l3lab/ntp-mathlib huggingface dataset:
/-- Bourbaki GT III.6.5 Theorem I:
ℤ-bilinear continuous maps from dense images into a complete Hausdorff group extend by continuity.
Note: Bourbaki assumes that α and β are also complete Hausdorff, but this is not necessary. -/
theorem extend_Z_bilin : Continuous (extend (de.prod df) (fun p : β × δ => φ p.1 p.2)) := by
  refine' continuous_extend_of_cauchy _ _
  rintro ⟨x₀, y₀⟩
  constructor
  · apply NeBot.map
    apply comap_neBot
    intro U h
    rcases mem_closure_iff_nhds.1 ((de.prod df).dense (x₀, y₀)) U h with ⟨x, x_in, ⟨z, z_x⟩⟩
    exists z
    aesop
  · suffices map (fun p : (β × δ) × β × δ => (fun p : β × δ => φ p.1 p.2) p.2 -
      (fun p : β × δ => φ p.1 p.2) p.1)
        (comap (fun p : (β × δ) × β × δ => ((e p.1.1, f p.1.2), (e p.2.1, f p.2.2)))
        (𝓝 (x₀, y₀) ×ˢ 𝓝 (x₀, y₀))) ≤ 𝓝 0 by
      rwa [uniformity_eq_comap_nhds_zero G, prod_map_map_eq, ← map_le_iff_le_comap, Filter.map_map,
        prod_comap_comap_eq]
    intro W' W'_nhd
    have key := extend_Z_bilin_key de df hφ W'_nhd x₀ y₀
    rcases key with ⟨U, U_nhd, V, V_nhd, h⟩
    rw [mem_comap] at U_nhd
    rcases U_nhd with ⟨U', U'_nhd, U'_sub⟩
    rw [mem_comap] at V_nhd
    rcases V_nhd with ⟨V', V'_nhd, V'_sub⟩
    rw [mem_map, mem_comap, nhds_prod_eq]
    exists (U' ×ˢ V') ×ˢ U' ×ˢ V'
    rw [mem_prod_same_iff]
    simp only [exists_prop]
    constructor
    · have := prod_mem_prod U'_nhd V'_nhd
      tauto
    · intro p h'
      simp only [Set.mem_preimage, Set.prod_mk_mem_set_prod_eq] at h'
      rcases p with ⟨⟨x, y⟩, ⟨x', y'⟩⟩
      apply h <;> 

Proof state:
case mk.right.intro.intro.intro.intro.intro.intro.intro.intro.right.mk.mk.mk.x
α : Type u_1
β : Type u_2
γ : Type u_3
δ : Type u_4
G : Type u_5
inst✝¹⁶ : TopologicalSpace α
inst✝¹⁵ : AddCommGroup α
inst✝¹⁴ : TopologicalAddGroup α
inst✝¹³ : TopologicalSpace β
inst✝¹² : AddCommGroup β
inst✝¹¹ : TopologicalAddGroup β
inst✝¹⁰ : TopologicalSpace γ
inst✝⁹ : AddCommGroup γ
inst✝⁸ : TopologicalAddGroup γ
inst✝⁷ : TopologicalSpace δ
inst✝⁶ : AddCommGroup δ
inst✝⁵ : TopologicalAddGroup δ
inst✝⁴ : UniformSpace G
inst✝³ : AddCommGroup G
inst✝² : UniformAddGroup G
inst✝¹ : SeparatedSpace G
inst✝ : CompleteSpace G
e : β →+ α
de : DenseInducing ⇑e
f : δ →+ γ
df : DenseInducing ⇑f
φ : β →+ δ →+ G
hφ : Continuous fun p => (φ p.1) p.2
W'✝ : Set G
W'_nhd✝ : W'✝ ∈ 𝓝 0
x₀ : α
y₀ : γ
W' : Set G
W'_nhd : W' ∈ 𝓝 0
U : Set β
V : Set δ
h : ∀ x ∈ U, ∀ x' ∈ U, ∀ y ∈ V, ∀ y' ∈ V, (fun p => (φ p.1) p.2) (x', y') - (fun p => (φ p.1) p.2) (x, y) ∈ W'
U' : Set α
U'_nhd : U' ∈ 𝓝 x₀
U'_sub : ⇑e ⁻¹' U' ⊆ U
V' : Set γ
V'_nhd : V' ∈ 𝓝 y₀
V'_sub : ⇑f ⁻¹' V' ⊆ V
x : β
y : δ
x' : β
y' : δ
h' :
  (e ((x, y), x', y').1.1 ∈ U' ∧ f ((x, y), x', y').1.2 ∈ V') ∧
    e ((x, y), x', y').2.1 ∈ U' ∧ f ((x, y), x', y').2.2 ∈ V'
⊢ y' ∈ V

