Sr Examen
Expresión AB⇔Av¬C⇒¬A⇔Bv¬C(¬AvB)
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Solución
Ha introducido
[src]
(a∧b)⇔((a∨(¬c))⇒(¬a))⇔(b∨((¬c)∧(b∨(¬a))))
$$\left(a \wedge b\right) ⇔ \left(\left(a \vee \neg c\right) \Rightarrow \neg a\right) ⇔ \left(b \vee \left(\neg c \wedge \left(b \vee \neg a\right)\right)\right)$$
Solución detallada
$$\left(a \vee \neg c\right) \Rightarrow \neg a = \neg a$$
$$b \vee \left(\neg c \wedge \left(b \vee \neg a\right)\right) = b \vee \left(\neg a \wedge \neg c\right)$$
$$\left(a \wedge b\right) ⇔ \left(\left(a \vee \neg c\right) \Rightarrow \neg a\right) ⇔ \left(b \vee \left(\neg c \wedge \left(b \vee \neg a\right)\right)\right) = a \wedge \neg b$$
$$b \vee \left(\neg c \wedge \left(b \vee \neg a\right)\right) = b \vee \left(\neg a \wedge \neg c\right)$$
$$\left(a \wedge b\right) ⇔ \left(\left(a \vee \neg c\right) \Rightarrow \neg a\right) ⇔ \left(b \vee \left(\neg c \wedge \left(b \vee \neg a\right)\right)\right) = a \wedge \neg b$$
Tabla de verdad
+---+---+---+--------+ | a | b | c | result | +===+===+===+========+ | 0 | 0 | 0 | 0 | +---+---+---+--------+ | 0 | 0 | 1 | 0 | +---+---+---+--------+ | 0 | 1 | 0 | 0 | +---+---+---+--------+ | 0 | 1 | 1 | 0 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 0 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+