4dd6cead83
It was not a good idea to use heterogeneous equality as the default equality in Lean.
It creates the following problems.
- Heterogeneous equality does not propagate constraints in the elaborator.
For example, suppose that l has type (List Int), then the expression
l = nil
will not propagate the type (List Int) to nil.
- It is easy to write false. For example, suppose x has type Real, and the user
writes x = 0. This is equivalent to false, since 0 has type Nat. The elaborator cannot introduce
the coercion since x = 0 is a type correct expression.
Homogeneous equality does not suffer from the problems above.
We keep heterogeneous equality because it is useful for generating proof terms.
Signed-off-by: Leonardo de Moura <leonardo@microsoft.com>
118 lines
4.2 KiB
Text
118 lines
4.2 KiB
Text
Set: pp::colors
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Set: pp::unicode
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Assumed: N
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Assumed: h
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Proved: CongrH
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Set: lean::pp::implicit
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Theorem CongrH {a1 a2 b1 b2 : N} (H1 : eq::explicit N a1 b1) (H2 : eq::explicit N a2 b2) : eq::explicit
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N
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(h a1 a2)
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(h b1 b2) :=
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Congr::explicit
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N
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(λ x : N, N)
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(h a1)
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(h b1)
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a2
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b2
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(Congr::explicit N (λ x : N, N → N) h h a1 b1 (Refl::explicit (N → N → N) h) H1)
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H2
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Theorem CongrH::explicit (a1 a2 b1 b2 : N) (H1 : a1 = b1) (H2 : a2 = b2) : (h a1 a2) = (h b1 b2) := CongrH H1 H2
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Set: lean::pp::implicit
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Theorem CongrH {a1 a2 b1 b2 : N} (H1 : a1 = b1) (H2 : a2 = b2) : (h a1 a2) = (h b1 b2) := Congr (Congr (Refl h) H1) H2
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Theorem CongrH::explicit (a1 a2 b1 b2 : N) (H1 : a1 = b1) (H2 : a2 = b2) : (h a1 a2) = (h b1 b2) := CongrH H1 H2
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Proved: Example1
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Set: lean::pp::implicit
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Theorem Example1 (a b c d : N)
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(H : eq::explicit N a b ∧ eq::explicit N b c ∨ eq::explicit N a d ∧ eq::explicit N d c) : eq::explicit
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N
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(h a b)
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(h c b) :=
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DisjCases::explicit
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(eq::explicit N a b ∧ eq::explicit N b c)
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(eq::explicit N a d ∧ eq::explicit N d c)
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((h a b) == (h c b))
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H
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(λ H1 : eq::explicit N a b ∧ eq::explicit N b c,
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CongrH::explicit
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a
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b
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c
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b
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(Trans::explicit
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N
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a
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b
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c
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(Conjunct1::explicit (eq::explicit N a b) (eq::explicit N b c) H1)
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(Conjunct2::explicit (eq::explicit N a b) (eq::explicit N b c) H1))
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(Refl::explicit N b))
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(λ H1 : eq::explicit N a d ∧ eq::explicit N d c,
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CongrH::explicit
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a
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b
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c
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b
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(Trans::explicit
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N
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a
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d
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c
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(Conjunct1::explicit (eq::explicit N a d) (eq::explicit N d c) H1)
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(Conjunct2::explicit (eq::explicit N a d) (eq::explicit N d c) H1))
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(Refl::explicit N b))
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Proved: Example2
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Set: lean::pp::implicit
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Theorem Example2 (a b c d : N)
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(H : eq::explicit N a b ∧ eq::explicit N b c ∨ eq::explicit N a d ∧ eq::explicit N d c) : eq::explicit
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N
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(h a b)
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(h c b) :=
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DisjCases::explicit
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(eq::explicit N a b ∧ eq::explicit N b c)
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(eq::explicit N a d ∧ eq::explicit N d c)
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((h a b) == (h c b))
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H
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(λ H1 : eq::explicit N a b ∧ eq::explicit N b c,
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CongrH::explicit
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a
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b
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c
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b
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(Trans::explicit
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N
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a
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b
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c
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(Conjunct1::explicit (a == b) (b == c) H1)
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(Conjunct2::explicit (a == b) (b == c) H1))
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(Refl::explicit N b))
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(λ H1 : eq::explicit N a d ∧ eq::explicit N d c,
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CongrH::explicit
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a
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b
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c
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b
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(Trans::explicit
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N
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a
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d
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c
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(Conjunct1::explicit (a == d) (d == c) H1)
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(Conjunct2::explicit (a == d) (d == c) H1))
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(Refl::explicit N b))
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Proved: Example3
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Set: lean::pp::implicit
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Theorem Example3 (a b c d e : N) (H : a = b ∧ b = e ∧ b = c ∨ a = d ∧ d = c) : (h a b) = (h c b) :=
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DisjCases
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H
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(λ H1 : a = b ∧ b = e ∧ b = c, CongrH (Trans (Conjunct1 H1) (Conjunct2 (Conjunct2 H1))) (Refl b))
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(λ H1 : a = d ∧ d = c, CongrH (Trans (Conjunct1 H1) (Conjunct2 H1)) (Refl b))
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Proved: Example4
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Set: lean::pp::implicit
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Theorem Example4 (a b c d e : N) (H : a = b ∧ b = e ∧ b = c ∨ a = d ∧ d = c) : (h a c) = (h c a) :=
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DisjCases
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H
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(λ H1 : a = b ∧ b = e ∧ b = c,
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let AeqC := Trans (Conjunct1 H1) (Conjunct2 (Conjunct2 H1)) in CongrH AeqC (Symm AeqC))
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(λ H1 : a = d ∧ d = c, let AeqC := Trans (Conjunct1 H1) (Conjunct2 H1) in CongrH AeqC (Symm AeqC))
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