Sr Examen
Lógica matemática/ avbv(c&(avb'vc')')

Expresión avbv(c&(avb'vc')')

El profesor se sorprenderá mucho al ver tu solución correcta😉

⌨

    Solución

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