Sr Examen
Expresión d^¬a^(b∨c)^(b∨¬c∨¬d∨¬c^¬d)^(a∨¬b∨c)
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
d∧(¬a)∧(b∨c)∧(a∨c∨(¬b))∧(b∨(¬c)∨(¬d)∨((¬c)∧(¬d)))
$$d \wedge \neg a \wedge \left(b \vee c\right) \wedge \left(a \vee c \vee \neg b\right) \wedge \left(b \vee \left(\neg c \wedge \neg d\right) \vee \neg c \vee \neg d\right)$$
Solución detallada
$$b \vee \left(\neg c \wedge \neg d\right) \vee \neg c \vee \neg d = b \vee \neg c \vee \neg d$$
$$d \wedge \neg a \wedge \left(b \vee c\right) \wedge \left(a \vee c \vee \neg b\right) \wedge \left(b \vee \left(\neg c \wedge \neg d\right) \vee \neg c \vee \neg d\right) = b \wedge c \wedge d \wedge \neg a$$
$$d \wedge \neg a \wedge \left(b \vee c\right) \wedge \left(a \vee c \vee \neg b\right) \wedge \left(b \vee \left(\neg c \wedge \neg d\right) \vee \neg c \vee \neg d\right) = b \wedge c \wedge d \wedge \neg a$$
Tabla de verdad
+---+---+---+---+--------+ | a | b | c | d | result | +===+===+===+===+========+ | 0 | 0 | 0 | 0 | 0 | +---+---+---+---+--------+ | 0 | 0 | 0 | 1 | 0 | +---+---+---+---+--------+ | 0 | 0 | 1 | 0 | 0 | +---+---+---+---+--------+ | 0 | 0 | 1 | 1 | 0 | +---+---+---+---+--------+ | 0 | 1 | 0 | 0 | 0 | +---+---+---+---+--------+ | 0 | 1 | 0 | 1 | 0 | +---+---+---+---+--------+ | 0 | 1 | 1 | 0 | 0 | +---+---+---+---+--------+ | 0 | 1 | 1 | 1 | 1 | +---+---+---+---+--------+ | 1 | 0 | 0 | 0 | 0 | +---+---+---+---+--------+ | 1 | 0 | 0 | 1 | 0 | +---+---+---+---+--------+ | 1 | 0 | 1 | 0 | 0 | +---+---+---+---+--------+ | 1 | 0 | 1 | 1 | 0 | +---+---+---+---+--------+ | 1 | 1 | 0 | 0 | 0 | +---+---+---+---+--------+ | 1 | 1 | 0 | 1 | 0 | +---+---+---+---+--------+ | 1 | 1 | 1 | 0 | 0 | +---+---+---+---+--------+ | 1 | 1 | 1 | 1 | 0 | +---+---+---+---+--------+