Next tactic:
tauto
/-- The quotient `G ⧸ N` of a complete first countable topological group `G` by a normal subgroup
is itself complete. [N. Bourbaki, *General Topology*, IX.3.1 Proposition 4][bourbaki1966b]

Because a topological group is not equipped with a `UniformSpace` instance by default, we must
explicitly provide it in order to consider completeness. See `QuotientGroup.completeSpace` for a
version in which `G` is already equipped with a uniform structure. -/
@[to_additive "The quotient `G ⧸ N` of a complete first countable topological additive group
`G` by a normal additive subgroup is itself complete. Consequently, quotients of Banach spaces by
subspaces are complete. [N. Bourbaki, *General Topology*, IX.3.1 Proposition 4][bourbaki1966b]

Because an additive topological group is not equipped with a `UniformSpace` instance by default,
we must explicitly provide it in order to consider completeness. See
`QuotientAddGroup.completeSpace` for a version in which `G` is already equipped with a uniform
structure."]
instance QuotientGroup.completeSpace' (G : Type u) [Group G] [TopologicalSpace G]
    [TopologicalGroup G] [FirstCountableTopology G] (N : Subgroup G) [N.Normal]
    [@CompleteSpace G (TopologicalGroup.toUniformSpace G)] :
    @CompleteSpace (G ⧸ N) (TopologicalGroup.toUniformSpace (G ⧸ N)) := by
  /- Since `G ⧸ N` is a topological group it is a uniform space, and since `G` is first countable
    the uniformities of both `G` and `G ⧸ N` are countably generated. Moreover, we may choose a
    sequential antitone neighborhood basis `u` for `𝓝 (1 : G)` so that `(u (n + 1)) ^ 2 ⊆ u n`, and
    this descends to an antitone neighborhood basis `v` for `𝓝 (1 : G ⧸ N)`. Since `𝓤 (G ⧸ N)` is
    countably generated, it suffices to show any Cauchy sequence `x` converges. -/
  

Proof state:
G : Type u
inst✝⁵ : Group G
inst✝⁴ : TopologicalSpace G
inst✝³ : TopologicalGroup G
inst✝² : FirstCountableTopology G
N : Subgroup G
inst✝¹ : Subgroup.Normal N
inst✝ : CompleteSpace G
⊢ CompleteSpace (G ⧸ N)

Next tactic:
letI : UniformSpace (G ⧸ N) := TopologicalGroup.toUniformSpace (G ⧸ N)
/-- The quotient `G ⧸ N` of a complete first countable topological group `G` by a normal subgroup
is itself complete. [N. Bourbaki, *General Topology*, IX.3.1 Proposition 4][bourbaki1966b]

Because a topological group is not equipped with a `UniformSpace` instance by default, we must
explicitly provide it in order to consider completeness. See `QuotientGroup.completeSpace` for a
version in which `G` is already equipped with a uniform structure. -/
@[to_additive "The quotient `G ⧸ N` of a complete first countable topological additive group
`G` by a normal additive subgroup is itself complete. Consequently, quotients of Banach spaces by
subspaces are complete. [N. Bourbaki, *General Topology*, IX.3.1 Proposition 4][bourbaki1966b]

Because an additive topological group is not equipped with a `UniformSpace` instance by default,
we must explicitly provide it in order to consider completeness. See
`QuotientAddGroup.completeSpace` for a version in which `G` is already equipped with a uniform
structure."]
instance QuotientGroup.completeSpace' (G : Type u) [Group G] [TopologicalSpace G]
    [TopologicalGroup G] [FirstCountableTopology G] (N : Subgroup G) [N.Normal]
    [@CompleteSpace G (TopologicalGroup.toUniformSpace G)] :
    @CompleteSpace (G ⧸ N) (TopologicalGroup.toUniformSpace (G ⧸ N)) := by
  /- Since `G ⧸ N` is a topological group it is a uniform space, and since `G` is first countable
    the uniformities of both `G` and `G ⧸ N` are countably generated. Moreover, we may choose a
    sequential antitone neighborhood basis `u` for `𝓝 (1 : G)` so that `(u (n + 1)) ^ 2 ⊆ u n`, and
    this descends to an antitone neighborhood basis `v` for `𝓝 (1 : G ⧸ N)`. Since `𝓤 (G ⧸ N)` is
    countably generated, it suffices to show any Cauchy sequence `x` converges. -/
  letI : UniformSpace (G ⧸ N) := TopologicalGroup.toUniformSpace (G ⧸ N)
  

