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 ∧ S ∈ L → A ⁡ S = 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 ∧ S ∈ L → W ∈ AssAlg
5 2 subrgss ⊢ S ∈ SubRing ⁡ W → S ⊆ V
6 5 3ad2ant2 ⊢ W ∈ AssAlg ∧ S ∈ SubRing ⁡ W ∧ S ∈ L → S ⊆ V
7 1 2 3 aspval ⊢ W ∈ AssAlg ∧ S ⊆ V → A ⁡ S = ⋂ t ∈ SubRing ⁡ W ∩ L | S ⊆ t
8 4 6 7 syl2anc ⊢ W ∈ AssAlg ∧ S ∈ SubRing ⁡ W ∧ S ∈ L → A ⁡ S = ⋂ t ∈ SubRing ⁡ W ∩ L | S ⊆ t
9 3simpc ⊢ W ∈ AssAlg ∧ S ∈ SubRing ⁡ W ∧ S ∈ L → S ∈ SubRing ⁡ W ∧ S ∈ L
10 elin ⊢ S ∈ SubRing ⁡ W ∩ L ↔ S ∈ SubRing ⁡ W ∧ S ∈ L
11 9 10 sylibr ⊢ W ∈ AssAlg ∧ S ∈ SubRing ⁡ W ∧ S ∈ L → S ∈ SubRing ⁡ W ∩ L
12 intmin ⊢ S ∈ SubRing ⁡ W ∩ L → ⋂ t ∈ SubRing ⁡ W ∩ L | S ⊆ t = S
13 11 12 syl ⊢ W ∈ AssAlg ∧ S ∈ SubRing ⁡ W ∧ S ∈ L → ⋂ t ∈ SubRing ⁡ W ∩ L | S ⊆ t = S
14 8 13 eqtrd ⊢ W ∈ AssAlg ∧ S ∈ SubRing ⁡ W ∧ S ∈ L → A ⁡ S = S