Sr Examen
Expresión xy+(xyz∨(¬(¬xyz)→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)∨((¬(y∧z∧(¬x)))⇒(x∧(¬y)∧(¬z)))
$$\left(x \wedge y\right) \vee \left(x \wedge y \wedge z\right) \vee \left(\neg \left(y \wedge z \wedge \neg x\right) \Rightarrow \left(x \wedge \neg y \wedge \neg z\right)\right)$$
Solución detallada
$$\neg \left(y \wedge z \wedge \neg x\right) = x \vee \neg y \vee \neg z$$
$$\neg \left(y \wedge z \wedge \neg x\right) \Rightarrow \left(x \wedge \neg y \wedge \neg z\right) = \left(x \vee y\right) \wedge \left(x \vee z\right) \wedge \left(y \vee \neg z\right) \wedge \left(z \vee \neg y\right) \wedge \left(\neg x \vee \neg y\right) \wedge \left(\neg x \vee \neg z\right)$$
$$\left(x \wedge y\right) \vee \left(x \wedge y \wedge z\right) \vee \left(\neg \left(y \wedge z \wedge \neg x\right) \Rightarrow \left(x \wedge \neg y \wedge \neg z\right)\right) = \left(x \wedge \neg z\right) \vee \left(y \wedge z\right)$$
$$\neg \left(y \wedge z \wedge \neg x\right) \Rightarrow \left(x \wedge \neg y \wedge \neg z\right) = \left(x \vee y\right) \wedge \left(x \vee z\right) \wedge \left(y \vee \neg z\right) \wedge \left(z \vee \neg y\right) \wedge \left(\neg x \vee \neg y\right) \wedge \left(\neg x \vee \neg z\right)$$
$$\left(x \wedge y\right) \vee \left(x \wedge y \wedge z\right) \vee \left(\neg \left(y \wedge z \wedge \neg x\right) \Rightarrow \left(x \wedge \neg y \wedge \neg z\right)\right) = \left(x \wedge \neg z\right) \vee \left(y \wedge z\right)$$
Tabla de verdad
+---+---+---+--------+ | x | y | z | result | +===+===+===+========+ | 0 | 0 | 0 | 0 | +---+---+---+--------+ | 0 | 0 | 1 | 0 | +---+---+---+--------+ | 0 | 1 | 0 | 0 | +---+---+---+--------+ | 0 | 1 | 1 | 1 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 0 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+
FNC
[src]
$$\left(x \vee y\right) \wedge \left(x \vee z\right) \wedge \left(y \vee \neg z\right) \wedge \left(z \vee \neg z\right)$$
(x∨y)∧(x∨z)∧(y∨(¬z))∧(z∨(¬z))
FND
[src]
Ya está reducido a FND
$$\left(x \wedge \neg z\right) \vee \left(y \wedge z\right)$$
(y∧z)∨(x∧(¬z))