Sr Examen
Expresión ¬((A↓B)+¬(A→C))↔¬A+C
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Solución
Ha introducido
[src]
(c∨(¬a))⇔(¬((a↓b)∨(¬(a⇒c))))
$$\neg \left(\left(a ↓ b\right) \vee a \not\Rightarrow c\right) ⇔ \left(c \vee \neg a\right)$$
Solución detallada
$$a ↓ b = \neg a \wedge \neg b$$
$$a \Rightarrow c = c \vee \neg a$$
$$a \not\Rightarrow c = a \wedge \neg c$$
$$\left(a ↓ b\right) \vee a \not\Rightarrow c = \left(a \wedge \neg c\right) \vee \left(\neg a \wedge \neg b\right)$$
$$\neg \left(\left(a ↓ b\right) \vee a \not\Rightarrow c\right) = \left(a \wedge c\right) \vee \left(b \wedge \neg a\right)$$
$$\neg \left(\left(a ↓ b\right) \vee a \not\Rightarrow c\right) ⇔ \left(c \vee \neg a\right) = a \vee b$$
$$a \Rightarrow c = c \vee \neg a$$
$$a \not\Rightarrow c = a \wedge \neg c$$
$$\left(a ↓ b\right) \vee a \not\Rightarrow c = \left(a \wedge \neg c\right) \vee \left(\neg a \wedge \neg b\right)$$
$$\neg \left(\left(a ↓ b\right) \vee a \not\Rightarrow c\right) = \left(a \wedge c\right) \vee \left(b \wedge \neg a\right)$$
$$\neg \left(\left(a ↓ b\right) \vee a \not\Rightarrow c\right) ⇔ \left(c \vee \neg a\right) = a \vee b$$
Tabla de verdad
+---+---+---+--------+ | a | b | c | result | +===+===+===+========+ | 0 | 0 | 0 | 0 | +---+---+---+--------+ | 0 | 0 | 1 | 0 | +---+---+---+--------+ | 0 | 1 | 0 | 1 | +---+---+---+--------+ | 0 | 1 | 1 | 1 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+