Theorem dvhvscacbv 41958

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 6881	. . . 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 4844	. . 3 ⊢ (𝑠 = 𝑡 → ⟨(𝑠‘(1st ‘𝑓)), (𝑠 ∘ (2nd ‘𝑓))⟩ = ⟨(𝑡‘(1st ‘𝑓)), (𝑡 ∘ (2nd ‘𝑓))⟩)
5		2fveq3 6887	. . . 4 ⊢ (𝑓 = 𝑔 → (𝑡‘(1st ‘𝑓)) = (𝑡‘(1st ‘𝑔)))
6		fveq2 6882	. . . . 5 ⊢ (𝑓 = 𝑔 → (2nd ‘𝑓) = (2nd ‘𝑔))
7	6	coeq2d 5846	. . . 4 ⊢ (𝑓 = 𝑔 → (𝑡 ∘ (2nd ‘𝑓)) = (𝑡 ∘ (2nd ‘𝑔)))
8	5, 7	opeq12d 4844	. . 3 ⊢ (𝑓 = 𝑔 → ⟨(𝑡‘(1st ‘𝑓)), (𝑡 ∘ (2nd ‘𝑓))⟩ = ⟨(𝑡‘(1st ‘𝑔)), (𝑡 ∘ (2nd ‘𝑔))⟩)
9	4, 8	cbvmpov 7511	. 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 2785	1 ⊢ · = (𝑡 ∈ 𝐸, 𝑔 ∈ (𝑇 × 𝐸) ↦ ⟨(𝑡‘(1st ‘𝑔)), (𝑡 ∘ (2nd ‘𝑔))⟩)

This proof depends on syntax axioms:   = wceq 1570  ⟨cop 4593   × cxp 5657   ∘ ccom 5663  ‘cfv 6537   ∈ cmpo 7418  1^st c1st 7987  2^nd c2nd 7988

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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734

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 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3415  df-v 3455  df-dif 3905  df-un 3907  df-ss 3919  df-nul 4283  df-if 4486  df-sn 4588  df-pr 4590  df-op 4594  df-uni 4871  df-br 5108  df-opab 5172  df-co 5668  df-iota 6493  df-fv 6545  df-oprab 7420  df-mpo 7421

This theorem is used by:  dvhvscaval  41959