Metamath Proof Explorer


Theorem 0psubN

Description: The empty set is a projective subspace. Remark below Definition 15.1 of MaedaMaeda p. 61. (Contributed by NM, 13-Oct-2011) (New usage is discouraged.)

Ref Expression
Hypothesis 0psub.s 𝑆 = ( PSubSp ‘ 𝐾 )
Assertion 0psubN ( 𝐾𝑉 → ∅ ∈ 𝑆 )

Proof

Step Hyp Ref Expression
1 0psub.s 𝑆 = ( PSubSp ‘ 𝐾 )
2 0ss ∅ ⊆ ( Atoms ‘ 𝐾 )
3 ral0 𝑝 ∈ ∅ ∀ 𝑞 ∈ ∅ ∀ 𝑟 ∈ ( Atoms ‘ 𝐾 ) ( 𝑟 ( le ‘ 𝐾 ) ( 𝑝 ( join ‘ 𝐾 ) 𝑞 ) → 𝑟 ∈ ∅ )
4 2 3 pm3.2i ( ∅ ⊆ ( Atoms ‘ 𝐾 ) ∧ ∀ 𝑝 ∈ ∅ ∀ 𝑞 ∈ ∅ ∀ 𝑟 ∈ ( Atoms ‘ 𝐾 ) ( 𝑟 ( le ‘ 𝐾 ) ( 𝑝 ( join ‘ 𝐾 ) 𝑞 ) → 𝑟 ∈ ∅ ) )
5 eqid ( le ‘ 𝐾 ) = ( le ‘ 𝐾 )
6 eqid ( join ‘ 𝐾 ) = ( join ‘ 𝐾 )
7 eqid ( Atoms ‘ 𝐾 ) = ( Atoms ‘ 𝐾 )
8 5 6 7 1 ispsubsp ( 𝐾𝑉 → ( ∅ ∈ 𝑆 ↔ ( ∅ ⊆ ( Atoms ‘ 𝐾 ) ∧ ∀ 𝑝 ∈ ∅ ∀ 𝑞 ∈ ∅ ∀ 𝑟 ∈ ( Atoms ‘ 𝐾 ) ( 𝑟 ( le ‘ 𝐾 ) ( 𝑝 ( join ‘ 𝐾 ) 𝑞 ) → 𝑟 ∈ ∅ ) ) ) )
9 4 8 mpbiri ( 𝐾𝑉 → ∅ ∈ 𝑆 )