Sr Examen
Expresión ¬(((x↓y)→z)≡x)
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Solución detallada
$$x ↓ y = \neg x \wedge \neg y$$
$$\left(x ↓ y\right) \Rightarrow z = x \vee y \vee z$$
$$x ⇔ \left(\left(x ↓ y\right) \Rightarrow z\right) = x \vee \left(\neg y \wedge \neg z\right)$$
$$x \not\equiv \left(\left(x ↓ y\right) \Rightarrow z\right) = \neg x \wedge \left(y \vee z\right)$$
$$\left(x ↓ y\right) \Rightarrow z = x \vee y \vee z$$
$$x ⇔ \left(\left(x ↓ y\right) \Rightarrow z\right) = x \vee \left(\neg y \wedge \neg z\right)$$
$$x \not\equiv \left(\left(x ↓ y\right) \Rightarrow z\right) = \neg x \wedge \left(y \vee z\right)$$
Tabla de verdad
+---+---+---+--------+ | x | y | z | result | +===+===+===+========+ | 0 | 0 | 0 | 0 | +---+---+---+--------+ | 0 | 0 | 1 | 1 | +---+---+---+--------+ | 0 | 1 | 0 | 1 | +---+---+---+--------+ | 0 | 1 | 1 | 1 | +---+---+---+--------+ | 1 | 0 | 0 | 0 | +---+---+---+--------+ | 1 | 0 | 1 | 0 | +---+---+---+--------+ | 1 | 1 | 0 | 0 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+