Theorem sbceq2g 4384

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 4377	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3873	. . 3 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌𝐵 = 𝐵)
3	2	eqeq1d 2767	. 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 2146  [wsbc 3746  ⦋csb 3854

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 2148  ax-9 2156  ax-10 2179  ax-11 2195  ax-12 2216  ax-ext 2737

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 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-nfc 2914  df-v 3459  df-sbc 3747  df-csb 3855

This theorem is used by:  csbsng  4676  csbmpt12  5544  mpof1o2d  8127  opsbc2ie  32895  f1od2  33136  bj-snsetex  37658  csbmpo123  38036  csbfinxpg  38093  poimirlem26  38356  cdlemkid3N  41767  cdlemkid4  41768  brtrclfv2  44513  frege116  44765