Theorem sbcnestg 4389

Description: Nest the composition of two substitutions. Usage of this theorem is discouraged because it depends on ax-13 2404. Use the weaker sbcnestgw 4384 when possible. (Contributed by NM, 27-Nov-2005.) (Proof shortened by Mario Carneiro, 11-Nov-2016.) (New usage is discouraged.)

Assertion

Ref	Expression
sbcnestg	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥][𝐵 / 𝑦]𝜑 ↔ [⦋𝐴 / 𝑥⦌𝐵 / 𝑦]𝜑))

Distinct variable group:   𝜑,𝑥

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

Proof of Theorem sbcnestg

Step	Hyp	Ref	Expression
1		nfv 1949	. . 3 ⊢ Ⅎ𝑥𝜑
2	1	ax-gen 1828	. 2 ⊢ ∀𝑦Ⅎ𝑥𝜑
3		sbcnestgf 4387	. 2 ⊢ ((𝐴 ∈ 𝑉 ∧ ∀𝑦Ⅎ𝑥𝜑) → ([𝐴 / 𝑥][𝐵 / 𝑦]𝜑 ↔ [⦋𝐴 / 𝑥⦌𝐵 / 𝑦]𝜑))
4	2, 3	mpan2 704	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥][𝐵 / 𝑦]𝜑 ↔ [⦋𝐴 / 𝑥⦌𝐵 / 𝑦]𝜑))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1568  Ⅎwnf 1816   ∈ wcel 2147  [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-13 2404  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:  sbcco3g  4391