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Lógica matemática/ P∧(Q→R)

Expresión P∧(Q→R)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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