====Lösung der Aufgabe 6====

a) 

^$X$ ^$Y$ ^$X \rightarrow Y$ ^$\overline{X}$ ^$\overline{X} \rightarrow Y$  ^$\left( X \rightarrow Y \right) \land \left( \overline{X} \rightarrow Y \right)$ ^$\left( X \rightarrow Y \right) \land \left( \overline{X} \rightarrow Y \right)\rightarrow X$ |
|1 |1 |1 |0 |1 |1 |1 |
|1 |0 |0 |0 |1 |0 |1 |
|0 |1 |1 |1 |1 |1 |0 |
|0 |0 |1 |1 |0 |0 |1 |

b) 

^$X$ ^$Y$ ^$X \rightarrow Y$ ^$\left( X \rightarrow Y \right) \rightarrow X$  ^$\left( \left( X \rightarrow Y \right) \rightarrow X \right) \rightarrow X$ ^$\overline{\left( \left( X \rightarrow Y \right) \rightarrow X \right) \rightarrow X}$ |
|1 |1 |1 |1 |1 |0 |
|1 |0 |0 |1 |1 |0 |
|0 |1 |1 |0 |1 |0 |
|0 |0 |1 |0 |1 |0 |

c)

^$X$ ^$Y$ ^$\overline{X}$ ^$\overline{Y}$ ^$\overline{X} \land \overline{Y}$ ^$\overline{ \overline{X} \land \overline{Y}}$ ^$X \lor Y$  ^$\left( \overline{ \overline{X} \land \overline{Y}} \right) \leftrightarrow \left( X \lor Y \right)$ |
|1 |1 |0 |0 |0 |1 |1 |1 |
|1 |0 |0 |1 |0 |1 |1 |1 |
|0 |1 |1 |0 |0 |1 |1 |1 |
|0 |0 |1 |1 |1 |0 |0 |1 |

d)

^$X$ ^$Y$ ^$Z$ ^$\overline{X}$ ^$\overline{Y}$ ^$\overline{X} \land  \overline{Y}$ ^$Z \rightarrow Y$  ^$\left( \overline{X} \land  \overline{Y} \right) \rightarrow \left( Z \rightarrow Y \right)$ |
|1 |1 |1 |0 |0 |0 |1 |1 |
|1 |1 |0 |0 |0 |0 |1 |1 |
|1 |0 |1 |0 |1 |0 |0 |1 |
|1 |0 |0 |0 |1 |0 |1 |1 |
|0 |1 |1 |1 |0 |0 |1 |1 |
|0 |1 |0 |1 |0 |0 |1 |1 |
|0 |0 |1 |1 |1 |1 |0 |0 |
|0 |0 |0 |1 |1 |1 |1 |1 |

e)

^$X$ ^$Y$ ^$X \rightarrow Y$ ^$Y \rightarrow X$ ^$X \leftrightarrow Y$ ^$\left( Y \rightarrow X \right) \rightarrow \left( X \leftrightarrow Y \right)$ ^$\left( X \rightarrow Y \right) \rightarrow \left( \left( Y \rightarrow X \right) \rightarrow \left( X \leftrightarrow Y \right) \right)$ |
|1 |1 |1 |1 |1 |1 |1 |
|1 |0 |0 |1 |0 |0 |1 |
|0 |1 |1 |0 |0 |1 |1 |
|0 |0 |1 |1 |1 |1 |1 |