Sr Examen
Expresión ((a⇒b)&(avc)&(c⇒¬d)&d)⇒b
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Solución
Ha introducido
[src]
(d∧(a⇒b)∧(a∨c)∧(c⇒(¬d)))⇒b
$$\left(d \wedge \left(a \Rightarrow b\right) \wedge \left(c \Rightarrow \neg d\right) \wedge \left(a \vee c\right)\right) \Rightarrow b$$
Solución detallada
$$a \Rightarrow b = b \vee \neg a$$
$$c \Rightarrow \neg d = \neg c \vee \neg d$$
$$d \wedge \left(a \Rightarrow b\right) \wedge \left(c \Rightarrow \neg d\right) \wedge \left(a \vee c\right) = a \wedge b \wedge d \wedge \neg c$$
$$\left(d \wedge \left(a \Rightarrow b\right) \wedge \left(c \Rightarrow \neg d\right) \wedge \left(a \vee c\right)\right) \Rightarrow b = 1$$
$$c \Rightarrow \neg d = \neg c \vee \neg d$$
$$d \wedge \left(a \Rightarrow b\right) \wedge \left(c \Rightarrow \neg d\right) \wedge \left(a \vee c\right) = a \wedge b \wedge d \wedge \neg c$$
$$\left(d \wedge \left(a \Rightarrow b\right) \wedge \left(c \Rightarrow \neg d\right) \wedge \left(a \vee c\right)\right) \Rightarrow b = 1$$
Tabla de verdad
+---+---+---+---+--------+ | a | b | c | d | result | +===+===+===+===+========+ | 0 | 0 | 0 | 0 | 1 | +---+---+---+---+--------+ | 0 | 0 | 0 | 1 | 1 | +---+---+---+---+--------+ | 0 | 0 | 1 | 0 | 1 | +---+---+---+---+--------+ | 0 | 0 | 1 | 1 | 1 | +---+---+---+---+--------+ | 0 | 1 | 0 | 0 | 1 | +---+---+---+---+--------+ | 0 | 1 | 0 | 1 | 1 | +---+---+---+---+--------+ | 0 | 1 | 1 | 0 | 1 | +---+---+---+---+--------+ | 0 | 1 | 1 | 1 | 1 | +---+---+---+---+--------+ | 1 | 0 | 0 | 0 | 1 | +---+---+---+---+--------+ | 1 | 0 | 0 | 1 | 1 | +---+---+---+---+--------+ | 1 | 0 | 1 | 0 | 1 | +---+---+---+---+--------+ | 1 | 0 | 1 | 1 | 1 | +---+---+---+---+--------+ | 1 | 1 | 0 | 0 | 1 | +---+---+---+---+--------+ | 1 | 1 | 0 | 1 | 1 | +---+---+---+---+--------+ | 1 | 1 | 1 | 0 | 1 | +---+---+---+---+--------+ | 1 | 1 | 1 | 1 | 1 | +---+---+---+---+--------+