Proof state:
G : Type u
inst✝⁵ : Group G
inst✝⁴ : TopologicalSpace G
inst✝³ : TopologicalGroup G
inst✝² : FirstCountableTopology G
N : Subgroup G
inst✝¹ : Subgroup.Normal N
inst✝ : CompleteSpace G
this : UniformSpace (G ⧸ N) := TopologicalGroup.toUniformSpace (G ⧸ N)
⊢ CompleteSpace (G ⧸ N)

Next tactic:
letI : UniformSpace G := TopologicalGroup.toUniformSpace G
/-- The quotient `G ⧸ N` of a complete first countable topological group `G` by a normal subgroup
is itself complete. [N. Bourbaki, *General Topology*, IX.3.1 Proposition 4][bourbaki1966b]

Because a topological group is not equipped with a `UniformSpace` instance by default, we must
explicitly provide it in order to consider completeness. See `QuotientGroup.completeSpace` for a
version in which `G` is already equipped with a uniform structure. -/
@[to_additive "The quotient `G ⧸ N` of a complete first countable topological additive group
`G` by a normal additive subgroup is itself complete. Consequently, quotients of Banach spaces by
subspaces are complete. [N. Bourbaki, *General Topology*, IX.3.1 Proposition 4][bourbaki1966b]

Because an additive topological group is not equipped with a `UniformSpace` instance by default,
we must explicitly provide it in order to consider completeness. See
`QuotientAddGroup.completeSpace` for a version in which `G` is already equipped with a uniform
structure."]
instance QuotientGroup.completeSpace' (G : Type u) [Group G] [TopologicalSpace G]
    [TopologicalGroup G] [FirstCountableTopology G] (N : Subgroup G) [N.Normal]
    [@CompleteSpace G (TopologicalGroup.toUniformSpace G)] :
    @CompleteSpace (G ⧸ N) (TopologicalGroup.toUniformSpace (G ⧸ N)) := by
  /- Since `G ⧸ N` is a topological group it is a uniform space, and since `G` is first countable
    the uniformities of both `G` and `G ⧸ N` are countably generated. Moreover, we may choose a
    sequential antitone neighborhood basis `u` for `𝓝 (1 : G)` so that `(u (n + 1)) ^ 2 ⊆ u n`, and
    this descends to an antitone neighborhood basis `v` for `𝓝 (1 : G ⧸ N)`. Since `𝓤 (G ⧸ N)` is
    countably generated, it suffices to show any Cauchy sequence `x` converges. -/
  letI : UniformSpace (G ⧸ N) := TopologicalGroup.toUniformSpace (G ⧸ N)
  letI : UniformSpace G := TopologicalGroup.toUniformSpace G
  

Proof state:
G : Type u
inst✝⁵ : Group G
inst✝⁴ : TopologicalSpace G
inst✝³ : TopologicalGroup G
inst✝² : FirstCountableTopology G
N : Subgroup G
inst✝¹ : Subgroup.Normal N
inst✝ : CompleteSpace G
this✝ : UniformSpace (G ⧸ N) := TopologicalGroup.toUniformSpace (G ⧸ N)
this : UniformSpace G := TopologicalGroup.toUniformSpace G
⊢ CompleteSpace (G ⧸ N)

Next tactic:
haveI : (𝓤 (G ⧸ N)).IsCountablyGenerated := comap.isCountablyGenerated _ _
/-- The quotient `G ⧸ N` of a complete first countable topological group `G` by a normal subgroup
is itself complete. [N. Bourbaki, *General Topology*, IX.3.1 Proposition 4][bourbaki1966b]

Because a topological group is not equipped with a `UniformSpace` instance by default, we must
explicitly provide it in order to consider completeness. See `QuotientGroup.completeSpace` for a
version in which `G` is already equipped with a uniform structure. -/
@[to_additive "The quotient `G ⧸ N` of a complete first countable topological additive group
`G` by a normal additive subgroup is itself complete. Consequently, quotients of Banach spaces by
subspaces are complete. [N. Bourbaki, *General Topology*, IX.3.1 Proposition 4][bourbaki1966b]

