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Lógica matemática/ ¬(A⇔B)∧C⇔B

Expresión ¬(A⇔B)∧C⇔B

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    Solución

    Ha introducido [src] 
    b⇔(c∧(¬(a⇔b)))
    $$b ⇔ \left(c \wedge a \not\equiv b\right)$$
    Solución detallada
    $$a ⇔ b = \left(a \wedge b\right) \vee \left(\neg a \wedge \neg b\right)$$
    $$a \not\equiv b = \left(a \wedge \neg b\right) \vee \left(b \wedge \neg a\right)$$
    $$c \wedge a \not\equiv b = c \wedge \left(a \vee b\right) \wedge \left(\neg a \vee \neg b\right)$$
    $$b ⇔ \left(c \wedge a \not\equiv b\right) = \left(c \wedge \neg a\right) \vee \left(\neg b \wedge \neg c\right)$$
    Simplificación [src]
    $$\left(c \wedge \neg a\right) \vee \left(\neg b \wedge \neg c\right)$$ 
    (c∧(¬a))∨((¬b)∧(¬c))
    Tabla de verdad 
    +---+---+---+--------+
| a | b | c | result |
+===+===+===+========+
| 0 | 0 | 0 | 1      |
+---+---+---+--------+
| 0 | 0 | 1 | 1      |
+---+---+---+--------+
| 0 | 1 | 0 | 0      |
+---+---+---+--------+
| 0 | 1 | 1 | 1      |
+---+---+---+--------+
| 1 | 0 | 0 | 1      |
+---+---+---+--------+
| 1 | 0 | 1 | 0      |
+---+---+---+--------+
| 1 | 1 | 0 | 0      |
+---+---+---+--------+
| 1 | 1 | 1 | 0      |
+---+---+---+--------+
    FNC [src]
    $$\left(c \vee \neg b\right) \wedge \left(c \vee \neg c\right) \wedge \left(\neg a \vee \neg b\right) \wedge \left(\neg a \vee \neg c\right)$$ 
    (c∨(¬b))∧(c∨(¬c))∧((¬a)∨(¬b))∧((¬a)∨(¬c))
    FNDP [src]
    $$\left(c \wedge \neg a\right) \vee \left(\neg b \wedge \neg c\right)$$ 
    (c∧(¬a))∨((¬b)∧(¬c))
    FND [src]
    Ya está reducido a FND
    $$\left(c \wedge \neg a\right) \vee \left(\neg b \wedge \neg c\right)$$ 
    (c∧(¬a))∨((¬b)∧(¬c))
    FNCD [src]
    $$\left(c \vee \neg b\right) \wedge \left(\neg a \vee \neg c\right)$$ 
    (c∨(¬b))∧((¬a)∨(¬c))
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