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