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Expresión ¬C⇒(¬A∨B)∧B⇔C
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Solución
Ha introducido
[src]
c⇔((¬c)⇒(b∧(b∨(¬a))))
$$c ⇔ \left(\neg c \Rightarrow \left(b \wedge \left(b \vee \neg a\right)\right)\right)$$
Solución detallada
$$b \wedge \left(b \vee \neg a\right) = b$$
$$\neg c \Rightarrow \left(b \wedge \left(b \vee \neg a\right)\right) = b \vee c$$
$$c ⇔ \left(\neg c \Rightarrow \left(b \wedge \left(b \vee \neg a\right)\right)\right) = c \vee \neg b$$
$$\neg c \Rightarrow \left(b \wedge \left(b \vee \neg a\right)\right) = b \vee c$$
$$c ⇔ \left(\neg c \Rightarrow \left(b \wedge \left(b \vee \neg a\right)\right)\right) = c \vee \neg b$$
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 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 0 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+