Metamath Proof Explorer


Theorem resvlemOLD

Description: Obsolete version of resvlem as of 31-Oct-2024. Other elements of a structure restriction. (Contributed by Thierry Arnoux, 6-Sep-2018) (New usage is discouraged.) (Proof modification is discouraged.)

Ref Expression
Hypotheses resvlemOLD.r 𝑅 = ( 𝑊v 𝐴 )
resvlemOLD.e 𝐶 = ( 𝐸𝑊 )
resvlemOLD.f 𝐸 = Slot 𝑁
resvlemOLD.n 𝑁 ∈ ℕ
resvlemOLD.b 𝑁 ≠ 5
Assertion resvlemOLD ( 𝐴𝑉𝐶 = ( 𝐸𝑅 ) )

Proof

Step Hyp Ref Expression
1 resvlemOLD.r 𝑅 = ( 𝑊v 𝐴 )
2 resvlemOLD.e 𝐶 = ( 𝐸𝑊 )
3 resvlemOLD.f 𝐸 = Slot 𝑁
4 resvlemOLD.n 𝑁 ∈ ℕ
5 resvlemOLD.b 𝑁 ≠ 5
6 eqid ( Scalar ‘ 𝑊 ) = ( Scalar ‘ 𝑊 )
7 eqid ( Base ‘ ( Scalar ‘ 𝑊 ) ) = ( Base ‘ ( Scalar ‘ 𝑊 ) )
8 1 6 7 resvid2 ( ( ( Base ‘ ( Scalar ‘ 𝑊 ) ) ⊆ 𝐴𝑊 ∈ V ∧ 𝐴𝑉 ) → 𝑅 = 𝑊 )
9 8 fveq2d ( ( ( Base ‘ ( Scalar ‘ 𝑊 ) ) ⊆ 𝐴𝑊 ∈ V ∧ 𝐴𝑉 ) → ( 𝐸𝑅 ) = ( 𝐸𝑊 ) )
10 9 3expib ( ( Base ‘ ( Scalar ‘ 𝑊 ) ) ⊆ 𝐴 → ( ( 𝑊 ∈ V ∧ 𝐴𝑉 ) → ( 𝐸𝑅 ) = ( 𝐸𝑊 ) ) )
11 1 6 7 resvval2 ( ( ¬ ( Base ‘ ( Scalar ‘ 𝑊 ) ) ⊆ 𝐴𝑊 ∈ V ∧ 𝐴𝑉 ) → 𝑅 = ( 𝑊 sSet ⟨ ( Scalar ‘ ndx ) , ( ( Scalar ‘ 𝑊 ) ↾s 𝐴 ) ⟩ ) )
12 11 fveq2d ( ( ¬ ( Base ‘ ( Scalar ‘ 𝑊 ) ) ⊆ 𝐴𝑊 ∈ V ∧ 𝐴𝑉 ) → ( 𝐸𝑅 ) = ( 𝐸 ‘ ( 𝑊 sSet ⟨ ( Scalar ‘ ndx ) , ( ( Scalar ‘ 𝑊 ) ↾s 𝐴 ) ⟩ ) ) )
13 3 4 ndxid 𝐸 = Slot ( 𝐸 ‘ ndx )
14 3 4 ndxarg ( 𝐸 ‘ ndx ) = 𝑁
15 14 5 eqnetri ( 𝐸 ‘ ndx ) ≠ 5
16 scandx ( Scalar ‘ ndx ) = 5
17 15 16 neeqtrri ( 𝐸 ‘ ndx ) ≠ ( Scalar ‘ ndx )
18 13 17 setsnid ( 𝐸𝑊 ) = ( 𝐸 ‘ ( 𝑊 sSet ⟨ ( Scalar ‘ ndx ) , ( ( Scalar ‘ 𝑊 ) ↾s 𝐴 ) ⟩ ) )
19 12 18 eqtr4di ( ( ¬ ( Base ‘ ( Scalar ‘ 𝑊 ) ) ⊆ 𝐴𝑊 ∈ V ∧ 𝐴𝑉 ) → ( 𝐸𝑅 ) = ( 𝐸𝑊 ) )
20 19 3expib ( ¬ ( Base ‘ ( Scalar ‘ 𝑊 ) ) ⊆ 𝐴 → ( ( 𝑊 ∈ V ∧ 𝐴𝑉 ) → ( 𝐸𝑅 ) = ( 𝐸𝑊 ) ) )
21 10 20 pm2.61i ( ( 𝑊 ∈ V ∧ 𝐴𝑉 ) → ( 𝐸𝑅 ) = ( 𝐸𝑊 ) )
22 reldmresv Rel dom ↾v
23 22 ovprc1 ( ¬ 𝑊 ∈ V → ( 𝑊v 𝐴 ) = ∅ )
24 1 23 eqtrid ( ¬ 𝑊 ∈ V → 𝑅 = ∅ )
25 24 fveq2d ( ¬ 𝑊 ∈ V → ( 𝐸𝑅 ) = ( 𝐸 ‘ ∅ ) )
26 3 str0 ∅ = ( 𝐸 ‘ ∅ )
27 25 26 eqtr4di ( ¬ 𝑊 ∈ V → ( 𝐸𝑅 ) = ∅ )
28 fvprc ( ¬ 𝑊 ∈ V → ( 𝐸𝑊 ) = ∅ )
29 27 28 eqtr4d ( ¬ 𝑊 ∈ V → ( 𝐸𝑅 ) = ( 𝐸𝑊 ) )
30 29 adantr ( ( ¬ 𝑊 ∈ V ∧ 𝐴𝑉 ) → ( 𝐸𝑅 ) = ( 𝐸𝑊 ) )
31 21 30 pm2.61ian ( 𝐴𝑉 → ( 𝐸𝑅 ) = ( 𝐸𝑊 ) )
32 2 31 eqtr4id ( 𝐴𝑉𝐶 = ( 𝐸𝑅 ) )