feat(builtin/num): define lt predicate, and prove basic theorems
Signed-off-by: Leonardo de Moura <leonardo@microsoft.com>
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@ -94,11 +94,73 @@ theorem induction {P : num → Bool} (H1 : P zero) (H2 : ∀ n, P n → P (succ
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show P a,
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from subst Qa abst_eq
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theorem induction_on {P : num → Bool} (a : num) (H1 : P zero) (H2 : ∀ n, P n → P (succ n)) : P a
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:= induction H1 H2 a
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definition lt (m n : num) := ∃ P, (∀ i, P (succ i) → P i) ∧ P m ∧ ¬ P n
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infix 50 < : lt
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theorem lt_elim {m n : num} {B : Bool} (H1 : m < n)
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(H2 : ∀ (P : num → Bool), (∀ i, P (succ i) → P i) → P m → ¬ P n → B)
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: B
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:= obtain P Pdef, from H1,
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H2 P (and_eliml Pdef) (and_eliml (and_elimr Pdef)) (and_elimr (and_elimr Pdef))
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theorem lt_intro {m n : num} {P : num → Bool} (H1 : ∀ i, P (succ i) → P i) (H2 : P m) (H3 : ¬ P n) : m < n
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:= exists_intro P (and_intro H1 (and_intro H2 H3))
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set_opaque lt true
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theorem lt_nrefl (n : num) : ¬ (n < n)
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:= not_intro
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(assume N : n < n,
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lt_elim N (λ P Pred Pn nPn, absurd Pn nPn))
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theorem lt_succ {m n : num} : succ m < n → m < n
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:= assume H : succ m < n,
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lt_elim H
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(λ (P : num → Bool) (Pred : ∀ i, P (succ i) → P i) (Psm : P (succ m)) (nPn : ¬ P n),
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have Pm : P m,
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from Pred m Psm,
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show m < n,
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from lt_intro Pred Pm nPn)
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theorem not_lt_zero (n : num) : ¬ (n < zero)
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:= induction_on n
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(show ¬ (zero < zero),
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from lt_nrefl zero)
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(λ (n : num) (iH : ¬ (n < zero)),
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show ¬ (succ n < zero),
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from not_intro
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(assume N : succ n < zero,
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have nLTzero : n < zero,
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from lt_succ N,
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show false,
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from absurd nLTzero iH))
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theorem z_lt_succ_z : zero < succ zero
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:= let P : num → Bool := λ x, x = zero
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in have Pred : ∀ i, P (succ i) → P i,
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from take i, assume Ps : P (succ i),
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have si_eq_z : succ i = zero,
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from Ps,
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have si_ne_z : succ i ≠ zero,
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from succ_nz i,
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show P i,
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from absurd_elim (P i) si_eq_z (succ_nz i),
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have Pz : P zero,
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from (refl zero),
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have nPs : ¬ P (succ zero),
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from succ_nz zero,
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show zero < succ zero,
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from lt_intro Pred Pz nPs
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set_opaque num true
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set_opaque Z true
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set_opaque S true
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set_opaque zero true
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set_opaque succ true
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set_opaque lt true
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end
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definition num := num::num
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