59 lines
1.4 KiB
Plaintext
59 lines
1.4 KiB
Plaintext
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/-
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# References
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1. Enderton, Herbert B. A Mathematical Introduction to Logic. 2nd ed. San Diego:
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Harcourt/Academic Press, 2001.
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-/
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import Mathlib.Tactic.Ring
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universe u
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/--[1]
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An n-tuple is defined recursively as:
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⟨x₁, ..., xₙ₊₁⟩ = ⟨⟨x₁, ..., xₙ⟩, xₙ₊₁⟩
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As [1] notes, it is also useful to define ⟨x⟩ = x.
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--/
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inductive Tuple (α : Type u) : Nat → Type u where
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| nil : Tuple α 0
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| cons : {n : Nat} → Tuple α n → α → Tuple α (n + 1)
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syntax (priority := high) "⟨" term,+ "⟩" : term
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macro_rules
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| `(⟨$x⟩) => `(Tuple.cons Tuple.nil $x)
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| `(⟨$xs:term,*, $x⟩) => `(Tuple.cons ⟨$xs,*⟩ $x)
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namespace Tuple
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/--
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Returns the value at the nth-index of the given tuple.
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-/
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def index (t : Tuple α n) (m : Nat) : 1 ≤ m ∧ m ≤ n → α := by
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intro h
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cases t
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· case nil =>
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have ff : 1 ≤ 0 := Nat.le_trans h.left h.right
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ring_nf at ff
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exact False.elim ff
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. case cons n' init last =>
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by_cases k : m = n' + 1
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· exact last
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· exact index init m (And.intro h.left (by
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have h₂ : m + 1 ≤ n' + 1 := Nat.lt_of_le_of_ne h.right k
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norm_num at h₂
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exact h₂))
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-- TODO: Prove the following theorem
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theorem eq_by_index (t₁ t₂ : Tuple α n)
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: (t₁ = t₂) ↔ (∀ i : Nat, 1 ≤ i ∧ i ≤ n → index t₁ i = index t₂ i) := by
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apply Iff.intro
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· sorry
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· sorry
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-- TODO: [1] Lemma 0A
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end Tuple
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