Theorem dvhvscacbv 42347

Description: Change bound variables to isolate them later. (Contributed by NM, 20-Nov-2013.)

Hypothesis

Ref	Expression
dvhvscaval.s	⊢ · = (𝑠 ∈ 𝐸, 𝑓 ∈ (𝑇 × 𝐸) ↦ ⟨(𝑠‘(1st ‘𝑓)), (𝑠 ∘ (2nd ‘𝑓))⟩)

Assertion

Ref	Expression
dvhvscacbv	⊢ · = (𝑡 ∈ 𝐸, 𝑔 ∈ (𝑇 × 𝐸) ↦ ⟨(𝑡‘(1st ‘𝑔)), (𝑡 ∘ (2nd ‘𝑔))⟩)

Distinct variable groups:   𝑓,𝑠,𝑡,𝑔,𝐸   𝑇,𝑠,𝑓,𝑡,𝑔

Allowed substitution hints:   · (𝑡, 𝑓, 𝑔, 𝑠)

Proof of Theorem dvhvscacbv

Step	Hyp	Ref	Expression
1		dvhvscaval.s	. 2 ⊢ · = (𝑠 ∈ 𝐸, 𝑓 ∈ (𝑇 × 𝐸) ↦ ⟨(𝑠‘(1st ‘𝑓)), (𝑠 ∘ (2nd ‘𝑓))⟩)
2		fveq1 6907	. . . 4 ⊢ (𝑠 = 𝑡 → (𝑠‘(1st ‘𝑓)) = (𝑡‘(1st ‘𝑓)))
3		coeq1 5841	. . . 4 ⊢ (𝑠 = 𝑡 → (𝑠 ∘ (2nd ‘𝑓)) = (𝑡 ∘ (2nd ‘𝑓)))
4	2, 3	opeq12d 4851	. . 3 ⊢ (𝑠 = 𝑡 → ⟨(𝑠‘(1st ‘𝑓)), (𝑠 ∘ (2nd ‘𝑓))⟩ = ⟨(𝑡‘(1st ‘𝑓)), (𝑡 ∘ (2nd ‘𝑓))⟩)
5		2fveq3 6913	. . . 4 ⊢ (𝑓 = 𝑔 → (𝑡‘(1st ‘𝑓)) = (𝑡‘(1st ‘𝑔)))
6		fveq2 6908	. . . . 5 ⊢ (𝑓 = 𝑔 → (2nd ‘𝑓) = (2nd ‘𝑔))
7	6	coeq2d 5846	. . . 4 ⊢ (𝑓 = 𝑔 → (𝑡 ∘ (2nd ‘𝑓)) = (𝑡 ∘ (2nd ‘𝑔)))
8	5, 7	opeq12d 4851	. . 3 ⊢ (𝑓 = 𝑔 → ⟨(𝑡‘(1st ‘𝑓)), (𝑡 ∘ (2nd ‘𝑓))⟩ = ⟨(𝑡‘(1st ‘𝑔)), (𝑡 ∘ (2nd ‘𝑔))⟩)
9	4, 8	cbvmpov 7538	. 2 ⊢ (𝑠 ∈ 𝐸, 𝑓 ∈ (𝑇 × 𝐸) ↦ ⟨(𝑠‘(1st ‘𝑓)), (𝑠 ∘ (2nd ‘𝑓))⟩) = (𝑡 ∈ 𝐸, 𝑔 ∈ (𝑇 × 𝐸) ↦ ⟨(𝑡‘(1st ‘𝑔)), (𝑡 ∘ (2nd ‘𝑔))⟩)
10	1, 9	eqtri 2786	1 ⊢ · = (𝑡 ∈ 𝐸, 𝑔 ∈ (𝑇 × 𝐸) ↦ ⟨(𝑡‘(1st ‘𝑔)), (𝑡 ∘ (2nd ‘𝑔))⟩)

This proof depends on syntax axioms:   = wceq 1570  ⟨cop 4838   × cxp 5653   ∘ ccom 5659  ‘cfv 6558   ∈ cmpo 7445  1^st c1st 8023  2^nd c2nd 8024

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-uni 4871  df-br 5107  df-opab 5171  df-co 5664  df-iota 6512  df-fv 6566  df-oprab 7447  df-mpo 7448

This theorem is used by:  dvhvscaval  42348