Sr Examen
Expresión ¬(A∧B)∨(C∧B∧A→¬(A∧¬C⇔B∧C))∧¬A∧C∧B∨A
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
a∨(¬(a∧b))∨(b∧c∧(¬a)∧((a∧b∧c)⇒(¬((b∧c)⇔(a∧(¬c))))))
$$a \vee \left(b \wedge c \wedge \left(\left(a \wedge b \wedge c\right) \Rightarrow \left(a \wedge \neg c\right) \not\equiv \left(b \wedge c\right)\right) \wedge \neg a\right) \vee \neg \left(a \wedge b\right)$$
Solución detallada
$$\neg \left(a \wedge b\right) = \neg a \vee \neg b$$
$$\left(a \wedge \neg c\right) ⇔ \left(b \wedge c\right) = \left(c \wedge \neg b\right) \vee \left(\neg a \wedge \neg c\right)$$
$$\left(a \wedge \neg c\right) \not\equiv \left(b \wedge c\right) = \left(a \wedge \neg c\right) \vee \left(b \wedge c\right)$$
$$\left(a \wedge b \wedge c\right) \Rightarrow \left(a \wedge \neg c\right) \not\equiv \left(b \wedge c\right) = 1$$
$$b \wedge c \wedge \left(\left(a \wedge b \wedge c\right) \Rightarrow \left(a \wedge \neg c\right) \not\equiv \left(b \wedge c\right)\right) \wedge \neg a = b \wedge c \wedge \neg a$$
$$a \vee \left(b \wedge c \wedge \left(\left(a \wedge b \wedge c\right) \Rightarrow \left(a \wedge \neg c\right) \not\equiv \left(b \wedge c\right)\right) \wedge \neg a\right) \vee \neg \left(a \wedge b\right) = 1$$
$$\left(a \wedge \neg c\right) ⇔ \left(b \wedge c\right) = \left(c \wedge \neg b\right) \vee \left(\neg a \wedge \neg c\right)$$
$$\left(a \wedge \neg c\right) \not\equiv \left(b \wedge c\right) = \left(a \wedge \neg c\right) \vee \left(b \wedge c\right)$$
$$\left(a \wedge b \wedge c\right) \Rightarrow \left(a \wedge \neg c\right) \not\equiv \left(b \wedge c\right) = 1$$
$$b \wedge c \wedge \left(\left(a \wedge b \wedge c\right) \Rightarrow \left(a \wedge \neg c\right) \not\equiv \left(b \wedge c\right)\right) \wedge \neg a = b \wedge c \wedge \neg a$$
$$a \vee \left(b \wedge c \wedge \left(\left(a \wedge b \wedge c\right) \Rightarrow \left(a \wedge \neg c\right) \not\equiv \left(b \wedge c\right)\right) \wedge \neg a\right) \vee \neg \left(a \wedge b\right) = 1$$
Tabla de verdad
+---+---+---+--------+ | a | b | c | 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 | +---+---+---+--------+