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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	⊢ ( 𝐾 ∈ 𝑉 → ∅ ∈ 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	⊢ ( 𝐾 ∈ 𝑉 → ( ∅ ∈ 𝑆 ↔ ( ∅ ⊆ ( Atoms ‘ 𝐾 ) ∧ ∀ 𝑝 ∈ ∅ ∀ 𝑞 ∈ ∅ ∀ 𝑟 ∈ ( Atoms ‘ 𝐾 ) ( 𝑟 ( le ‘ 𝐾 ) ( 𝑝 ( join ‘ 𝐾 ) 𝑞 ) → 𝑟 ∈ ∅ ) ) ) )
9	4 8	mpbiri	⊢ ( 𝐾 ∈ 𝑉 → ∅ ∈ 𝑆 )