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