Sr Examen
Lógica matemática/ (p&¬b&¬a)v(¬a&b&¬p)v(a&¬b&¬p)v(a&b&p)

Expresión (p&¬b&¬a)v(¬a&b&¬p)v(a&¬b&¬p)v(a&b&p)

El profesor se sorprenderá mucho al ver tu solución correcta😉

⌨

    Solución

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