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