Sr Examen
Expresión (((¬A∨B)⇒C)∨(¬D⇒((B∧C)∨A)))
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
((b∨(¬a))⇒c)∨((¬d)⇒(a∨(b∧c)))
$$\left(\neg d \Rightarrow \left(a \vee \left(b \wedge c\right)\right)\right) \vee \left(\left(b \vee \neg a\right) \Rightarrow c\right)$$
Solución detallada
$$\left(b \vee \neg a\right) \Rightarrow c = c \vee \left(a \wedge \neg b\right)$$
$$\neg d \Rightarrow \left(a \vee \left(b \wedge c\right)\right) = a \vee d \vee \left(b \wedge c\right)$$
$$\left(\neg d \Rightarrow \left(a \vee \left(b \wedge c\right)\right)\right) \vee \left(\left(b \vee \neg a\right) \Rightarrow c\right) = a \vee c \vee d$$
$$\neg d \Rightarrow \left(a \vee \left(b \wedge c\right)\right) = a \vee d \vee \left(b \wedge c\right)$$
$$\left(\neg d \Rightarrow \left(a \vee \left(b \wedge c\right)\right)\right) \vee \left(\left(b \vee \neg a\right) \Rightarrow c\right) = a \vee c \vee d$$
Tabla de verdad
+---+---+---+---+--------+ | a | b | c | d | result | +===+===+===+===+========+ | 0 | 0 | 0 | 0 | 0 | +---+---+---+---+--------+ | 0 | 0 | 0 | 1 | 1 | +---+---+---+---+--------+ | 0 | 0 | 1 | 0 | 1 | +---+---+---+---+--------+ | 0 | 0 | 1 | 1 | 1 | +---+---+---+---+--------+ | 0 | 1 | 0 | 0 | 0 | +---+---+---+---+--------+ | 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 | +---+---+---+---+--------+