Sr Examen
Expresión (x->y)<->¬(x->(y->z))
El profesor se sorprenderá mucho al ver tu solución correcta😉
⌨
Solución
Ha introducido
[src]
(x⇒y)⇔(¬(x⇒(y⇒z)))
$$\left(x \Rightarrow y\right) ⇔ x \not\Rightarrow \left(y \Rightarrow z\right)$$
Solución detallada
$$x \Rightarrow y = y \vee \neg x$$
$$y \Rightarrow z = z \vee \neg y$$
$$x \Rightarrow \left(y \Rightarrow z\right) = z \vee \neg x \vee \neg y$$
$$x \not\Rightarrow \left(y \Rightarrow z\right) = x \wedge y \wedge \neg z$$
$$\left(x \Rightarrow y\right) ⇔ x \not\Rightarrow \left(y \Rightarrow z\right) = x \wedge \left(\neg y \vee \neg z\right)$$
$$y \Rightarrow z = z \vee \neg y$$
$$x \Rightarrow \left(y \Rightarrow z\right) = z \vee \neg x \vee \neg y$$
$$x \not\Rightarrow \left(y \Rightarrow z\right) = x \wedge y \wedge \neg z$$
$$\left(x \Rightarrow y\right) ⇔ x \not\Rightarrow \left(y \Rightarrow z\right) = x \wedge \left(\neg y \vee \neg z\right)$$
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 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+