Sr Examen
Expresión T→(((P→Q)→Q)∧(~P∧(~Q∨P)))
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Solución
Ha introducido
[src]
t⇒((¬p)∧(p∨(¬q))∧((p⇒q)⇒q))
$$t \Rightarrow \left(\left(\left(p \Rightarrow q\right) \Rightarrow q\right) \wedge \neg p \wedge \left(p \vee \neg q\right)\right)$$
Solución detallada
$$p \Rightarrow q = q \vee \neg p$$
$$\left(p \Rightarrow q\right) \Rightarrow q = p \vee q$$
$$\left(\left(p \Rightarrow q\right) \Rightarrow q\right) \wedge \neg p \wedge \left(p \vee \neg q\right) = \text{False}$$
$$t \Rightarrow \left(\left(\left(p \Rightarrow q\right) \Rightarrow q\right) \wedge \neg p \wedge \left(p \vee \neg q\right)\right) = \neg t$$
$$\left(p \Rightarrow q\right) \Rightarrow q = p \vee q$$
$$\left(\left(p \Rightarrow q\right) \Rightarrow q\right) \wedge \neg p \wedge \left(p \vee \neg q\right) = \text{False}$$
$$t \Rightarrow \left(\left(\left(p \Rightarrow q\right) \Rightarrow q\right) \wedge \neg p \wedge \left(p \vee \neg q\right)\right) = \neg t$$
Tabla de verdad
+---+---+---+--------+ | p | q | t | result | +===+===+===+========+ | 0 | 0 | 0 | 1 | +---+---+---+--------+ | 0 | 0 | 1 | 0 | +---+---+---+--------+ | 0 | 1 | 0 | 1 | +---+---+---+--------+ | 0 | 1 | 1 | 0 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 0 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+