Sr Examen
Expresión ¯((x↔y)→(x→¬z))∨(x∨¬y&z),
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
¬(x∨(z∧(¬y))∨((x⇔y)⇒(x⇒(¬z))))
$$\neg \left(x \vee \left(z \wedge \neg y\right) \vee \left(\left(x ⇔ y\right) \Rightarrow \left(x \Rightarrow \neg z\right)\right)\right)$$
Solución detallada
$$x ⇔ y = \left(x \wedge y\right) \vee \left(\neg x \wedge \neg y\right)$$
$$x \Rightarrow \neg z = \neg x \vee \neg z$$
$$\left(x ⇔ y\right) \Rightarrow \left(x \Rightarrow \neg z\right) = \neg x \vee \neg y \vee \neg z$$
$$x \vee \left(z \wedge \neg y\right) \vee \left(\left(x ⇔ y\right) \Rightarrow \left(x \Rightarrow \neg z\right)\right) = 1$$
$$\neg \left(x \vee \left(z \wedge \neg y\right) \vee \left(\left(x ⇔ y\right) \Rightarrow \left(x \Rightarrow \neg z\right)\right)\right) = \text{False}$$
$$x \Rightarrow \neg z = \neg x \vee \neg z$$
$$\left(x ⇔ y\right) \Rightarrow \left(x \Rightarrow \neg z\right) = \neg x \vee \neg y \vee \neg z$$
$$x \vee \left(z \wedge \neg y\right) \vee \left(\left(x ⇔ y\right) \Rightarrow \left(x \Rightarrow \neg z\right)\right) = 1$$
$$\neg \left(x \vee \left(z \wedge \neg y\right) \vee \left(\left(x ⇔ y\right) \Rightarrow \left(x \Rightarrow \neg z\right)\right)\right) = \text{False}$$
Tabla de verdad
+---+---+---+--------+ | x | y | z | result | +===+===+===+========+ | 0 | 0 | 0 | 0 | +---+---+---+--------+ | 0 | 0 | 1 | 0 | +---+---+---+--------+ | 0 | 1 | 0 | 0 | +---+---+---+--------+ | 0 | 1 | 1 | 0 | +---+---+---+--------+ | 1 | 0 | 0 | 0 | +---+---+---+--------+ | 1 | 0 | 1 | 0 | +---+---+---+--------+ | 1 | 1 | 0 | 0 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+