Sr Examen
Expresión x∨(y⇒z)≡(x∨y)⇒(x∨z)
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
(x∨(y⇒z))⇔((x∨y)⇒(x∨z))
$$\left(\left(x \vee y\right) \Rightarrow \left(x \vee z\right)\right) ⇔ \left(x \vee \left(y \Rightarrow z\right)\right)$$
Solución detallada
$$y \Rightarrow z = z \vee \neg y$$
$$x \vee \left(y \Rightarrow z\right) = x \vee z \vee \neg y$$
$$\left(x \vee y\right) \Rightarrow \left(x \vee z\right) = x \vee z \vee \neg y$$
$$\left(\left(x \vee y\right) \Rightarrow \left(x \vee z\right)\right) ⇔ \left(x \vee \left(y \Rightarrow z\right)\right) = 1$$
$$x \vee \left(y \Rightarrow z\right) = x \vee z \vee \neg y$$
$$\left(x \vee y\right) \Rightarrow \left(x \vee z\right) = x \vee z \vee \neg y$$
$$\left(\left(x \vee y\right) \Rightarrow \left(x \vee z\right)\right) ⇔ \left(x \vee \left(y \Rightarrow z\right)\right) = 1$$
Tabla de verdad
+---+---+---+--------+ | x | y | z | result | +===+===+===+========+ | 0 | 0 | 0 | 1 | +---+---+---+--------+ | 0 | 0 | 1 | 1 | +---+---+---+--------+ | 0 | 1 | 0 | 1 | +---+---+---+--------+ | 0 | 1 | 1 | 1 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+