Sr Examen
Expresión ((a+(-b))->ac)->(-(a->(-a))+b(-c))
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Solución
Ha introducido
[src]
((a∨(¬b))⇒(a∧c))⇒((b∧(¬c))∨(¬(a⇒(¬a))))
$$\left(\left(a \vee \neg b\right) \Rightarrow \left(a \wedge c\right)\right) \Rightarrow \left(\left(b \wedge \neg c\right) \vee a \not\Rightarrow \neg a\right)$$
Solución detallada
$$\left(a \vee \neg b\right) \Rightarrow \left(a \wedge c\right) = \left(a \wedge c\right) \vee \left(b \wedge \neg a\right)$$
$$a \Rightarrow \neg a = \neg a$$
$$a \not\Rightarrow \neg a = a$$
$$\left(b \wedge \neg c\right) \vee a \not\Rightarrow \neg a = a \vee \left(b \wedge \neg c\right)$$
$$\left(\left(a \vee \neg b\right) \Rightarrow \left(a \wedge c\right)\right) \Rightarrow \left(\left(b \wedge \neg c\right) \vee a \not\Rightarrow \neg a\right) = a \vee \neg b \vee \neg c$$
$$a \Rightarrow \neg a = \neg a$$
$$a \not\Rightarrow \neg a = a$$
$$\left(b \wedge \neg c\right) \vee a \not\Rightarrow \neg a = a \vee \left(b \wedge \neg c\right)$$
$$\left(\left(a \vee \neg b\right) \Rightarrow \left(a \wedge c\right)\right) \Rightarrow \left(\left(b \wedge \neg c\right) \vee a \not\Rightarrow \neg a\right) = a \vee \neg b \vee \neg c$$
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 | +---+---+---+--------+