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Lógica matemática/ ¬(b⇒a)∧(c⇒d)

Expresión ¬(b⇒a)∧(c⇒d)

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    Solución

    Ha introducido [src] 
    (c⇒d)∧(¬(b⇒a))
    (c⇒d)∧b⇏a\left(c \Rightarrow d\right) \wedge b \not\Rightarrow a
    Solución detallada
    c⇒d=d∨¬cc \Rightarrow d = d \vee \neg c
    b⇒a=a∨¬bb \Rightarrow a = a \vee \neg b
    b⇏a=b∧¬ab \not\Rightarrow a = b \wedge \neg a
    (c⇒d)∧b⇏a=b∧¬a∧(d∨¬c)\left(c \Rightarrow d\right) \wedge b \not\Rightarrow a = b \wedge \neg a \wedge \left(d \vee \neg c\right)
    Simplificación [src]
    b∧¬a∧(d∨¬c)b \wedge \neg a \wedge \left(d \vee \neg c\right) 
    b∧(¬a)∧(d∨(¬c))
    Tabla de verdad 
    +---+---+---+---+--------+
| a | b | c | d | result |
+===+===+===+===+========+
| 0 | 0 | 0 | 0 | 0      |
+---+---+---+---+--------+
| 0 | 0 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 0 | 0 | 1 | 0 | 0      |
+---+---+---+---+--------+
| 0 | 0 | 1 | 1 | 0      |
+---+---+---+---+--------+
| 0 | 1 | 0 | 0 | 1      |
+---+---+---+---+--------+
| 0 | 1 | 0 | 1 | 1      |
+---+---+---+---+--------+
| 0 | 1 | 1 | 0 | 0      |
+---+---+---+---+--------+
| 0 | 1 | 1 | 1 | 1      |
+---+---+---+---+--------+
| 1 | 0 | 0 | 0 | 0      |
+---+---+---+---+--------+
| 1 | 0 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 1 | 0 | 1 | 0 | 0      |
+---+---+---+---+--------+
| 1 | 0 | 1 | 1 | 0      |
+---+---+---+---+--------+
| 1 | 1 | 0 | 0 | 0      |
+---+---+---+---+--------+
| 1 | 1 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 1 | 1 | 1 | 0 | 0      |
+---+---+---+---+--------+
| 1 | 1 | 1 | 1 | 0      |
+---+---+---+---+--------+
    FNCD [src]
    b∧¬a∧(d∨¬c)b \wedge \neg a \wedge \left(d \vee \neg c\right) 
    b∧(¬a)∧(d∨(¬c))
    FND [src]
    (b∧d∧¬a)∨(b∧¬a∧¬c)\left(b \wedge d \wedge \neg a\right) \vee \left(b \wedge \neg a \wedge \neg c\right) 
    (b∧d∧(¬a))∨(b∧(¬a)∧(¬c))
    FNDP [src]
    (b∧d∧¬a)∨(b∧¬a∧¬c)\left(b \wedge d \wedge \neg a\right) \vee \left(b \wedge \neg a \wedge \neg c\right) 
    (b∧d∧(¬a))∨(b∧(¬a)∧(¬c))
    FNC [src]
    Ya está reducido a FNC
    b∧¬a∧(d∨¬c)b \wedge \neg a \wedge \left(d \vee \neg c\right) 
    b∧(¬a)∧(d∨(¬c))
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