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Lógica matemática/ (c∨d)→(b→(¬(¬(c∨d)∨¬b)))

Expresión (c∨d)→(b→(¬(¬(c∨d)∨¬b)))

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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