Theorem sbcco3gw 4386

Description: Composition of two substitutions. Version of sbcco3g 4391 with a disjoint variable condition, which does not require ax-13 2404. (Contributed by NM, 27-Nov-2005.) Avoid ax-13 2404. (Revised by GG, 26-Jan-2024.)

Hypothesis

Ref	Expression
sbcco3gw.1	⊢ (𝑥 = 𝐴 → 𝐵 = 𝐶)

Assertion

Ref	Expression
sbcco3gw	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥][𝐵 / 𝑦]𝜑 ↔ [𝐶 / 𝑦]𝜑))

Distinct variable groups:   𝑥,𝐴   𝜑,𝑥   𝑥,𝐶   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑦)   𝐴(𝑦)   𝐵(𝑥, 𝑦)   𝐶(𝑦)   𝑉(𝑥, 𝑦)

Proof of Theorem sbcco3gw

Step	Hyp	Ref	Expression
1		sbcnestgw 4384	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥][𝐵 / 𝑦]𝜑 ↔ [⦋𝐴 / 𝑥⦌𝐵 / 𝑦]𝜑))
2		elex 3475	. . 3 ⊢ (𝐴 ∈ 𝑉 → 𝐴 ∈ V)
3		nfcvd 2927	. . . 4 ⊢ (𝐴 ∈ V → Ⅎ𝑥𝐶)
4		sbcco3gw.1	. . . 4 ⊢ (𝑥 = 𝐴 → 𝐵 = 𝐶)
5	3, 4	csbiegf 3884	. . 3 ⊢ (𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐵 = 𝐶)
6		dfsbcq 3744	. . 3 ⊢ (⦋𝐴 / 𝑥⦌𝐵 = 𝐶 → ([⦋𝐴 / 𝑥⦌𝐵 / 𝑦]𝜑 ↔ [𝐶 / 𝑦]𝜑))
7	2, 5, 6	3syl 19	. 2 ⊢ (𝐴 ∈ 𝑉 → ([⦋𝐴 / 𝑥⦌𝐵 / 𝑦]𝜑 ↔ [𝐶 / 𝑦]𝜑))
8	1, 7	bitrd 282	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥][𝐵 / 𝑦]𝜑 ↔ [𝐶 / 𝑦]𝜑))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1570   ∈ wcel 2147  Vcvv 3454  [wsbc 3742  ⦋csb 3851

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-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-v 3456  df-sbc 3743  df-csb 3852

This theorem is used by:  fzshftral  13776  2rexfrabdioph  43996  3rexfrabdioph  43997  4rexfrabdioph  43998  6rexfrabdioph  43999  7rexfrabdioph  44000