Expresión (∼(p→q)∨~(q→p))∧(∼(q→r)∨~(r→q))
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Solución
Solución detallada
$$p \Rightarrow q = q \vee \neg p$$
$$q \Rightarrow p = p \vee \neg q$$
$$q \not\Rightarrow p = q \wedge \neg p$$
$$\left(p \Rightarrow q\right) \vee q \not\Rightarrow p = q \vee \neg p$$
$$\neg \left(\left(p \Rightarrow q\right) \vee q \not\Rightarrow p\right) = p \wedge \neg q$$
$$q \Rightarrow r = r \vee \neg q$$
$$r \Rightarrow q = q \vee \neg r$$
$$r \not\Rightarrow q = r \wedge \neg q$$
$$\left(q \Rightarrow r\right) \vee r \not\Rightarrow q = r \vee \neg q$$
$$\neg \left(\left(q \Rightarrow r\right) \vee r \not\Rightarrow q\right) = q \wedge \neg r$$
$$\neg \left(\left(p \Rightarrow q\right) \vee q \not\Rightarrow p\right) \wedge \neg \left(\left(q \Rightarrow r\right) \vee r \not\Rightarrow q\right) = \text{False}$$
Tabla de verdad
+---+---+---+--------+
| p | q | r | result |
+===+===+===+========+
| 0 | 0 | 0 | 0 |
+---+---+---+--------+
| 0 | 0 | 1 | 0 |
+---+---+---+--------+
| 0 | 1 | 0 | 0 |
+---+---+---+--------+
| 0 | 1 | 1 | 0 |
+---+---+---+--------+
| 1 | 0 | 0 | 0 |
+---+---+---+--------+
| 1 | 0 | 1 | 0 |
+---+---+---+--------+
| 1 | 1 | 0 | 0 |
+---+---+---+--------+
| 1 | 1 | 1 | 0 |
+---+---+---+--------+