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sav07_homework_4_solution [2007/06/16 21:52] vkuncak |
sav07_homework_4_solution [2007/06/16 22:12] vkuncak |
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| This is a sketch of the solution of [[SAV07 Homework 4]] | This is a sketch of the solution of [[SAV07 Homework 4]] | ||
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| Note that we have $R(assume(false)) \circ R(assert(false)) = R(assume(false))$. | Note that we have $R(assume(false)) \circ R(assert(false)) = R(assume(false))$. | ||
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| + | Basically, $R(assert(false))$ and $R(assume(false))$ both act as a left zeros of $\circ$ for the relations of the form $R(c)$. | ||
| ==== Part 2 ==== | ==== Part 2 ==== | ||
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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**: $skip;c \equiv c;skip \equiv c$. | ||
| $c_1 \sqsubseteq c_2\ \rightarrow\ c_1;c_3 \sqsubseteq c_2;c_3$: | $c_1 \sqsubseteq c_2\ \rightarrow\ c_1;c_3 \sqsubseteq c_2;c_3$: | ||