) = \left(a \wedge \neg c\right) \vee \left(b \wedge c\right)$$
$$\left(b \wedge \neg a\right) ⇔ \left(\left(a \wedge \neg c\right) \vee \left(b \wedge \left(\left(a \vee \neg c\right) \Rightarrow \left(a \wedge b \wedge \left(\left(a \wedge c\right) \vee \neg a \vee \neg b\right)\right)\right)\right)\right) = \left(c \wedge \neg a\right) \vee \left(c \wedge \neg b\right) \vee \left(\neg a \wedge \neg b\right)$$
$$a \wedge b \wedge \left(\left(a \wedge c\right) \vee \neg a \vee \neg b\right) = a \wedge b \wedge c$$
$$\left(a \vee \neg c\right) \Rightarrow \left(a \wedge b \wedge \left(\left(a \wedge c\right) \vee \neg a \vee \neg b\right)\right) = c \wedge \left(b \vee \neg a\right)$$
$$b \wedge \left(\left(a \vee \neg c\right) \Rightarrow \left(a \wedge b \wedge \left(\left(a \wedge c\right) \vee \neg a \vee \neg b\right)\right)\right) = b \wedge c$$
$$\left(a \wedge \neg c\right) \vee \left(b \wedge \left(\left(a \vee \neg c\right) \Rightarrow \left(a \wedge b \wedge \left(\left(a \wedge c\right) \vee \neg a \vee \neg b\right)\right)\right)\right) = \left(a \wedge \neg c\right) \vee \left(b \wedge c\right)$$
$$\left(b \wedge \neg a\right) ⇔ \left(\left(a \wedge \neg c\right) \vee \left(b \wedge \left(\left(a \vee \neg c\right) \Rightarrow \left(a \wedge b \wedge \left(\left(a \wedge c\right) \vee \neg a \vee \neg b\right)\right)\right)\right)\right) = \left(c \wedge \neg a\right) \vee \left(c \wedge \neg b\right) \vee \left(\neg a \wedge \neg b\right)$$
Simplificación
[src]
$$\left(c \wedge \neg a\right) \vee \left(c \wedge \neg b\right) \vee \left(\neg a \wedge \neg b\right)$$
(c∧(¬a))∨(c∧(¬b))∨((¬a)∧(¬b))
Tabla de verdad
+---+---+---+--------+ | a | b | c | result | +===+===+===+========+ | 0 | 0 | 0 | 1 | +---+---+---+--------+ | 0 | 0 | 1 | 1 | +---+---+---+--------+ | 0 | 1 | 0 | 0 | +---+---+---+--------+ | 0 | 1 | 1 | 1 | +---+---+---+--------+ | 1 | 0 | 0 | 0 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 0 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+
FNC
[src]
$$\left(c \vee \neg a\right) \wedge \left(c \vee \neg b\right) \wedge \left(\neg a \vee \neg b\right) \wedge \left(c \vee \neg a \vee \neg b\right)$$
(c∨(¬a))∧(c∨(¬b))∧((¬a)∨(¬b))∧(c∨(¬a)∨(¬b))
FNDP
[src]
$$\left(c \wedge \neg a\right) \vee \left(c \wedge \neg b\right) \vee \left(\neg a \wedge \neg b\right)$$
(c∧(¬a))∨(c∧(¬b))∨((¬a)∧(¬b))
FND
[src]
Ya está reducido a FND
$$\left(c \wedge \neg a\right) \vee \left(c \wedge \neg b\right) \vee \left(\neg a \wedge \neg b\right)$$
(c∧(¬a))∨(c∧(¬b))∨((¬a)∧(¬b))
FNCD
[src]
$$\left(c \vee \neg a\right) \wedge \left(c \vee \neg b\right) \wedge \left(\neg a \vee \neg b\right)$$
(c∨(¬a))∧(c∨(¬b))∧((¬a)∨(¬b))