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Lógica matemática/ ((¬B⇒A∨C))∨((¬A∨B)⇒C)

Expresión ((¬B⇒A∨C))∨((¬A∨B)⇒C)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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