Because an additive topological group is not equipped with a `UniformSpace` instance by default,
we must explicitly provide it in order to consider completeness. See
`QuotientAddGroup.completeSpace` for a version in which `G` is already equipped with a uniform
structure."]
instance QuotientGroup.completeSpace' (G : Type u) [Group G] [TopologicalSpace G]
    [TopologicalGroup G] [FirstCountableTopology G] (N : Subgroup G) [N.Normal]
    [@CompleteSpace G (TopologicalGroup.toUniformSpace G)] :
    @CompleteSpace (G ⧸ N) (TopologicalGroup.toUniformSpace (G ⧸ N)) := by
  /- Since `G ⧸ N` is a topological group it is a uniform space, and since `G` is first countable
    the uniformities of both `G` and `G ⧸ N` are countably generated. Moreover, we may choose a
    sequential antitone neighborhood basis `u` for `𝓝 (1 : G)` so that `(u (n + 1)) ^ 2 ⊆ u n`, and
    this descends to an antitone neighborhood basis `v` for `𝓝 (1 : G ⧸ N)`. Since `𝓤 (G ⧸ N)` is
    countably generated, it suffices to show any Cauchy sequence `x` converges. -/
  letI : UniformSpace (G ⧸ N) := TopologicalGroup.toUniformSpace (G ⧸ N)
  letI : UniformSpace G := TopologicalGroup.toUniformSpace G
  haveI : (𝓤 (G ⧸ N)).IsCountablyGenerated := comap.isCountablyGenerated _ _
  

Proof state:
G : Type u
inst✝⁵ : Group G
inst✝⁴ : TopologicalSpace G
inst✝³ : TopologicalGroup G
inst✝² : FirstCountableTopology G
N : Subgroup G
inst✝¹ : Subgroup.Normal N
inst✝ : CompleteSpace G
this✝¹ : UniformSpace (G ⧸ N) := TopologicalGroup.toUniformSpace (G ⧸ N)
this✝ : UniformSpace G := TopologicalGroup.toUniformSpace G
this : IsCountablyGenerated (𝓤 (G ⧸ N))
⊢ CompleteSpace (G ⧸ N)

Next tactic:
obtain ⟨u, hu, u_mul⟩ := TopologicalGroup.exists_antitone_basis_nhds_one G
/-- The quotient `G ⧸ N` of a complete first countable topological group `G` by a normal subgroup
is itself complete. [N. Bourbaki, *General Topology*, IX.3.1 Proposition 4][bourbaki1966b]

Because a topological group is not equipped with a `UniformSpace` instance by default, we must
explicitly provide it in order to consider completeness. See `QuotientGroup.completeSpace` for a
version in which `G` is already equipped with a uniform structure. -/
@[to_additive "The quotient `G ⧸ N` of a complete first countable topological additive group
`G` by a normal additive subgroup is itself complete. Consequently, quotients of Banach spaces by
subspaces are complete. [N. Bourbaki, *General Topology*, IX.3.1 Proposition 4][bourbaki1966b]

Because an additive topological group is not equipped with a `UniformSpace` instance by default,
we must explicitly provide it in order to consider completeness. See
`QuotientAddGroup.completeSpace` for a version in which `G` is already equipped with a uniform
structure."]
instance QuotientGroup.completeSpace' (G : Type u) [Group G] [TopologicalSpace G]
    [TopologicalGroup G] [FirstCountableTopology G] (N : Subgroup G) [N.Normal]
    [@CompleteSpace G (TopologicalGroup.toUniformSpace G)] :
    @CompleteSpace (G ⧸ N) (TopologicalGroup.toUniformSpace (G ⧸ N)) := by
  /- Since `G ⧸ N` is a topological group it is a uniform space, and since `G` is first countable
    the uniformities of both `G` and `G ⧸ N` are countably generated. Moreover, we may choose a
    sequential antitone neighborhood basis `u` for `𝓝 (1 : G)` so that `(u (n + 1)) ^ 2 ⊆ u n`, and
    this descends to an antitone neighborhood basis `v` for `𝓝 (1 : G ⧸ N)`. Since `𝓤 (G ⧸ N)` is
    countably generated, it suffices to show any Cauchy sequence `x` converges. -/
  letI : UniformSpace (G ⧸ N) := TopologicalGroup.toUniformSpace (G ⧸ N)
  letI : UniformSpace G := TopologicalGroup.toUniformSpace G
  haveI : (𝓤 (G ⧸ N)).IsCountablyGenerated := comap.isCountablyGenerated _ _
  obtain ⟨u, hu, u_mul⟩ := TopologicalGroup.exists_antitone_basis_nhds_one G
  

