Expresión (x⇔¬yz)⇒(x⇔¬yt)
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
$$x ⇔ \left(z \wedge \neg y\right) = \left(y \wedge \neg x\right) \vee \left(\neg x \wedge \neg z\right) \vee \left(x \wedge z \wedge \neg y\right)$$
$$x ⇔ \left(t \wedge \neg y\right) = \left(y \wedge \neg x\right) \vee \left(\neg t \wedge \neg x\right) \vee \left(t \wedge x \wedge \neg y\right)$$
$$\left(x ⇔ \left(z \wedge \neg y\right)\right) \Rightarrow \left(x ⇔ \left(t \wedge \neg y\right)\right) = y \vee \left(t \wedge x\right) \vee \left(t \wedge z\right) \vee \left(x \wedge \neg z\right) \vee \left(z \wedge \neg x\right) \vee \left(\neg t \wedge \neg x\right) \vee \left(\neg t \wedge \neg z\right)$$
$$y \vee \left(t \wedge x\right) \vee \left(t \wedge z\right) \vee \left(x \wedge \neg z\right) \vee \left(z \wedge \neg x\right) \vee \left(\neg t \wedge \neg x\right) \vee \left(\neg t \wedge \neg z\right)$$
y∨(t∧x)∨(t∧z)∨(x∧(¬z))∨(z∧(¬x))∨((¬t)∧(¬x))∨((¬t)∧(¬z))
Tabla de verdad
+---+---+---+---+--------+
| t | x | y | z | result |
+===+===+===+===+========+
| 0 | 0 | 0 | 0 | 1 |
+---+---+---+---+--------+
| 0 | 0 | 0 | 1 | 1 |
+---+---+---+---+--------+
| 0 | 0 | 1 | 0 | 1 |
+---+---+---+---+--------+
| 0 | 0 | 1 | 1 | 1 |
+---+---+---+---+--------+
| 0 | 1 | 0 | 0 | 1 |
+---+---+---+---+--------+
| 0 | 1 | 0 | 1 | 0 |
+---+---+---+---+--------+
| 0 | 1 | 1 | 0 | 1 |
+---+---+---+---+--------+
| 0 | 1 | 1 | 1 | 1 |
+---+---+---+---+--------+
| 1 | 0 | 0 | 0 | 0 |
+---+---+---+---+--------+
| 1 | 0 | 0 | 1 | 1 |
+---+---+---+---+--------+
| 1 | 0 | 1 | 0 | 1 |
+---+---+---+---+--------+
| 1 | 0 | 1 | 1 | 1 |
+---+---+---+---+--------+
| 1 | 1 | 0 | 0 | 1 |
+---+---+---+---+--------+
| 1 | 1 | 0 | 1 | 1 |
+---+---+---+---+--------+
| 1 | 1 | 1 | 0 | 1 |
+---+---+---+---+--------+
| 1 | 1 | 1 | 1 | 1 |
+---+---+---+---+--------+
$$\left(t \vee y \vee \neg x \vee \neg z\right) \wedge \left(x \vee y \vee z \vee \neg t\right) \wedge \left(t \vee x \vee y \vee z \vee \neg t\right) \wedge \left(t \vee x \vee y \vee \neg t \vee \neg x\right) \wedge \left(t \vee x \vee y \vee \neg x \vee \neg z\right) \wedge \left(t \vee y \vee z \vee \neg t \vee \neg z\right) \wedge \left(t \vee y \vee z \vee \neg x \vee \neg z\right) \wedge \left(t \vee y \vee \neg t \vee \neg x \vee \neg z\right) \wedge \left(x \vee y \vee z \vee \neg t \vee \neg x\right) \wedge \left(x \vee y \vee z \vee \neg t \vee \neg z\right) \wedge \left(x \vee y \vee z \vee \neg x \vee \neg z\right) \wedge \left(t \vee x \vee y \vee z \vee \neg t \vee \neg x\right) \wedge \left(t \vee x \vee y \vee z \vee \neg t \vee \neg z\right) \wedge \left(t \vee x \vee y \vee z \vee \neg x \vee \neg z\right) \wedge \left(t \vee x \vee y \vee \neg t \vee \neg x \vee \neg z\right) \wedge \left(t \vee y \vee z \vee \neg t \vee \neg x \vee \neg z\right) \wedge \left(x \vee y \vee z \vee \neg t \vee \neg x \vee \neg z\right) \wedge \left(t \vee x \vee y \vee z \vee \neg t \vee \neg x \vee \neg z\right)$$
(x∨y∨z∨(¬t))∧(t∨y∨(¬x)∨(¬z))∧(t∨x∨y∨z∨(¬t))∧(t∨x∨y∨(¬t)∨(¬x))∧(t∨x∨y∨(¬x)∨(¬z))∧(t∨y∨z∨(¬t)∨(¬z))∧(t∨y∨z∨(¬x)∨(¬z))∧(x∨y∨z∨(¬t)∨(¬x))∧(x∨y∨z∨(¬t)∨(¬z))∧(x∨y∨z∨(¬x)∨(¬z))∧(t∨y∨(¬t)∨(¬x)∨(¬z))∧(t∨x∨y∨z∨(¬t)∨(¬x))∧(t∨x∨y∨z∨(¬t)∨(¬z))∧(t∨x∨y∨z∨(¬x)∨(¬z))∧(t∨x∨y∨(¬t)∨(¬x)∨(¬z))∧(t∨y∨z∨(¬t)∨(¬x)∨(¬z))∧(x∨y∨z∨(¬t)∨(¬x)∨(¬z))∧(t∨x∨y∨z∨(¬t)∨(¬x)∨(¬z))
$$\left(t \vee y \vee \neg x \vee \neg z\right) \wedge \left(x \vee y \vee z \vee \neg t\right)$$
(x∨y∨z∨(¬t))∧(t∨y∨(¬x)∨(¬z))
Ya está reducido a FND
$$y \vee \left(t \wedge x\right) \vee \left(t \wedge z\right) \vee \left(x \wedge \neg z\right) \vee \left(z \wedge \neg x\right) \vee \left(\neg t \wedge \neg x\right) \vee \left(\neg t \wedge \neg z\right)$$
y∨(t∧x)∨(t∧z)∨(x∧(¬z))∨(z∧(¬x))∨((¬t)∧(¬x))∨((¬t)∧(¬z))
$$y \vee \left(t \wedge x\right) \vee \left(t \wedge z\right) \vee \left(x \wedge \neg z\right) \vee \left(z \wedge \neg x\right) \vee \left(\neg t \wedge \neg x\right) \vee \left(\neg t \wedge \neg z\right)$$
y∨(t∧x)∨(t∧z)∨(x∧(¬z))∨(z∧(¬x))∨((¬t)∧(¬x))∨((¬t)∧(¬z))