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Lógica matemática/ ((P∨Q)∧(P→R)∧(Q→R))→R

Expresión ((P∨Q)∧(P→R)∧(Q→R))→R

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    Solución

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