Metamath Proof Explorer


Theorem lshplss

Description: A hyperplane is a subspace. (Contributed by NM, 3-Jul-2014)

Ref Expression
Hypotheses lshplss.s 𝑆 = ( LSubSp ‘ 𝑊 )
lshplss.h 𝐻 = ( LSHyp ‘ 𝑊 )
lshplss.w ( 𝜑𝑊 ∈ LMod )
lshplss.u ( 𝜑𝑈𝐻 )
Assertion lshplss ( 𝜑𝑈𝑆 )

Proof

Step Hyp Ref Expression
1 lshplss.s 𝑆 = ( LSubSp ‘ 𝑊 )
2 lshplss.h 𝐻 = ( LSHyp ‘ 𝑊 )
3 lshplss.w ( 𝜑𝑊 ∈ LMod )
4 lshplss.u ( 𝜑𝑈𝐻 )
5 eqid ( Base ‘ 𝑊 ) = ( Base ‘ 𝑊 )
6 eqid ( LSpan ‘ 𝑊 ) = ( LSpan ‘ 𝑊 )
7 5 6 1 2 islshp ( 𝑊 ∈ LMod → ( 𝑈𝐻 ↔ ( 𝑈𝑆𝑈 ≠ ( Base ‘ 𝑊 ) ∧ ∃ 𝑣 ∈ ( Base ‘ 𝑊 ) ( ( LSpan ‘ 𝑊 ) ‘ ( 𝑈 ∪ { 𝑣 } ) ) = ( Base ‘ 𝑊 ) ) ) )
8 3 7 syl ( 𝜑 → ( 𝑈𝐻 ↔ ( 𝑈𝑆𝑈 ≠ ( Base ‘ 𝑊 ) ∧ ∃ 𝑣 ∈ ( Base ‘ 𝑊 ) ( ( LSpan ‘ 𝑊 ) ‘ ( 𝑈 ∪ { 𝑣 } ) ) = ( Base ‘ 𝑊 ) ) ) )
9 4 8 mpbid ( 𝜑 → ( 𝑈𝑆𝑈 ≠ ( Base ‘ 𝑊 ) ∧ ∃ 𝑣 ∈ ( Base ‘ 𝑊 ) ( ( LSpan ‘ 𝑊 ) ‘ ( 𝑈 ∪ { 𝑣 } ) ) = ( Base ‘ 𝑊 ) ) )
10 9 simp1d ( 𝜑𝑈𝑆 )