Sr Examen
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∨r))⇔((p⇒q)∨(p⇒r))
$$\left(p \Rightarrow \left(q \vee r\right)\right) ⇔ \left(\left(p \Rightarrow q\right) \vee \left(p \Rightarrow r\right)\right)$$
Solución detallada
$$p \Rightarrow \left(q \vee r\right) = q \vee r \vee \neg p$$
$$p \Rightarrow q = q \vee \neg p$$
$$p \Rightarrow r = r \vee \neg p$$
$$\left(p \Rightarrow q\right) \vee \left(p \Rightarrow r\right) = q \vee r \vee \neg p$$
$$\left(p \Rightarrow \left(q \vee r\right)\right) ⇔ \left(\left(p \Rightarrow q\right) \vee \left(p \Rightarrow r\right)\right) = 1$$
$$p \Rightarrow q = q \vee \neg p$$
$$p \Rightarrow r = r \vee \neg p$$
$$\left(p \Rightarrow q\right) \vee \left(p \Rightarrow r\right) = q \vee r \vee \neg p$$
$$\left(p \Rightarrow \left(q \vee r\right)\right) ⇔ \left(\left(p \Rightarrow q\right) \vee \left(p \Rightarrow r\right)\right) = 1$$
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 | +---+---+---+--------+