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Lógica matemática/ Р˄(Q↔R)≡(Р˄Q)↔(Р˄R)

Expresión Р˄(Q↔R)≡(Р˄Q)↔(Р˄R)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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