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

Expresión T→(((P→Q)→Q)∧(~P∧(~Q∨P)))

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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