Add new simplification rule
This adds the rule “(not F or G) === (F -> G)” to the simplification rule tableau.
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@ -430,6 +430,42 @@ struct SimplificationRuleDeMorganForConjunctions
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////////////////////////////////////////////////////////////////////////////////////////////////////
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struct SimplificationRuleImplicationFromDisjunction
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{
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static constexpr const auto Description = "(not F or G) === (F -> G)";
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static SimplificationResult apply(ast::Formula &formula)
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{
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if (!formula.is<ast::Or>())
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return SimplificationResult::Unchanged;
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auto &or_ = formula.get<ast::Or>();
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if (or_.arguments.size() != 2)
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return SimplificationResult::Unchanged;
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const auto leftIsNot = or_.arguments[0].is<ast::Not>();
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const auto rightIsNot = or_.arguments[1].is<ast::Not>();
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if (leftIsNot == rightIsNot)
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return SimplificationResult::Unchanged;
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auto &negativeSide = leftIsNot ? or_.arguments[0] : or_.arguments[1];
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auto &positiveSide = leftIsNot ? or_.arguments[1] : or_.arguments[0];
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assert(negativeSide.is<ast::Not>());
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assert(!positiveSide.is<ast::Not>());
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auto &negativeSideArgument = negativeSide.get<ast::Not>().argument;
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formula = ast::Formula::make<ast::Implies>(std::move(negativeSideArgument), std::move(positiveSide));
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return SimplificationResult::Simplified;
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}
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};
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////////////////////////////////////////////////////////////////////////////////////////////////////
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const auto simplifyWithDefaultRules =
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simplify
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<
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@ -442,7 +478,8 @@ const auto simplifyWithDefaultRules =
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SimplificationRuleExistsWithoutQuantifiedVariables,
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SimplificationRuleInWithPrimitiveArguments,
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SimplificationRuleSubsumptionInBiconditionals,
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SimplificationRuleDeMorganForConjunctions
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SimplificationRuleDeMorganForConjunctions,
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SimplificationRuleImplicationFromDisjunction
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>;
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////////////////////////////////////////////////////////////////////////////////////////////////////
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@ -153,7 +153,7 @@ TEST_CASE("[completion] Rules are completed", "[completion]")
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CHECK(output.str() ==
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"forall V1 (covered(V1) <-> exists U1 in(V1, U1))\n"
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"forall V2, V3 (in(V2, V3) -> (V2 in 1..n and V3 in 1..r))\n"
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"forall U2 (not U2 in 1..n or covered(U2))\n"
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"forall U2 (U2 in 1..n -> covered(U2))\n"
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"forall U3, U4, U5 (not in(U3, U4) or not in(U5, U4) or not exists X1 (X1 in (U3 + U5) and in(X1, U4)))\n");
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}
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@ -168,7 +168,7 @@ TEST_CASE("[hidden predicate elimination] Hidden predicates are correctly elimin
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CHECK(output.str() ==
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"(s -> not #false)\n"
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"(t -> not #false)\n"
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"(not s or t)\n"
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"(s -> t)\n"
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"(#false or #false or not #false)\n");
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}
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@ -56,7 +56,7 @@ TEST_CASE("[placeholders] Programs with placeholders are correctly completed", "
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CHECK(output.str() ==
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"forall V1, V2 (color(V1, V2) -> (vertex(V1) and color(V2)))\n"
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"forall U1 (not vertex(U1) or exists U2 color(U1, U2))\n"
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"forall U1 (vertex(U1) -> exists U2 color(U1, U2))\n"
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"forall U3, U4, U5 (not color(U3, U4) or not color(U5, U4) or not edge(U3, U5))\n");
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}
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}
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