Theorem bj-sbeqALT 35012

Description: Substitution in an equality (use the more general version bj-sbeq 35013 instead, without disjoint variable condition). (Contributed by BJ, 6-Oct-2018.) (New usage is discouraged.) (Proof modification is discouraged.)

Assertion

Ref	Expression
bj-sbeqALT	⊢ ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 = ⦋𝑦 / 𝑥⦌𝐵)

Distinct variable group:   𝑥,𝑦

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

Proof of Theorem bj-sbeqALT

Step	Hyp	Ref	Expression
1		nfcsb1v 3853	. . 3 ⊢ Ⅎ𝑥⦋𝑦 / 𝑥⦌𝐴
2		nfcsb1v 3853	. . 3 ⊢ Ⅎ𝑥⦋𝑦 / 𝑥⦌𝐵
3	1, 2	nfeq 2919	. 2 ⊢ Ⅎ𝑥⦋𝑦 / 𝑥⦌𝐴 = ⦋𝑦 / 𝑥⦌𝐵
4		csbeq1a 3842	. . 3 ⊢ (𝑥 = 𝑦 → 𝐴 = ⦋𝑦 / 𝑥⦌𝐴)
5		csbeq1a 3842	. . 3 ⊢ (𝑥 = 𝑦 → 𝐵 = ⦋𝑦 / 𝑥⦌𝐵)
6	4, 5	eqeq12d 2754	. 2 ⊢ (𝑥 = 𝑦 → (𝐴 = 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 = ⦋𝑦 / 𝑥⦌𝐵))
7	3, 6	sbiev 2312	1 ⊢ ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 = ⦋𝑦 / 𝑥⦌𝐵)

Syntax hints:   ↔ wb 205   = wceq 1539  [wsb 2068  ⦋csb 3828

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-10 2139  ax-11 2156  ax-12 2173  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-ex 1784  df-nf 1788  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-nfc 2888  df-sbc 3712  df-csb 3829

This theorem is referenced by: (None)