import macros

-- Well-founded relation definition
-- We are essentially saying that a relation R is well-founded
--    if every non-empty "set" P, has a R-minimal element
definition wf {A : (Type U)} (R : A → A → Bool) : Bool
:= ∀ P, (∃ w, P w) → ∃ min, P min ∧ ∀ b, R b min → ¬ P b

-- Well-founded induction theorem
theorem wf_induction {A : (Type U)} {R : A → A → Bool} {P : A → Bool} (Hwf : wf R) (iH : ∀ x, (∀ y, R y x → P y) → P x)
                     : ∀ x, P x
:= by_contradiction (assume N : ¬ ∀ x, P x,
     obtain (w : A) (Hw : ¬ P w), from not_forall_elim N,
         -- The main "trick" is to define Q x as ¬ P x.
         -- Since R is well-founded, there must be a R-minimal element r s.t. Q r (which is ¬ P r)
         let Q : A → Bool  := λ x, ¬ P x in
         have Qw  : ∃ w, Q w,
           from exists_intro w Hw,
         have Qwf : ∃ min, Q min ∧ ∀ b, R b min → ¬ Q b,
           from Hwf Q Qw,
         obtain (r : A) (Hr : Q r ∧ ∀ b, R b r → ¬ Q b),
           from Qwf,
         -- Using the inductive hypothesis iH and Hr, we show P r, and derive the contradiction.
         have s1 : ∀ b, R b r → P b,
           from take b : A, assume H : R b r,
                  -- We are using Hr to derive ¬ ¬ P b
                  not_not_elim (and_elimr Hr b H),
         have s2 : P r,
           from iH r s1,
         have s3 : ¬ P r,
           from and_eliml Hr,
         show false,
           from absurd s2 s3)

-- More compact proof
theorem wf_induction2 {A : (Type U)} {R : A → A → Bool} {P : A → Bool} (Hwf : wf R) (iH : ∀ x, (∀ y, R y x → P y) → P x)
                     : ∀ x, P x
:= by_contradiction (assume N : ¬ ∀ x, P x,
     obtain (w : A) (Hw : ¬ P w), from not_forall_elim N,
         -- The main "trick" is to define Q x as ¬ P x.
         -- Since R is well-founded, there must be a R-minimal element r s.t. Q r (which is ¬ P r)
         let Q : A → Bool  := λ x, ¬ P x in
         obtain (r : A) (Hr : Q r ∧ ∀ b, R b r → ¬ Q b),
           from Hwf Q (exists_intro w Hw),
         -- Using the inductive hypothesis iH and Hr, we show P r, and derive the contradiction.
         have s1 : ∀ b, R b r → P b,
           from take b : A, assume H : R b r,
                  -- We are using Hr to derive ¬ ¬ P b
                  not_not_elim (and_elimr Hr b H),
         absurd (iH r s1) (and_eliml Hr))