Proof state:
case intro.intro
G : Type u
inst✝⁵ : Group G
inst✝⁴ : TopologicalSpace G
inst✝³ : TopologicalGroup G
inst✝² : FirstCountableTopology G
N : Subgroup G
inst✝¹ : Subgroup.Normal N
inst✝ : CompleteSpace G
this✝¹ : UniformSpace (G ⧸ N) := TopologicalGroup.toUniformSpace (G ⧸ N)
this✝ : UniformSpace G := TopologicalGroup.toUniformSpace G
this : IsCountablyGenerated (𝓤 (G ⧸ N))
u : ℕ → Set G
hu : HasAntitoneBasis (𝓝 1) u
u_mul : ∀ (n : ℕ), u (n + 1) * u (n + 1) ⊆ u n
⊢ CompleteSpace (G ⧸ N)

Next tactic:
obtain ⟨hv, v_anti⟩ := @HasAntitoneBasis.map _ _ _ _ _ _ ((↑) : G → G ⧸ N) hu
/-- The quotient `G ⧸ N` of a complete first countable topological group `G` by a normal subgroup
is itself complete. [N. Bourbaki, *General Topology*, IX.3.1 Proposition 4][bourbaki1966b]

Because a topological group is not equipped with a `UniformSpace` instance by default, we must
explicitly provide it in order to consider completeness. See `QuotientGroup.completeSpace` for a
version in which `G` is already equipped with a uniform structure. -/
@[to_additive "The quotient `G ⧸ N` of a complete first countable topological additive group
`G` by a normal additive subgroup is itself complete. Consequently, quotients of Banach spaces by
subspaces are complete. [N. Bourbaki, *General Topology*, IX.3.1 Proposition 4][bourbaki1966b]

Because an additive topological group is not equipped with a `UniformSpace` instance by default,
we must explicitly provide it in order to consider completeness. See
`QuotientAddGroup.completeSpace` for a version in which `G` is already equipped with a uniform
structure."]
instance QuotientGroup.completeSpace' (G : Type u) [Group G] [TopologicalSpace G]
    [TopologicalGroup G] [FirstCountableTopology G] (N : Subgroup G) [N.Normal]
    [@CompleteSpace G (TopologicalGroup.toUniformSpace G)] :
    @CompleteSpace (G ⧸ N) (TopologicalGroup.toUniformSpace (G ⧸ N)) := by
  /- Since `G ⧸ N` is a topological group it is a uniform space, and since `G` is first countable
    the uniformities of both `G` and `G ⧸ N` are countably generated. Moreover, we may choose a
    sequential antitone neighborhood basis `u` for `𝓝 (1 : G)` so that `(u (n + 1)) ^ 2 ⊆ u n`, and
    this descends to an antitone neighborhood basis `v` for `𝓝 (1 : G ⧸ N)`. Since `𝓤 (G ⧸ N)` is
    countably generated, it suffices to show any Cauchy sequence `x` converges. -/
  letI : UniformSpace (G ⧸ N) := TopologicalGroup.toUniformSpace (G ⧸ N)
  letI : UniformSpace G := TopologicalGroup.toUniformSpace G
  haveI : (𝓤 (G ⧸ N)).IsCountablyGenerated := comap.isCountablyGenerated _ _
  obtain ⟨u, hu, u_mul⟩ := TopologicalGroup.exists_antitone_basis_nhds_one G
  obtain ⟨hv, v_anti⟩ := @HasAntitoneBasis.map _ _ _ _ _ _ ((↑) : G → G ⧸ N) hu
  

