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Metamath Proof Explorer

Theorem tendoinvcl

Description: Closure of multiplicative inverse for endomorphism. We use the scalar inverse of the vector space since it is much simpler than the direct inverse of cdleml8 . (Contributed by NM, 10-Apr-2014) (Revised by Mario Carneiro, 23-Jun-2014)

Ref Expression
Hypotheses tendoinv.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
tendoinv.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
tendoinv.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
tendoinv.e ⊢ 𝐸 = ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 )
tendoinv.o ⊢ 𝑂 = ( ℎ ∈ 𝑇 ↦ ( I ↾ 𝐵 ) )
tendoinv.u ⊢ 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
tendoinv.f ⊢ 𝐹 = ( Scalar ‘ 𝑈 )
tendoinv.n ⊢ 𝑁 = ( invr ‘ 𝐹 )
Assertion tendoinvcl ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑆 ∈ 𝐸 ∧ 𝑆 ≠ 𝑂 ) → ( ( 𝑁 ‘ 𝑆 ) ∈ 𝐸 ∧ ( 𝑁 ‘ 𝑆 ) ≠ 𝑂 ) )

Proof

Step Hyp Ref Expression
1 tendoinv.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
2 tendoinv.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
3 tendoinv.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
4 tendoinv.e ⊢ 𝐸 = ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 )
5 tendoinv.o ⊢ 𝑂 = ( ℎ ∈ 𝑇 ↦ ( I ↾ 𝐵 ) )
6 tendoinv.u ⊢ 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
7 tendoinv.f ⊢ 𝐹 = ( Scalar ‘ 𝑈 )
8 tendoinv.n ⊢ 𝑁 = ( invr ‘ 𝐹 )
9 eqid ⊢ ( ( EDRing ‘ 𝐾 ) ‘ 𝑊 ) = ( ( EDRing ‘ 𝐾 ) ‘ 𝑊 )
10 2 9 6 7 dvhsca ⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → 𝐹 = ( ( EDRing ‘ 𝐾 ) ‘ 𝑊 ) )
11 2 9 erngdv ⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → ( ( EDRing ‘ 𝐾 ) ‘ 𝑊 ) ∈ DivRing )
12 10 11 eqeltrd ⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → 𝐹 ∈ DivRing )
13 12 3ad2ant1 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑆 ∈ 𝐸 ∧ 𝑆 ≠ 𝑂 ) → 𝐹 ∈ DivRing )
14 simp2 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑆 ∈ 𝐸 ∧ 𝑆 ≠ 𝑂 ) → 𝑆 ∈ 𝐸 )
15 eqid ⊢ ( Base ‘ 𝐹 ) = ( Base ‘ 𝐹 )
16 2 4 6 7 15 dvhbase ⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → ( Base ‘ 𝐹 ) = 𝐸 )
17 16 3ad2ant1 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑆 ∈ 𝐸 ∧ 𝑆 ≠ 𝑂 ) → ( Base ‘ 𝐹 ) = 𝐸 )
18 14 17 eleqtrrd ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑆 ∈ 𝐸 ∧ 𝑆 ≠ 𝑂 ) → 𝑆 ∈ ( Base ‘ 𝐹 ) )
19 simp3 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑆 ∈ 𝐸 ∧ 𝑆 ≠ 𝑂 ) → 𝑆 ≠ 𝑂 )
20 10 fveq2d ⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → ( 0g ‘ 𝐹 ) = ( 0g ‘ ( ( EDRing ‘ 𝐾 ) ‘ 𝑊 ) ) )
21 eqid ⊢ ( 0g ‘ ( ( EDRing ‘ 𝐾 ) ‘ 𝑊 ) ) = ( 0g ‘ ( ( EDRing ‘ 𝐾 ) ‘ 𝑊 ) )
22 1 2 3 9 5 21 erng0g ⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → ( 0g ‘ ( ( EDRing ‘ 𝐾 ) ‘ 𝑊 ) ) = 𝑂 )
23 20 22 eqtrd ⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → ( 0g ‘ 𝐹 ) = 𝑂 )
24 23 3ad2ant1 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑆 ∈ 𝐸 ∧ 𝑆 ≠ 𝑂 ) → ( 0g ‘ 𝐹 ) = 𝑂 )
25 19 24 neeqtrrd ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑆 ∈ 𝐸 ∧ 𝑆 ≠ 𝑂 ) → 𝑆 ≠ ( 0g ‘ 𝐹 ) )
26 eqid ⊢ ( 0g ‘ 𝐹 ) = ( 0g ‘ 𝐹 )
27 15 26 8 drnginvrcl ⊢ ( ( 𝐹 ∈ DivRing ∧ 𝑆 ∈ ( Base ‘ 𝐹 ) ∧ 𝑆 ≠ ( 0g ‘ 𝐹 ) ) → ( 𝑁 ‘ 𝑆 ) ∈ ( Base ‘ 𝐹 ) )
28 13 18 25 27 syl3anc ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑆 ∈ 𝐸 ∧ 𝑆 ≠ 𝑂 ) → ( 𝑁 ‘ 𝑆 ) ∈ ( Base ‘ 𝐹 ) )
29 28 17 eleqtrd ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑆 ∈ 𝐸 ∧ 𝑆 ≠ 𝑂 ) → ( 𝑁 ‘ 𝑆 ) ∈ 𝐸 )
30 15 26 8 drnginvrn0 ⊢ ( ( 𝐹 ∈ DivRing ∧ 𝑆 ∈ ( Base ‘ 𝐹 ) ∧ 𝑆 ≠ ( 0g ‘ 𝐹 ) ) → ( 𝑁 ‘ 𝑆 ) ≠ ( 0g ‘ 𝐹 ) )
31 13 18 25 30 syl3anc ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑆 ∈ 𝐸 ∧ 𝑆 ≠ 𝑂 ) → ( 𝑁 ‘ 𝑆 ) ≠ ( 0g ‘ 𝐹 ) )
32 31 24 neeqtrd ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑆 ∈ 𝐸 ∧ 𝑆 ≠ 𝑂 ) → ( 𝑁 ‘ 𝑆 ) ≠ 𝑂 )
33 29 32 jca ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑆 ∈ 𝐸 ∧ 𝑆 ≠ 𝑂 ) → ( ( 𝑁 ‘ 𝑆 ) ∈ 𝐸 ∧ ( 𝑁 ‘ 𝑆 ) ≠ 𝑂 ) )