e3dc552c39
The example tests/lua/simp1.lua demonstrates the issue.
The higher-order matcher matches closed terms that are definitionally equal.
So, given a definition
definition a := 1
it will match 'a' with '1' since they are definitionally equal.
Then, if we have a theorem
theorem a_eq_1 : a = 1
as a rewrite rule, it was triggering the following infinite loop when simplifying the expression "a"
a --> 1 --> 1 --> 1 ...
The first simplification is expected. The other ones are not.
The problem is that "1" is definitionally equal to "a", and they match.
The rewrite_rule_set manager accepts the rule a --> 1 since the left-hand-side does not occur in the right-hand-side.
To avoid this loop, we test if the new expression is not equal to the previous one.
Signed-off-by: Leonardo de Moura <leonardo@microsoft.com>
44 lines
1 KiB
Lua
44 lines
1 KiB
Lua
add_rewrite_rules({"Nat", "add_zerol"})
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add_rewrite_rules({"Nat", "add_zeror"})
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parse_lean_cmds([[
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variable f : Nat -> Nat -> Nat
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variable g : Nat -> Nat
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variable b : Nat
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definition a := 1
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theorem a_eq_1 : a = 1
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:= refl a
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definition c := 1
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set_opaque a true
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axiom f_id (x : Nat) : f x 1 = 2*x
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axiom g_g_x (x : Nat) : (not (x = 0)) -> g (g x) = 0
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]])
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add_rewrite_rules("a_eq_1")
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add_rewrite_rules("f_id")
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add_rewrite_rules("eq_id")
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-- set_option({"lean", "pp", "implicit"}, true)
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e, pr = simplify(parse_lean('fun x, f (f x (0 + a)) (g (b + 0))'))
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print(e)
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print(pr)
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local env = get_environment()
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print(env:type_check(pr))
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e, pr = simplify(parse_lean('forall x, let d := a + 1 in f x a >= d'))
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print(e)
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print(pr)
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local env = get_environment()
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print(env:type_check(pr))
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e, pr = simplify(parse_lean('(fun x, f (f x (0 + a)) (g (b + 0))) b'))
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print(e)
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print(pr)
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local env = get_environment()
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print(env:type_check(pr))
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e, pr = simplify(parse_lean('(fun x y, f x y) = f'))
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print(e)
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print(pr)
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local env = get_environment()
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print(env:type_check(pr))
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