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Expresión а⇒b=¬a+b&¬(a⇒b)=a⊕¬b
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Solución
Ha introducido
[src]
(¬b)⊕(a⇔(a⇒b)⇔((¬a)∨(b∧(¬(a⇒b)))))
$$\neg b ⊕ \left(a ⇔ \left(a \Rightarrow b\right) ⇔ \left(\left(b \wedge a \not\Rightarrow b\right) \vee \neg a\right)\right)$$
Solución detallada
$$a \Rightarrow b = b \vee \neg a$$
$$a \not\Rightarrow b = a \wedge \neg b$$
$$b \wedge a \not\Rightarrow b = \text{False}$$
$$\left(b \wedge a \not\Rightarrow b\right) \vee \neg a = \neg a$$
$$a ⇔ \left(a \Rightarrow b\right) ⇔ \left(\left(b \wedge a \not\Rightarrow b\right) \vee \neg a\right) = \text{False}$$
$$\neg b ⊕ \left(a ⇔ \left(a \Rightarrow b\right) ⇔ \left(\left(b \wedge a \not\Rightarrow b\right) \vee \neg a\right)\right) = \neg b$$
$$a \not\Rightarrow b = a \wedge \neg b$$
$$b \wedge a \not\Rightarrow b = \text{False}$$
$$\left(b \wedge a \not\Rightarrow b\right) \vee \neg a = \neg a$$
$$a ⇔ \left(a \Rightarrow b\right) ⇔ \left(\left(b \wedge a \not\Rightarrow b\right) \vee \neg a\right) = \text{False}$$
$$\neg b ⊕ \left(a ⇔ \left(a \Rightarrow b\right) ⇔ \left(\left(b \wedge a \not\Rightarrow b\right) \vee \neg a\right)\right) = \neg b$$
Tabla de verdad
+---+---+--------+ | a | b | result | +===+===+========+ | 0 | 0 | 1 | +---+---+--------+ | 0 | 1 | 0 | +---+---+--------+ | 1 | 0 | 1 | +---+---+--------+ | 1 | 1 | 0 | +---+---+--------+