Metamath Proof Explorer


Theorem aspid

Description: The algebraic span of a subalgebra is itself. ( spanid analog.) (Contributed by Mario Carneiro, 7-Jan-2015)

Ref Expression
Hypotheses aspval.a 𝐴 = ( AlgSpan ‘ 𝑊 )
aspval.v 𝑉 = ( Base ‘ 𝑊 )
aspval.l 𝐿 = ( LSubSp ‘ 𝑊 )
Assertion aspid ( ( 𝑊 ∈ AssAlg ∧ 𝑆 ∈ ( SubRing ‘ 𝑊 ) ∧ 𝑆𝐿 ) → ( 𝐴𝑆 ) = 𝑆 )

Proof

Step Hyp Ref Expression
1 aspval.a 𝐴 = ( AlgSpan ‘ 𝑊 )
2 aspval.v 𝑉 = ( Base ‘ 𝑊 )
3 aspval.l 𝐿 = ( LSubSp ‘ 𝑊 )
4 simp1 ( ( 𝑊 ∈ AssAlg ∧ 𝑆 ∈ ( SubRing ‘ 𝑊 ) ∧ 𝑆𝐿 ) → 𝑊 ∈ AssAlg )
5 2 subrgss ( 𝑆 ∈ ( SubRing ‘ 𝑊 ) → 𝑆𝑉 )
6 5 3ad2ant2 ( ( 𝑊 ∈ AssAlg ∧ 𝑆 ∈ ( SubRing ‘ 𝑊 ) ∧ 𝑆𝐿 ) → 𝑆𝑉 )
7 1 2 3 aspval ( ( 𝑊 ∈ AssAlg ∧ 𝑆𝑉 ) → ( 𝐴𝑆 ) = { 𝑡 ∈ ( ( SubRing ‘ 𝑊 ) ∩ 𝐿 ) ∣ 𝑆𝑡 } )
8 4 6 7 syl2anc ( ( 𝑊 ∈ AssAlg ∧ 𝑆 ∈ ( SubRing ‘ 𝑊 ) ∧ 𝑆𝐿 ) → ( 𝐴𝑆 ) = { 𝑡 ∈ ( ( SubRing ‘ 𝑊 ) ∩ 𝐿 ) ∣ 𝑆𝑡 } )
9 3simpc ( ( 𝑊 ∈ AssAlg ∧ 𝑆 ∈ ( SubRing ‘ 𝑊 ) ∧ 𝑆𝐿 ) → ( 𝑆 ∈ ( SubRing ‘ 𝑊 ) ∧ 𝑆𝐿 ) )
10 elin ( 𝑆 ∈ ( ( SubRing ‘ 𝑊 ) ∩ 𝐿 ) ↔ ( 𝑆 ∈ ( SubRing ‘ 𝑊 ) ∧ 𝑆𝐿 ) )
11 9 10 sylibr ( ( 𝑊 ∈ AssAlg ∧ 𝑆 ∈ ( SubRing ‘ 𝑊 ) ∧ 𝑆𝐿 ) → 𝑆 ∈ ( ( SubRing ‘ 𝑊 ) ∩ 𝐿 ) )
12 intmin ( 𝑆 ∈ ( ( SubRing ‘ 𝑊 ) ∩ 𝐿 ) → { 𝑡 ∈ ( ( SubRing ‘ 𝑊 ) ∩ 𝐿 ) ∣ 𝑆𝑡 } = 𝑆 )
13 11 12 syl ( ( 𝑊 ∈ AssAlg ∧ 𝑆 ∈ ( SubRing ‘ 𝑊 ) ∧ 𝑆𝐿 ) → { 𝑡 ∈ ( ( SubRing ‘ 𝑊 ) ∩ 𝐿 ) ∣ 𝑆𝑡 } = 𝑆 )
14 8 13 eqtrd ( ( 𝑊 ∈ AssAlg ∧ 𝑆 ∈ ( SubRing ‘ 𝑊 ) ∧ 𝑆𝐿 ) → ( 𝐴𝑆 ) = 𝑆 )