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