lean2/examples/lean/dep_if.lean

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import macros
import tactic
-- "Dependent" if-then-else (dep_if c t e)
-- The then-branch t gets a proof for c, and the else-branch e gets a proof for ¬ c
-- We also prove that
-- 1) given H : c, (dep_if c t e) = t H
-- 2) given H : ¬ c, (dep_if c t e) = e H
-- We define the "dependent" if-then-else using Hilbert's choice operator ε.
-- Note that ε is only applicable to non-empty types. Thus, we first
-- prove the following auxiliary theorem.
theorem nonempty_resolve {A : TypeU} {c : Bool} (t : c → A) (e : ¬ c → A) : nonempty A
:= or_elim (em c)
(λ Hc, nonempty_range (nonempty_intro t) Hc)
(λ Hnc, nonempty_range (nonempty_intro e) Hnc)
-- The actual definition
definition dep_if {A : TypeU} (c : Bool) (t : c → A) (e : ¬ c → A) : A
:= ε (nonempty_resolve t e) (λ r, (∀ Hc : c, r = t Hc) ∧ (∀ Hnc : ¬ c, r = e Hnc))
theorem then_simp (A : TypeU) (c : Bool) (r : A) (t : c → A) (e : ¬ c → A) (H : c)
: (∀ Hc : c, r = t Hc) ∧ (∀ Hnc : ¬ c, r = e Hnc) ↔ r = t H
:= let s1 : (∀ Hc : c, r = t Hc) ↔ r = t H
:= iff_intro
(assume Hl : (∀ Hc : c, r = t Hc),
Hl H)
(assume Hr : r = t H,
λ Hc : c, subst Hr (proof_irrel H Hc)),
s2 : (∀ Hnc : ¬ c, r = e Hnc) ↔ true
:= eqt_intro (λ Hnc : ¬ c, absurd_elim (r = e Hnc) H Hnc)
in by simp
-- Given H : c, (dep_if c t e) = t H
theorem dep_if_true {A : TypeU} (c : Bool) (t : c → A) (e : ¬ c → A) (H : c) : dep_if c t e = t H
:= let s1 : (λ r, (∀ Hc : c, r = t Hc) ∧ (∀ Hnc : ¬ c, r = e Hnc)) = (λ r, r = t H)
:= funext (λ r, then_simp A c r t e H)
in calc dep_if c t e = ε (nonempty_resolve t e) (λ r, (∀ Hc : c, r = t Hc) ∧ (∀ Hnc : ¬ c, r = e Hnc)) : refl (dep_if c t e)
... = ε (nonempty_resolve t e) (λ r, r = t H) : { s1 }
... = t H : eps_singleton (nonempty_resolve t e) (t H)
theorem else_simp (A : TypeU) (c : Bool) (r : A) (t : c → A) (e : ¬ c → A) (H : ¬ c)
: (∀ Hc : c, r = t Hc) ∧ (∀ Hnc : ¬ c, r = e Hnc) ↔ r = e H
:= let s1 : (∀ Hc : c, r = t Hc) ↔ true
:= eqt_intro (λ Hc : c, absurd_elim (r = t Hc) Hc H),
s2 : (∀ Hnc : ¬ c, r = e Hnc) ↔ r = e H
:= iff_intro
(assume Hl : (∀ Hnc : ¬ c, r = e Hnc),
Hl H)
(assume Hr : r = e H,
λ Hnc : ¬ c, subst Hr (proof_irrel H Hnc))
in by simp
-- Given H : ¬ c, (dep_if c t e) = e H
theorem dep_if_false {A : TypeU} (c : Bool) (t : c → A) (e : ¬ c → A) (H : ¬ c) : dep_if c t e = e H
:= let s1 : (λ r, (∀ Hc : c, r = t Hc) ∧ (∀ Hnc : ¬ c, r = e Hnc)) = (λ r, r = e H)
:= funext (λ r, else_simp A c r t e H)
in calc dep_if c t e = ε (nonempty_resolve t e) (λ r, (∀ Hc : c, r = t Hc) ∧ (∀ Hnc : ¬ c, r = e Hnc)) : refl (dep_if c t e)
... = ε (nonempty_resolve t e) (λ r, r = e H) : { s1 }
... = e H : eps_singleton (nonempty_resolve t e) (e H)
import cast
theorem dep_if_congr {A : TypeM} (c1 c2 : Bool)
(t1 : c1 → A) (t2 : c2 → A)
(e1 : ¬ c1 → A) (e2 : ¬ c2 → A)
(Hc : c1 = c2)
(Ht : t1 = cast (by simp) t2)
(He : e1 = cast (by simp) e2)
: dep_if c1 t1 e1 = dep_if c2 t2 e2
:= by simp
scope
-- Here is an example where dep_if is useful
-- Suppose we have a (div s t H) where H is a proof for t ≠ 0
variable div (s : Nat) (t : Nat) (H : t ≠ 0) : Nat
-- Now, we want to define a function that
-- returns 0 if x = 0
-- and div 10 x _ otherwise
-- We can't use the standard if-the-else, because we don't have a way to synthesize the proof for x ≠ 0
check λ x, dep_if (x = 0) (λ H, 0) (λ H : ¬ x = 0, div 10 x H)
pop_scope
-- If the dependent then/else branches do not use the proofs Hc : c and Hn : ¬ c, then we
-- can reduce the dependent-if to a regular if
theorem dep_if_if {A : TypeU} (c : Bool) (t e : A) : dep_if c (λ Hc, t) (λ Hn, e) = if c then t else e
:= or_elim (em c)
(assume Hc : c, calc dep_if c (λ Hc, t) (λ Hn, e) = (λ Hc, t) Hc : dep_if_true _ _ _ Hc
... = if c then t else e : by simp)
(assume Hn : ¬ c, calc dep_if c (λ Hc, t) (λ Hn, e) = (λ Hn, e) Hn : dep_if_false _ _ _ Hn
... = if c then t else e : by simp)