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author | Gregor Kleen <gkleen@yggdrasil.li> | 2016-05-15 21:41:48 +0200 |
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committer | Gregor Kleen <gkleen@yggdrasil.li> | 2016-05-15 21:41:48 +0200 |
commit | 0195d2cc23edcbaff79c4fee936276954a042d41 (patch) | |
tree | 706032546ce6026eb64f010dd0dd061c9b1d01fd /ss2016/lds/03/H3-3.md | |
parent | dc76e09714cb0392ec71d4fee18fdc1e01cab338 (diff) | |
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lds 03
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diff --git a/ss2016/lds/03/H3-3.md b/ss2016/lds/03/H3-3.md new file mode 100644 index 0000000..69368cc --- /dev/null +++ b/ss2016/lds/03/H3-3.md | |||
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1 | Es ist $T \circ S = \{ (a, c) \in A \times C \,|\, \exists b \in B \ldotp (a, b) \in S \land (b, c) \in T \}$ | ||
2 | |||
3 | Es sei $a \in A$, $c_1 \in C$ und $c_2 \in C$. | ||
4 | Es gelte zudem $(a, c_1) \in T \circ S \land (a, c_2) \in T \circ S$. | ||
5 | |||
6 | Nach der Definition von $T \circ S$ existieren daher $b_1 \in B$ und $b_2 \in B$ sodass $(a, b_1) \in S$, $(a, b_2) \in S$, $(b_1, c_1) \in T$ und $(b_2, c_2) \in T$. | ||
7 | Da $S$ rechts-eindeutig ist gilt $b_1 = b_2$ und daher, wegen der rechts-Eindeutigkeit von $T$, auch $c_1 = c_2$. | ||
8 | |||
9 | Wegen der freien Wahl von $a$, $c_1$ und $c_2$ ist $T \circ S$ rechts-eindeutig. | ||