Sr Examen
Expresión a→c=(aV(b&c))→(aVb)&C
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Solución
Ha introducido
[src]
(a⇒c)⇔((a∨(b∧c))⇒(c∧(a∨b)))
$$\left(a \Rightarrow c\right) ⇔ \left(\left(a \vee \left(b \wedge c\right)\right) \Rightarrow \left(c \wedge \left(a \vee b\right)\right)\right)$$
Solución detallada
$$a \Rightarrow c = c \vee \neg a$$
$$\left(a \vee \left(b \wedge c\right)\right) \Rightarrow \left(c \wedge \left(a \vee b\right)\right) = c \vee \neg a$$
$$\left(a \Rightarrow c\right) ⇔ \left(\left(a \vee \left(b \wedge c\right)\right) \Rightarrow \left(c \wedge \left(a \vee b\right)\right)\right) = 1$$
$$\left(a \vee \left(b \wedge c\right)\right) \Rightarrow \left(c \wedge \left(a \vee b\right)\right) = c \vee \neg a$$
$$\left(a \Rightarrow c\right) ⇔ \left(\left(a \vee \left(b \wedge c\right)\right) \Rightarrow \left(c \wedge \left(a \vee b\right)\right)\right) = 1$$
Tabla de verdad
+---+---+---+--------+ | a | b | c | result | +===+===+===+========+ | 0 | 0 | 0 | 1 | +---+---+---+--------+ | 0 | 0 | 1 | 1 | +---+---+---+--------+ | 0 | 1 | 0 | 1 | +---+---+---+--------+ | 0 | 1 | 1 | 1 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+