Proof state:
case intro.intro.mk
G : Type u
inst✝⁵ : Group G
inst✝⁴ : TopologicalSpace G
inst✝³ : TopologicalGroup G
inst✝² : FirstCountableTopology G
N : Subgroup G
inst✝¹ : Subgroup.Normal N
inst✝ : CompleteSpace G
this✝¹ : UniformSpace (G ⧸ N) := TopologicalGroup.toUniformSpace (G ⧸ N)
this✝ : UniformSpace G := TopologicalGroup.toUniformSpace G
this : IsCountablyGenerated (𝓤 (G ⧸ N))
u : ℕ → Set G
hu : HasAntitoneBasis (𝓝 1) u
u_mul : ∀ (n : ℕ), u (n + 1) * u (n + 1) ⊆ u n
hv : HasBasis (Filter.map mk (𝓝 1)) (fun x => True) fun n => mk '' u n
v_anti : Antitone fun n => mk '' u n
⊢ CompleteSpace (G ⧸ N)

Next tactic:
rw [← QuotientGroup.nhds_eq N 1, QuotientGroup.mk_one] at hv
/-- The quotient `G ⧸ N` of a complete first countable topological group `G` by a normal subgroup
is itself complete. [N. Bourbaki, *General Topology*, IX.3.1 Proposition 4][bourbaki1966b]

Because a topological group is not equipped with a `UniformSpace` instance by default, we must
explicitly provide it in order to consider completeness. See `QuotientGroup.completeSpace` for a
version in which `G` is already equipped with a uniform structure. -/
@[to_additive "The quotient `G ⧸ N` of a complete first countable topological additive group
`G` by a normal additive subgroup is itself complete. Consequently, quotients of Banach spaces by
subspaces are complete. [N. Bourbaki, *General Topology*, IX.3.1 Proposition 4][bourbaki1966b]

Because an additive topological group is not equipped with a `UniformSpace` instance by default,
we must explicitly provide it in order to consider completeness. See
`QuotientAddGroup.completeSpace` for a version in which `G` is already equipped with a uniform
structure."]
instance QuotientGroup.completeSpace' (G : Type u) [Group G] [TopologicalSpace G]
    [TopologicalGroup G] [FirstCountableTopology G] (N : Subgroup G) [N.Normal]
    [@CompleteSpace G (TopologicalGroup.toUniformSpace G)] :
    @CompleteSpace (G ⧸ N) (TopologicalGroup.toUniformSpace (G ⧸ N)) := by
  /- Since `G ⧸ N` is a topological group it is a uniform space, and since `G` is first countable
    the uniformities of both `G` and `G ⧸ N` are countably generated. Moreover, we may choose a
    sequential antitone neighborhood basis `u` for `𝓝 (1 : G)` so that `(u (n + 1)) ^ 2 ⊆ u n`, and
    this descends to an antitone neighborhood basis `v` for `𝓝 (1 : G ⧸ N)`. Since `𝓤 (G ⧸ N)` is
    countably generated, it suffices to show any Cauchy sequence `x` converges. -/
  letI : UniformSpace (G ⧸ N) := TopologicalGroup.toUniformSpace (G ⧸ N)
  letI : UniformSpace G := TopologicalGroup.toUniformSpace G
  haveI : (𝓤 (G ⧸ N)).IsCountablyGenerated := comap.isCountablyGenerated _ _
  obtain ⟨u, hu, u_mul⟩ := TopologicalGroup.exists_antitone_basis_nhds_one G
  obtain ⟨hv, v_anti⟩ := @HasAntitoneBasis.map _ _ _ _ _ _ ((↑) : G → G ⧸ N) hu
  rw [← QuotientGroup.nhds_eq N 1, QuotientGroup.mk_one] at hv
  

Proof state:
case intro.intro.mk
G : Type u
inst✝⁵ : Group G
inst✝⁴ : TopologicalSpace G
inst✝³ : TopologicalGroup G
inst✝² : FirstCountableTopology G
N : Subgroup G
inst✝¹ : Subgroup.Normal N
inst✝ : CompleteSpace G
this✝¹ : UniformSpace (G ⧸ N) := TopologicalGroup.toUniformSpace (G ⧸ N)
this✝ : UniformSpace G := TopologicalGroup.toUniformSpace G
this : IsCountablyGenerated (𝓤 (G ⧸ N))
u : ℕ → Set G
hu : HasAntitoneBasis (𝓝 1) u
u_mul : ∀ (n : ℕ), u (n + 1) * u (n + 1) ⊆ u n
hv : HasBasis (𝓝 1) (fun x => True) fun n => mk '' u n
v_anti : Antitone fun n => mk '' u n
⊢ CompleteSpace (G ⧸ N)

