Sr Examen
Lógica matemática/ ((A∧¬B)→D)∨¬(D∨E)

Expresión ((A∧¬B)→D)∨¬(D∨E)

El profesor se sorprenderá mucho al ver tu solución correcta😉

⌨

    Solución

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