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sav07_homework_4_solution [2007/06/16 21:58] vkuncak |
sav07_homework_4_solution [2007/06/16 22:12] (current) vkuncak |
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$X \rightarrow Y$: Let $s.c_2 \subseteq G$. Let $Q = s.c_2$. Then $s.c_2 \subseteq Q$, so also $s.c_1 \subseteq Q$. Thus $s.c_1 \subseteq s.c_2$. | $X \rightarrow Y$: Let $s.c_2 \subseteq G$. Let $Q = s.c_2$. Then $s.c_2 \subseteq Q$, so also $s.c_1 \subseteq Q$. Thus $s.c_1 \subseteq s.c_2$. | ||
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==== Part 5 ==== | ==== Part 5 ==== | ||
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+ | **Lemma**: $skip;c \equiv c;skip \equiv c$. | ||
$c_1 \sqsubseteq c_2\ \rightarrow\ c_1;c_3 \sqsubseteq c_2;c_3$: | $c_1 \sqsubseteq c_2\ \rightarrow\ c_1;c_3 \sqsubseteq c_2;c_3$: |