Next tactic:
refine' UniformSpace.complete_of_cauchySeq_tendsto fun x hx => _
/-- The quotient `G ⧸ N` of a complete first countable uniform group `G` by a normal subgroup
is itself complete. In contrast to `QuotientGroup.completeSpace'`, in this version `G` is
already equipped with a uniform structure.
[N. Bourbaki, *General Topology*, IX.3.1 Proposition 4][bourbaki1966b]

Even though `G` is equipped with a uniform structure, the quotient `G ⧸ N` does not inherit a
uniform structure, so it is still provided manually via `TopologicalGroup.toUniformSpace`.
In the most common use cases, this coincides (definitionally) with the uniform structure on the
quotient obtained via other means.  -/
@[to_additive "The quotient `G ⧸ N` of a complete first countable uniform additive group
`G` by a normal additive subgroup is itself complete. Consequently, quotients of Banach spaces by
subspaces are complete. In contrast to `QuotientAddGroup.completeSpace'`, in this version
`G` is already equipped with a uniform structure.
[N. Bourbaki, *General Topology*, IX.3.1 Proposition 4][bourbaki1966b]

Even though `G` is equipped with a uniform structure, the quotient `G ⧸ N` does not inherit a
uniform structure, so it is still provided manually via `TopologicalAddGroup.toUniformSpace`.
In the most common use case ─ quotients of normed additive commutative groups by subgroups ─
significant care was taken so that the uniform structure inherent in that setting coincides
(definitionally) with the uniform structure provided here."]
instance QuotientGroup.completeSpace (G : Type u) [Group G] [us : UniformSpace G] [UniformGroup G]
    [FirstCountableTopology G] (N : Subgroup G) [N.Normal] [hG : CompleteSpace G] :
    @CompleteSpace (G ⧸ N) (TopologicalGroup.toUniformSpace (G ⧸ N)) := by
  

Proof state:
G : Type u
inst✝³ : Group G
us : UniformSpace G
inst✝² : UniformGroup G
inst✝¹ : FirstCountableTopology G
N : Subgroup G
inst✝ : Subgroup.Normal N
hG : CompleteSpace G
⊢ CompleteSpace (G ⧸ N)

Next tactic:
rw [← @UniformGroup.toUniformSpace_eq _ us _ _] at hG
/-- The quotient `G ⧸ N` of a complete first countable uniform group `G` by a normal subgroup
is itself complete. In contrast to `QuotientGroup.completeSpace'`, in this version `G` is
already equipped with a uniform structure.
[N. Bourbaki, *General Topology*, IX.3.1 Proposition 4][bourbaki1966b]

Even though `G` is equipped with a uniform structure, the quotient `G ⧸ N` does not inherit a
uniform structure, so it is still provided manually via `TopologicalGroup.toUniformSpace`.
In the most common use cases, this coincides (definitionally) with the uniform structure on the
quotient obtained via other means.  -/
@[to_additive "The quotient `G ⧸ N` of a complete first countable uniform additive group
`G` by a normal additive subgroup is itself complete. Consequently, quotients of Banach spaces by
subspaces are complete. In contrast to `QuotientAddGroup.completeSpace'`, in this version
`G` is already equipped with a uniform structure.
[N. Bourbaki, *General Topology*, IX.3.1 Proposition 4][bourbaki1966b]

Even though `G` is equipped with a uniform structure, the quotient `G ⧸ N` does not inherit a
uniform structure, so it is still provided manually via `TopologicalAddGroup.toUniformSpace`.
In the most common use case ─ quotients of normed additive commutative groups by subgroups ─
significant care was taken so that the uniform structure inherent in that setting coincides
(definitionally) with the uniform structure provided here."]
instance QuotientGroup.completeSpace (G : Type u) [Group G] [us : UniformSpace G] [UniformGroup G]
    [FirstCountableTopology G] (N : Subgroup G) [N.Normal] [hG : CompleteSpace G] :
    @CompleteSpace (G ⧸ N) (TopologicalGroup.toUniformSpace (G ⧸ N)) := by
  rw [← @UniformGroup.toUniformSpace_eq _ us _ _] at hG
  

Proof state:
G : Type u
inst✝³ : Group G
us : UniformSpace G
inst✝² : UniformGroup G
inst✝¹ : FirstCountableTopology G
N : Subgroup G
inst✝ : Subgroup.Normal N
hG : CompleteSpace G
⊢ CompleteSpace (G ⧸ N)

Next tactic:
infer_instance