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Lógica matemática/ ((x->y)^(y->z))->(x->z)

Expresión ((x->y)^(y->z))->(x->z)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

    Ha introducido [src] 
    ((x⇒y)∧(y⇒z))⇒(x⇒z)
    ((x⇒y)∧(y⇒z))⇒(x⇒z)\left(\left(x \Rightarrow y\right) \wedge \left(y \Rightarrow z\right)\right) \Rightarrow \left(x \Rightarrow z\right)
    Solución detallada
    x⇒y=y∨¬xx \Rightarrow y = y \vee \neg x
    y⇒z=z∨¬yy \Rightarrow z = z \vee \neg y
    (x⇒y)∧(y⇒z)=(y∧z)∨(¬x∧¬y)\left(x \Rightarrow y\right) \wedge \left(y \Rightarrow z\right) = \left(y \wedge z\right) \vee \left(\neg x \wedge \neg y\right)
    x⇒z=z∨¬xx \Rightarrow z = z \vee \neg x
    ((x⇒y)∧(y⇒z))⇒(x⇒z)=1\left(\left(x \Rightarrow y\right) \wedge \left(y \Rightarrow z\right)\right) \Rightarrow \left(x \Rightarrow z\right) = 1
    Simplificación [src]
    1 
    1
    Tabla de verdad 
    +---+---+---+--------+
| x | y | z | 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      |
+---+---+---+--------+
    FNDP [src]
    1 
    1
    FNC [src]
    Ya está reducido a FNC
    1 
    1
    FNCD [src]
    1 
    1
    FND [src]
    Ya está reducido a FND
    1 
    1
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