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