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