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