Theorem sbcco3gw 4386

Description: Composition of two substitutions. Version of sbcco3g 4391 with a disjoint variable condition, which does not require ax-13 2403. (Contributed by NM, 27-Nov-2005.) Avoid ax-13 2403. (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 3474	. . 3 ⊢ (𝐴 ∈ 𝑉 → 𝐴 ∈ V)
3		nfcvd 2925	. . . 4 ⊢ (𝐴 ∈ V → Ⅎ𝑥𝐶)
4		sbcco3gw.1	. . . 4 ⊢ (𝑥 = 𝐴 → 𝐵 = 𝐶)
5	3, 4	csbiegf 3883	. . 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 2145  Vcvv 3453  [wsbc 3742  ⦋csb 3850

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-10 2178  ax-11 2194  ax-12 2215  ax-ext 2734

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 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-nfc 2911  df-v 3455  df-sbc 3743  df-csb 3851

This theorem is used by:  fzshftral  13672  2rexfrabdioph  43624  3rexfrabdioph  43625  4rexfrabdioph  43626  6rexfrabdioph  43627  7rexfrabdioph  43628