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Expresión s⇒(((¬p)∧q)∧(r))

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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