999f23cbc0
This is the correct fix for the id declaration pretty printing discrepancy reported by Daniel. TODO: decide whether we need another eq-mode where names are ignored. For example, in blast, it makes sense to increase sharing by ignoring binder names.
39 lines
2.2 KiB
Text
39 lines
2.2 KiB
Text
[fixed, fixed, eq, eq]
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λ (A : Type) (s : has_add A) (a a_1 : A) (e_3 : a = a_1) (a_2 a_3 : A) (e_4 : a_2 = a_3), congr (congr_arg add e_3) e_4
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:
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∀ (A : Type) (s : has_add A) (a a_1 : A), a = a_1 → (∀ (a_2 a_3 : A), a_2 = a_3 → a + a_2 = a_1 + a_3)
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[eq, cast, fixed, eq, eq]
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λ (c c_1 : Prop) (e_1 : c = c_1) (H : decidable c) (A : Type) (t t_1 : A) (e_4 : t = t_1) (e e_2 : A) (e_5 : e = e_2),
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eq.drec (eq.drec (eq.drec (eq.refl (ite c t e)) e_5) e_4) e_1
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:
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∀ (c c_1 : Prop) (e_1 : c = c_1) (H : decidable c) (A : Type) (t t_1 : A),
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t = t_1 → (∀ (e e_2 : A), e = e_2 → ite c t e = ite c_1 t_1 e_2)
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[fixed, eq, eq]
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λ (A : Type) (a a_1 : list A) (e_2 : a = a_1) (a_2 a_3 : list A) (e_3 : a_2 = a_3), congr (congr_arg perm e_2) e_3
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:
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∀ (A : Type) (a a_1 : list A), a = a_1 → (∀ (a_2 a_3 : list A), a_2 = a_3 → perm a a_2 = perm a_1 a_3)
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[eq, eq, cast, cast]
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λ (x x_1 : ℕ) (e_1 : x = x_1) (y y_1 : ℕ) (e_2 : y = y_1) (a : p x) (a_1 : q x y),
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eq.drec (eq.drec (eq.refl (f x y (eq.rec a (eq.refl x)) (eq.rec (eq.rec a_1 (eq.refl y)) (eq.refl x)))) e_2) e_1
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:
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∀ (x x_1 : ℕ) (e_1 : x = x_1) (y y_1 : ℕ) (e_2 : y = y_1) (a : p x) (a_1 : q x y),
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f x y a a_1 = f x_1 y_1 (eq.rec a e_1) (eq.rec (eq.rec a_1 e_2) e_1)
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[fixed, eq, eq, cast, cast, cast, cast, cast, cast]
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λ (A : Type) (n n_1 : A) (e_2 : n = n_1) (m m_1 : A) (e_3 : m = m_1) (H₁ : p n) (H₂ : p m) (H₃ : q n n H₁ H₁)
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(H₄ : q n m H₁ H₂) (H₅ : r n m H₁ H₂ H₄) (H₆ : r n n H₁ H₁ H₃),
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eq.drec
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(eq.drec
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(eq.refl
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(h A n m (eq.rec H₁ (eq.refl n)) (eq.rec H₂ (eq.refl m)) (eq.drec H₃ (eq.refl n))
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(eq.drec (eq.drec H₄ (eq.refl m)) (eq.refl n))
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(eq.drec (eq.drec H₅ (eq.refl m)) (eq.refl n))
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(eq.drec H₆ (eq.refl n))))
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e_3)
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e_2
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:
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∀ (A : Type) (n n_1 : A) (e_2 : n = n_1) (m m_1 : A) (e_3 : m = m_1) (H₁ : p n) (H₂ : p m)
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(H₃ : q n n H₁ H₁) (H₄ : q n m H₁ H₂) (H₅ : r n m H₁ H₂ H₄) (H₆ : r n n H₁ H₁ H₃),
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h A n m H₁ H₂ H₃ H₄ H₅ H₆ = h A n_1 m_1 (eq.rec H₁ e_2) (eq.rec H₂ e_3) (eq.drec H₃ e_2)
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(eq.drec (eq.drec H₄ e_3) e_2)
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(eq.drec (eq.drec H₅ e_3) e_2)
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(eq.drec H₆ e_2)
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