Sr Examen
Expresión А^В→(А+(В≡С))
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Solución
Ha introducido
[src]
(a∧b)⇒(a∨(b⇔c))
$$\left(a \wedge b\right) \Rightarrow \left(a \vee \left(b ⇔ c\right)\right)$$
Solución detallada
$$b ⇔ c = \left(b \wedge c\right) \vee \left(\neg b \wedge \neg c\right)$$
$$a \vee \left(b ⇔ c\right) = a \vee \left(b \wedge c\right) \vee \left(\neg b \wedge \neg c\right)$$
$$\left(a \wedge b\right) \Rightarrow \left(a \vee \left(b ⇔ c\right)\right) = 1$$
$$a \vee \left(b ⇔ c\right) = a \vee \left(b \wedge c\right) \vee \left(\neg b \wedge \neg c\right)$$
$$\left(a \wedge b\right) \Rightarrow \left(a \vee \left(b ⇔ c\right)\right) = 1$$
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 | +---+---+---+--------+