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Lógica matemática/ BandnotC↔AorB

Expresión BandnotC↔AorB

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    Solución

    Ha introducido [src] 
    (a∨b)⇔(b∧(¬c))
    $$\left(b \wedge \neg c\right) ⇔ \left(a \vee b\right)$$
    Solución detallada
    $$\left(b \wedge \neg c\right) ⇔ \left(a \vee b\right) = \left(b \wedge \neg c\right) \vee \left(\neg a \wedge \neg b\right)$$
    Simplificación [src]
    $$\left(b \wedge \neg c\right) \vee \left(\neg a \wedge \neg b\right)$$ 
    (b∧(¬c))∨((¬a)∧(¬b))
    Tabla de verdad 
    +---+---+---+--------+
| a | b | c | result |
+===+===+===+========+
| 0 | 0 | 0 | 1      |
+---+---+---+--------+
| 0 | 0 | 1 | 1      |
+---+---+---+--------+
| 0 | 1 | 0 | 1      |
+---+---+---+--------+
| 0 | 1 | 1 | 0      |
+---+---+---+--------+
| 1 | 0 | 0 | 0      |
+---+---+---+--------+
| 1 | 0 | 1 | 0      |
+---+---+---+--------+
| 1 | 1 | 0 | 1      |
+---+---+---+--------+
| 1 | 1 | 1 | 0      |
+---+---+---+--------+
    FNCD [src]
    $$\left(b \vee \neg a\right) \wedge \left(\neg b \vee \neg c\right)$$ 
    (b∨(¬a))∧((¬b)∨(¬c))
    FNC [src]
    $$\left(b \vee \neg a\right) \wedge \left(b \vee \neg b\right) \wedge \left(\neg a \vee \neg c\right) \wedge \left(\neg b \vee \neg c\right)$$ 
    (b∨(¬a))∧(b∨(¬b))∧((¬a)∨(¬c))∧((¬b)∨(¬c))
    FNDP [src]
    $$\left(b \wedge \neg c\right) \vee \left(\neg a \wedge \neg b\right)$$ 
    (b∧(¬c))∨((¬a)∧(¬b))
    FND [src]
    Ya está reducido a FND
    $$\left(b \wedge \neg c\right) \vee \left(\neg a \wedge \neg b\right)$$ 
    (b∧(¬c))∨((¬a)∧(¬b))
    © Sr Examen