Sr Examen
Expresión ((P∨Q)∧(P→R)∧(Q→R))→R
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
((p⇒r)∧(q⇒r)∧(p∨q))⇒r
$$\left(\left(p \Rightarrow r\right) \wedge \left(q \Rightarrow r\right) \wedge \left(p \vee q\right)\right) \Rightarrow r$$
Solución detallada
$$p \Rightarrow r = r \vee \neg p$$
$$q \Rightarrow r = r \vee \neg q$$
$$\left(p \Rightarrow r\right) \wedge \left(q \Rightarrow r\right) \wedge \left(p \vee q\right) = r \wedge \left(p \vee q\right)$$
$$\left(\left(p \Rightarrow r\right) \wedge \left(q \Rightarrow r\right) \wedge \left(p \vee q\right)\right) \Rightarrow r = 1$$
$$q \Rightarrow r = r \vee \neg q$$
$$\left(p \Rightarrow r\right) \wedge \left(q \Rightarrow r\right) \wedge \left(p \vee q\right) = r \wedge \left(p \vee q\right)$$
$$\left(\left(p \Rightarrow r\right) \wedge \left(q \Rightarrow r\right) \wedge \left(p \vee q\right)\right) \Rightarrow r = 1$$
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 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+