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

Expresión ((P→P)∨Q)∧(~Q∨(R∧Q))∧(P→(P∨~Q))

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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