Sr Examen
Lógica matemática/ ¬ab⇔a¬c∨b(a∨¬c⇒ab(¬a∨¬b∨ac))

Expresión ¬ab⇔a¬c∨b(a∨¬c⇒ab(¬a∨¬b∨ac))

El profesor se sorprenderá mucho al ver tu solución correcta😉

⌨

    Solución

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