2014-01-31 06:14:38 +00:00
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import macros
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import tactic
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-- "Dependent" if-then-else (dep_if c t e)
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-- The then-branch t gets a proof for c, and the else-branch e gets a proof for ¬ c
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-- We also prove that
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-- 1) given H : c, (dep_if c t e) = t H
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-- 2) given H : ¬ c, (dep_if c t e) = e H
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-- We define the "dependent" if-then-else using Hilbert's choice operator ε.
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-- Note that ε is only applicable to non-empty types. Thus, we first
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-- prove the following auxiliary theorem.
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theorem nonempty_resolve {A : TypeU} {c : Bool} (t : c → A) (e : ¬ c → A) : nonempty A
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:= or_elim (em c)
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(λ Hc, nonempty_range (nonempty_intro t) Hc)
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(λ Hnc, nonempty_range (nonempty_intro e) Hnc)
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-- The actual definition
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definition dep_if {A : TypeU} (c : Bool) (t : c → A) (e : ¬ c → A) : A
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:= ε (nonempty_resolve t e) (λ r, (∀ Hc : c, r = t Hc) ∧ (∀ Hnc : ¬ c, r = e Hnc))
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theorem then_simp (A : TypeU) (c : Bool) (r : A) (t : c → A) (e : ¬ c → A) (H : c)
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: (∀ Hc : c, r = t Hc) ∧ (∀ Hnc : ¬ c, r = e Hnc) ↔ r = t H
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:= let s1 : (∀ Hc : c, r = t Hc) ↔ r = t H
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:= iff_intro
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(assume Hl : (∀ Hc : c, r = t Hc),
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Hl H)
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(assume Hr : r = t H,
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λ Hc : c, subst Hr (proof_irrel H Hc)),
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s2 : (∀ Hnc : ¬ c, r = e Hnc) ↔ true
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:= eqt_intro (λ Hnc : ¬ c, absurd_elim (r = e Hnc) H Hnc)
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in by simp
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-- Given H : c, (dep_if c t e) = t H
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theorem dep_if_true {A : TypeU} (c : Bool) (t : c → A) (e : ¬ c → A) (H : c) : dep_if c t e = t H
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:= let s1 : (λ r, (∀ Hc : c, r = t Hc) ∧ (∀ Hnc : ¬ c, r = e Hnc)) = (λ r, r = t H)
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:= funext (λ r, then_simp A c r t e H)
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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)
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... = ε (nonempty_resolve t e) (λ r, r = t H) : { s1 }
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... = t H : eps_singleton (nonempty_resolve t e) (t H)
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theorem else_simp (A : TypeU) (c : Bool) (r : A) (t : c → A) (e : ¬ c → A) (H : ¬ c)
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: (∀ Hc : c, r = t Hc) ∧ (∀ Hnc : ¬ c, r = e Hnc) ↔ r = e H
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:= let s1 : (∀ Hc : c, r = t Hc) ↔ true
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:= eqt_intro (λ Hc : c, absurd_elim (r = t Hc) Hc H),
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s2 : (∀ Hnc : ¬ c, r = e Hnc) ↔ r = e H
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:= iff_intro
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(assume Hl : (∀ Hnc : ¬ c, r = e Hnc),
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Hl H)
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(assume Hr : r = e H,
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λ Hnc : ¬ c, subst Hr (proof_irrel H Hnc))
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in by simp
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-- Given H : ¬ c, (dep_if c t e) = e H
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theorem dep_if_false {A : TypeU} (c : Bool) (t : c → A) (e : ¬ c → A) (H : ¬ c) : dep_if c t e = e H
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:= let s1 : (λ r, (∀ Hc : c, r = t Hc) ∧ (∀ Hnc : ¬ c, r = e Hnc)) = (λ r, r = e H)
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:= funext (λ r, else_simp A c r t e H)
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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)
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... = ε (nonempty_resolve t e) (λ r, r = e H) : { s1 }
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... = e H : eps_singleton (nonempty_resolve t e) (e H)
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import cast
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theorem dep_if_congr {A : TypeM} (c1 c2 : Bool)
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(t1 : c1 → A) (t2 : c2 → A)
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(e1 : ¬ c1 → A) (e2 : ¬ c2 → A)
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(Hc : c1 = c2)
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(Ht : t1 = cast (by simp) t2)
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(He : e1 = cast (by simp) e2)
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: dep_if c1 t1 e1 = dep_if c2 t2 e2
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:= by simp
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2014-01-31 06:44:04 +00:00
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scope
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-- Here is an example where dep_if is useful
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-- Suppose we have a (div s t H) where H is a proof for t ≠ 0
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variable div (s : Nat) (t : Nat) (H : t ≠ 0) : Nat
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-- Now, we want to define a function that
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-- returns 0 if x = 0
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-- and div 10 x _ otherwise
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-- We can't use the standard if-the-else, because we don't have a way to synthesize the proof for x ≠ 0
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check λ x, dep_if (x = 0) (λ H, 0) (λ H : ¬ x = 0, div 10 x H)
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pop_scope
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-- If the dependent then/else branches do not use the proofs Hc : c and Hn : ¬ c, then we
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-- can reduce the dependent-if to a regular if
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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
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:= or_elim (em c)
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(assume Hc : c, calc dep_if c (λ Hc, t) (λ Hn, e) = (λ Hc, t) Hc : dep_if_true _ _ _ Hc
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... = if c then t else e : by simp)
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(assume Hn : ¬ c, calc dep_if c (λ Hc, t) (λ Hn, e) = (λ Hn, e) Hn : dep_if_false _ _ _ Hn
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... = if c then t else e : by simp)
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