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Lógica matemática/ pv(q^~p)и~p→(~qvp)

Expresión pv(q^~p)и~p→(~qvp)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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