test(tests/lean/run/nested_rec.lean): add nested recursion example based on well-founded recursion package
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tests/lean/run/nested_rec.lean
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41
tests/lean/run/nested_rec.lean
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open nat prod sigma
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-- We will define the following example by well-foudned recursion
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-- g 0 := 0
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-- g (succ x) := g (g x)
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definition g.F (x : nat) : (Π y, y < x → Σ r : nat, r ≤ y) → Σ r : nat, r ≤ x :=
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nat.cases_on x
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(λ f, (dpair zero (le.refl zero)))
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(λ x₁ (f : Π y, y < succ x₁ → Σ r : nat, r ≤ y),
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let p₁ := f x₁ (lt.base x₁) in
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let gx₁ := dpr₁ p₁ in
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let p₂ := f gx₁ (lt.of_le_of_lt (dpr₂ p₁) (lt.base x₁)) in
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let ggx₁ := dpr₁ p₂ in
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dpair ggx₁ (le.step (le.trans (dpr₂ p₂) (dpr₂ p₁))))
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definition g (x : nat) : nat :=
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dpr₁ (well_founded.fix g.F x)
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example : g 3 = 0 :=
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rfl
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example : g 6 = 0 :=
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rfl
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theorem g_zero : g 0 = 0 :=
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rfl
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theorem g_succ (a : nat) : g (succ a) = g (g a) :=
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have aux : well_founded.fix g.F (succ a) = dpair (g (g a)) _, from
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well_founded.fix_eq g.F (succ a),
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calc g (succ a) = dpr₁ (well_founded.fix g.F (succ a)) : rfl
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... = dpr₁ (dpair (g (g a)) _) : aux
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... = g (g a) : rfl
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theorem g_all_zero (a : nat) : g a = zero :=
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nat.induction_on a
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g_zero
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(λ a₁ (ih : g a₁ = zero), calc
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g (succ a₁) = g (g a₁) : g_succ
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... = g 0 : ih
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... = 0 : g_zero)
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