Theorem sbceq2g 4380

Description: Move proper substitution to second argument of an equality. (Contributed by NM, 30-Nov-2005.)

Assertion

Ref	Expression
sbceq2g	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ 𝐵 = ⦋𝐴 / 𝑥⦌𝐶))

Distinct variable group:   𝑥,𝐵

Allowed substitution hints:   𝐴(𝑥)   𝐶(𝑥)   𝑉(𝑥)

Proof of Theorem sbceq2g

Step	Hyp	Ref	Expression
1		sbceqg 4373	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3870	. . 3 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌𝐵 = 𝐵)
3	2	eqeq1d 2765	. 2 ⊢ (𝐴 ∈ 𝑉 → (⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵 = ⦋𝐴 / 𝑥⦌𝐶))
4	1, 3	bitrd 282	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ 𝐵 = ⦋𝐴 / 𝑥⦌𝐶))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1570   ∈ 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-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  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:  csbsng  4670  csbmpt12  5539  f1od2  8145  mpof1o2d  8174  opsbc2ie  33171  bj-snsetex  38091  csbmpo123  38522  csbfinxpg  38579  poimirlem26  38832  cdlemkid3N  42256  cdlemkid4  42257  brtrclfv2  44988  frege116  45240