Sr Examen
Expresión B¬A⇔AB∨¬(A⇒AB)
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Solución
Ha introducido
[src]
(b∧(¬a))⇔((a∧b)∨(¬(a⇒(a∧b))))
$$\left(b \wedge \neg a\right) ⇔ \left(\left(a \wedge b\right) \vee a \not\Rightarrow \left(a \wedge b\right)\right)$$
Solución detallada
$$a \Rightarrow \left(a \wedge b\right) = b \vee \neg a$$
$$a \not\Rightarrow \left(a \wedge b\right) = a \wedge \neg b$$
$$\left(a \wedge b\right) \vee a \not\Rightarrow \left(a \wedge b\right) = a$$
$$\left(b \wedge \neg a\right) ⇔ \left(\left(a \wedge b\right) \vee a \not\Rightarrow \left(a \wedge b\right)\right) = \neg a \wedge \neg b$$
$$a \not\Rightarrow \left(a \wedge b\right) = a \wedge \neg b$$
$$\left(a \wedge b\right) \vee a \not\Rightarrow \left(a \wedge b\right) = a$$
$$\left(b \wedge \neg a\right) ⇔ \left(\left(a \wedge b\right) \vee a \not\Rightarrow \left(a \wedge b\right)\right) = \neg a \wedge \neg b$$
Tabla de verdad
+---+---+--------+ | a | b | result | +===+===+========+ | 0 | 0 | 1 | +---+---+--------+ | 0 | 1 | 0 | +---+---+--------+ | 1 | 0 | 0 | +---+---+--------+ | 1 | 1 | 0 | +---+---+--------+