Theorem sbccow 3765

Description: A composition law for class substitution. Version of sbcco 3768 with a disjoint variable condition, which requires fewer axioms. (Contributed by NM, 26-Sep-2003.) Avoid ax-13 2404. (Revised by GG, 10-Jan-2024.)

Assertion

Ref	Expression
sbccow	⊢ ([𝐴 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑)

Distinct variable groups:   𝜑,𝑦   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥)   𝐴(𝑥, 𝑦)

Proof of Theorem sbccow

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sbcex 3752	. 2 ⊢ ([𝐴 / 𝑦][𝑦 / 𝑥]𝜑 → 𝐴 ∈ V)
2		sbcex 3752	. 2 ⊢ ([𝐴 / 𝑥]𝜑 → 𝐴 ∈ V)
3		dfsbcq 3744	. . 3 ⊢ (𝑧 = 𝐴 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑦][𝑦 / 𝑥]𝜑))
4		dfsbcq 3744	. . 3 ⊢ (𝑧 = 𝐴 → ([𝑧 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
5		sbsbc 3746	. . . . . 6 ⊢ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑)
6	5	sbbii 2115	. . . . 5 ⊢ ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑦 / 𝑥]𝜑)
7		sbco2vv 2138	. . . . 5 ⊢ ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑)
8		sbsbc 3746	. . . . 5 ⊢ ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑦 / 𝑥]𝜑)
9	6, 7, 8	3bitr3ri 305	. . . 4 ⊢ ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑)
10		sbsbc 3746	. . . 4 ⊢ ([𝑧 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑)
11	9, 10	bitri 278	. . 3 ⊢ ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑)
12	3, 4, 11	vtoclbg 3523	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
13	1, 2, 12	pm5.21nii 381	1 ⊢ ([𝐴 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑)

This proof depends on syntax axioms:   ↔ wb 209  [wsb 2101   ∈ wcel 2147  Vcvv 3454  [wsbc 3742

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-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-v 3456  df-sbc 3743

This theorem is used by:  sbc7  3774  sbccom  3822  sbcralt  3823  csbcow  3866  2nreu  4405  bnj62  35462  bnj610  35489  bnj976  35519  bnj1468  35587  sbccom2  39323  sbccom2f  39324  aomclem6  44327