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Asociatividad_del_supremo.lean
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Asociatividad_del_supremo.lean
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-- ---------------------------------------------------------------------
-- Ejercicio. Demostrar que en los retículos se verifica que
-- (x ⊔ y) ⊔ z = x ⊔ (y ⊔ z)
-- ---------------------------------------------------------------------
import order.lattice
variables {α : Type*} [lattice α]
variables x y z : α
-- 1ª demostración
-- ===============
example : (x ⊔ y) ⊔ z = x ⊔ (y ⊔ z) :=
begin
have h1 : (x ⊔ y) ⊔ z ≤ x ⊔ (y ⊔ z),
{ have h1a : x ⊔ y ≤ x ⊔ (y ⊔ z), by finish,
have h1b : z ≤ x ⊔ (y ⊔ z), by finish,
show (x ⊔ y) ⊔ z ≤ x ⊔ (y ⊔ z),
by exact sup_le h1a h1b, },
have h2 : x ⊔ (y ⊔ z) ≤ (x ⊔ y) ⊔ z,
{ have h2a : x ≤ (x ⊔ y) ⊔ z, by finish,
have h2b : y ⊔ z ≤ (x ⊔ y) ⊔ z, by finish,
show x ⊔ (y ⊔ z) ≤ (x ⊔ y) ⊔ z,
by exact sup_le h2a h2b, },
show (x ⊔ y) ⊔ z = x ⊔ (y ⊔ z),
by exact le_antisymm h1 h2,
end
-- 2ª demostración
-- ===============
example : (x ⊔ y) ⊔ z = x ⊔ (y ⊔ z) :=
begin
have h1 : (x ⊔ y) ⊔ z ≤ x ⊔ (y ⊔ z),
{ have h1a : x ⊔ y ≤ x ⊔ (y ⊔ z),
{ have h1a1 : x ≤ x ⊔ (y ⊔ z) :=
le_sup_left,
have h1a2 : y ≤ x ⊔ (y ⊔ z), calc
y ≤ y ⊔ z : le_sup_left
... ≤ x ⊔ (y ⊔ z) : le_sup_right,
show x ⊔ y ≤ x ⊔ (y ⊔ z),
by exact sup_le h1a1 h1a2, },
have h1b : z ≤ x ⊔ (y ⊔ z), calc
z ≤ y ⊔ z : le_sup_right
... ≤ x ⊔ (y ⊔ z) : le_sup_right,
show (x ⊔ y) ⊔ z ≤ x ⊔ (y ⊔ z),
by exact sup_le h1a h1b, },
have h2 : x ⊔ (y ⊔ z) ≤ (x ⊔ y) ⊔ z,
{ have h2a : x ≤ (x ⊔ y) ⊔ z, calc
x ≤ x ⊔ y : le_sup_left
... ≤ (x ⊔ y) ⊔ z : le_sup_left,
have h2b : y ⊔ z ≤ (x ⊔ y) ⊔ z,
{ have h2b1 : y ≤ (x ⊔ y) ⊔ z, calc
y ≤ x ⊔ y : le_sup_right
... ≤ (x ⊔ y) ⊔ z : le_sup_left,
have h2b2 : z ≤ (x ⊔ y) ⊔ z :=
le_sup_right,
show y ⊔ z ≤ (x ⊔ y) ⊔ z,
by exact sup_le h2b1 h2b2, },
show x ⊔ (y ⊔ z) ≤ (x ⊔ y) ⊔ z,
by exact sup_le h2a h2b, },
show (x ⊔ y) ⊔ z = x ⊔ (y ⊔ z),
by exact le_antisymm h1 h2,
end
-- 3ª demostración
-- ===============
example : (x ⊔ y) ⊔ z = x ⊔ (y ⊔ z) :=
begin
apply le_antisymm,
{ apply sup_le,
{ apply sup_le le_sup_left (le_sup_of_le_right le_sup_left)},
{ apply le_sup_of_le_right le_sup_right}},
{ apply sup_le,
{ apply le_sup_of_le_left le_sup_left},
{ apply sup_le (le_sup_of_le_left le_sup_right) le_sup_right}},
end
-- Su desarrollo es
--
-- ⊢ x ⊔ y ⊔ z = x ⊔ (y ⊔ z)
-- apply le_antisymm,
-- | ⊢ x ⊔ y ⊔ z ≤ x ⊔ (y ⊔ z)
-- | { apply sup_le,
-- | | ⊢ x ⊔ y ≤ x ⊔ (y ⊔ z)
-- | | { apply sup_le le_sup_left (le_sup_right_of_le le_sup_left)},
-- | | ⊢ z ≤ x ⊔ (y ⊔ z)
-- | | { apply le_sup_right_of_le le_sup_right}},
-- | ⊢ x ⊔ (y ⊔ z) ≤ x ⊔ y ⊔ z
-- | { apply sup_le,
-- | | ⊢ x ≤ x ⊔ y ⊔ z
-- | | { apply le_sup_left_of_le le_sup_left},
-- | | ⊢ y ⊔ z ≤ x ⊔ y ⊔ z
-- | | { apply sup_le (le_sup_left_of_le le_sup_right) le_sup_right}},
-- no goals
-- 4ª demostración
-- ===============
example : (x ⊔ y) ⊔ z = x ⊔ (y ⊔ z) :=
le_antisymm
(sup_le
(sup_le le_sup_left (le_sup_of_le_right le_sup_left))
(le_sup_of_le_right le_sup_right))
(sup_le
(le_sup_of_le_left le_sup_left)
(sup_le (le_sup_of_le_left le_sup_right) le_sup_right))
-- 5ª demostración
-- ===============
example : x ⊔ y ⊔ z = x ⊔ (y ⊔ z) :=
-- by library_search
sup_assoc
-- 6ª demostración
-- ===============
example : x ⊔ y ⊔ z = x ⊔ (y ⊔ z) :=
-- by hint
by finish