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cartesianproducts [2015/04/21 17:42] wikiadmin [Using sets] |
cartesianproducts [2015/04/21 17:43] wikiadmin |
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| The set of values for the scrutinee which **will** be matched by a given pattern can now be seen as a subset of $s$ expressed by constraints. Recall that each extractor $ex$ has a corresponding set $EX$ which defines all values which it matches. A pattern will successfully match a value if all extractors in it match what they are submitted with. In other words, for a pattern with calls to extractors $ex_1,ex_2,\ldots$ is a cartesian product of the same dimension as $s$, where the positions corresponding to the extractors are "replaced" with the proper sets: | The set of values for the scrutinee which **will** be matched by a given pattern can now be seen as a subset of $s$ expressed by constraints. Recall that each extractor $ex$ has a corresponding set $EX$ which defines all values which it matches. A pattern will successfully match a value if all extractors in it match what they are submitted with. In other words, for a pattern with calls to extractors $ex_1,ex_2,\ldots$ is a cartesian product of the same dimension as $s$, where the positions corresponding to the extractors are "replaced" with the proper sets: | ||
| - | \[ EX_1 \times B \times EX_2 \times \ldots \]. Note that by construction of the types $A,B,\ldots$, this is always a subset of $s$. | + | \begin{equation*} EX_1 \times B \times EX_2 \times \ldots \end{equation*}. Note that by construction of the types $A,B,\ldots$, this is always a subset of $s$. |
| === Completeness, disjointness and the like === | === Completeness, disjointness and the like === | ||
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| The constraints on the patterns are: | The constraints on the patterns are: | ||
| - | \[p_1 = ENode \] | + | \begin{equation*}p_1 = ENode \end{equation*} |
| - | \[p_2 = ELeaf \] | + | \begin{equation*}p_2 = ELeaf \end{equation*} |
| Completeness is expressed as $p_1 \cup p_2 \supseteq Tree$ and disjointness as $p_1 \cap p_2 = \emptyset$. Both are trivial with our axioms. | Completeness is expressed as $p_1 \cup p_2 \supseteq Tree$ and disjointness as $p_1 \cap p_2 = \emptyset$. Both are trivial with our axioms. | ||