Theorem cbvcsbw 3856

Description: Change bound variables in a class substitution. Interestingly, this does not require any bound variable conditions on 𝐴. Version of cbvcsb 3857 with a disjoint variable condition, which does not require ax-13 2401. (Contributed by Jeff Hankins, 13-Sep-2009.) Avoid ax-13 2401. (Revised by GG, 10-Jan-2024.)

Hypotheses

Ref	Expression
cbvcsbw.1	⊢ Ⅎ𝑦𝐶
cbvcsbw.2	⊢ Ⅎ𝑥𝐷
cbvcsbw.3	⊢ (𝑥 = 𝑦 → 𝐶 = 𝐷)

Assertion

Ref	Expression
cbvcsbw	⊢ ⦋𝐴 / 𝑥⦌𝐶 = ⦋𝐴 / 𝑦⦌𝐷

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝐴(𝑥, 𝑦)   𝐶(𝑥, 𝑦)   𝐷(𝑥, 𝑦)

Proof of Theorem cbvcsbw

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		cbvcsbw.1	. . . . 5 ⊢ Ⅎ𝑦𝐶
2	1	nfcri 2914	. . . 4 ⊢ Ⅎ𝑦 𝑧 ∈ 𝐶
3		cbvcsbw.2	. . . . 5 ⊢ Ⅎ𝑥𝐷
4	3	nfcri 2914	. . . 4 ⊢ Ⅎ𝑥 𝑧 ∈ 𝐷
5		cbvcsbw.3	. . . . 5 ⊢ (𝑥 = 𝑦 → 𝐶 = 𝐷)
6	5	eleq2d 2846	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑧 ∈ 𝐶 ↔ 𝑧 ∈ 𝐷))
7	2, 4, 6	cbvsbcw 3771	. . 3 ⊢ ([𝐴 / 𝑥]𝑧 ∈ 𝐶 ↔ [𝐴 / 𝑦]𝑧 ∈ 𝐷)
8	7	abbii 2827	. 2 ⊢ {𝑧 ∣ [𝐴 / 𝑥]𝑧 ∈ 𝐶} = {𝑧 ∣ [𝐴 / 𝑦]𝑧 ∈ 𝐷}
9		df-csb 3847	. 2 ⊢ ⦋𝐴 / 𝑥⦌𝐶 = {𝑧 ∣ [𝐴 / 𝑥]𝑧 ∈ 𝐶}
10		df-csb 3847	. 2 ⊢ ⦋𝐴 / 𝑦⦌𝐷 = {𝑧 ∣ [𝐴 / 𝑦]𝑧 ∈ 𝐷}
11	8, 9, 10	3eqtr4i 2793	1 ⊢ ⦋𝐴 / 𝑥⦌𝐶 = ⦋𝐴 / 𝑦⦌𝐷

This proof depends on syntax axioms:   → wi 4   = wceq 1569   ∈ wcel 2144  {cab 2738  Ⅎwnfc 2907  [wsbc 3738  ⦋csb 3846

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-11 2193  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-sbc 3739  df-csb 3847

This theorem is used by:  cbvsum  15867  cbvprod  16087  measiuns  34896  poimirlem26  38657  climinf2mpt  46817  climinfmpt  46818