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