Expresión ¬((a⇒b)∧¬c)⇒(a⊕¬c)
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Solución
Solución detallada
$$a \Rightarrow b = b \vee \neg a$$
$$\left(a \Rightarrow b\right) \wedge \neg c = \neg c \wedge \left(b \vee \neg a\right)$$
$$\neg \left(\left(a \Rightarrow b\right) \wedge \neg c\right) = c \vee \left(a \wedge \neg b\right)$$
$$a ⊕ \neg c = \left(a \wedge c\right) \vee \left(\neg a \wedge \neg c\right)$$
$$\neg \left(\left(a \Rightarrow b\right) \wedge \neg c\right) \Rightarrow \left(a ⊕ \neg c\right) = \left(a \wedge c\right) \vee \left(b \wedge \neg c\right) \vee \left(\neg a \wedge \neg c\right)$$
$$\left(a \wedge c\right) \vee \left(b \wedge \neg c\right) \vee \left(\neg a \wedge \neg c\right)$$
(a∧c)∨(b∧(¬c))∨((¬a)∧(¬c))
Tabla de verdad
+---+---+---+--------+
| a | b | c | result |
+===+===+===+========+
| 0 | 0 | 0 | 1 |
+---+---+---+--------+
| 0 | 0 | 1 | 0 |
+---+---+---+--------+
| 0 | 1 | 0 | 1 |
+---+---+---+--------+
| 0 | 1 | 1 | 0 |
+---+---+---+--------+
| 1 | 0 | 0 | 0 |
+---+---+---+--------+
| 1 | 0 | 1 | 1 |
+---+---+---+--------+
| 1 | 1 | 0 | 1 |
+---+---+---+--------+
| 1 | 1 | 1 | 1 |
+---+---+---+--------+
$$\left(a \vee \neg c\right) \wedge \left(b \vee c \vee \neg a\right)$$
$$\left(a \wedge c\right) \vee \left(b \wedge \neg c\right) \vee \left(\neg a \wedge \neg c\right)$$
(a∧c)∨(b∧(¬c))∨((¬a)∧(¬c))
$$\left(a \vee \neg c\right) \wedge \left(c \vee \neg c\right) \wedge \left(a \vee b \vee \neg a\right) \wedge \left(a \vee b \vee \neg c\right) \wedge \left(a \vee \neg a \vee \neg c\right) \wedge \left(b \vee c \vee \neg a\right) \wedge \left(b \vee c \vee \neg c\right) \wedge \left(c \vee \neg a \vee \neg c\right)$$
(a∨(¬c))∧(c∨(¬c))∧(a∨b∨(¬a))∧(a∨b∨(¬c))∧(b∨c∨(¬a))∧(b∨c∨(¬c))∧(a∨(¬a)∨(¬c))∧(c∨(¬a)∨(¬c))
Ya está reducido a FND
$$\left(a \wedge c\right) \vee \left(b \wedge \neg c\right) \vee \left(\neg a \wedge \neg c\right)$$
(a∧c)∨(b∧(¬c))∨((¬a)∧(¬c))