Metamath Proof Explorer


Theorem psubclsubN

Description: A closed projective subspace is a projective subspace. (Contributed by NM, 23-Jan-2012) (New usage is discouraged.)

Ref Expression
Hypotheses psubclsub.s ⊢ S = PSubSp ⁡ K
psubclsub.c ⊢ C = PSubCl ⁡ K
Assertion psubclsubN ⊢ K ∈ HL ∧ X ∈ C → X ∈ S

Proof

Step Hyp Ref Expression
1 psubclsub.s ⊢ S = PSubSp ⁡ K
2 psubclsub.c ⊢ C = PSubCl ⁡ K
3 eqid ⊢ ⊥ 𝑃 ⁡ K = ⊥ 𝑃 ⁡ K
4 3 2 psubcli2N ⊢ K ∈ HL ∧ X ∈ C → ⊥ 𝑃 ⁡ K ⁡ ⊥ 𝑃 ⁡ K ⁡ X = X
5 eqid ⊢ Atoms ⁡ K = Atoms ⁡ K
6 5 3 2 psubcliN ⊢ K ∈ HL ∧ X ∈ C → X ⊆ Atoms ⁡ K ∧ ⊥ 𝑃 ⁡ K ⁡ ⊥ 𝑃 ⁡ K ⁡ X = X
7 6 simpld ⊢ K ∈ HL ∧ X ∈ C → X ⊆ Atoms ⁡ K
8 5 1 3 polsubN ⊢ K ∈ HL ∧ X ⊆ Atoms ⁡ K → ⊥ 𝑃 ⁡ K ⁡ X ∈ S
9 7 8 syldan ⊢ K ∈ HL ∧ X ∈ C → ⊥ 𝑃 ⁡ K ⁡ X ∈ S
10 5 1 psubssat ⊢ K ∈ HL ∧ ⊥ 𝑃 ⁡ K ⁡ X ∈ S → ⊥ 𝑃 ⁡ K ⁡ X ⊆ Atoms ⁡ K
11 9 10 syldan ⊢ K ∈ HL ∧ X ∈ C → ⊥ 𝑃 ⁡ K ⁡ X ⊆ Atoms ⁡ K
12 5 1 3 polsubN ⊢ K ∈ HL ∧ ⊥ 𝑃 ⁡ K ⁡ X ⊆ Atoms ⁡ K → ⊥ 𝑃 ⁡ K ⁡ ⊥ 𝑃 ⁡ K ⁡ X ∈ S
13 11 12 syldan ⊢ K ∈ HL ∧ X ∈ C → ⊥ 𝑃 ⁡ K ⁡ ⊥ 𝑃 ⁡ K ⁡ X ∈ S
14 4 13 eqeltrrd ⊢ K ∈ HL ∧ X ∈ C → X ∈ S