Theorem bj-sbeqALT 37390

Description: Substitution in an equality (use the more general version bj-sbeq 37391 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 3877	. . 3 ⊢ Ⅎ𝑥⦋𝑦 / 𝑥⦌𝐴
2		nfcsb1v 3877	. . 3 ⊢ Ⅎ𝑥⦋𝑦 / 𝑥⦌𝐵
3	1, 2	nfeq 2938	. 2 ⊢ Ⅎ𝑥⦋𝑦 / 𝑥⦌𝐴 = ⦋𝑦 / 𝑥⦌𝐵
4		csbeq1a 3867	. . 3 ⊢ (𝑥 = 𝑦 → 𝐴 = ⦋𝑦 / 𝑥⦌𝐴)
5		csbeq1a 3867	. . 3 ⊢ (𝑥 = 𝑦 → 𝐵 = ⦋𝑦 / 𝑥⦌𝐵)
6	4, 5	eqeq12d 2779	. 2 ⊢ (𝑥 = 𝑦 → (𝐴 = 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 = ⦋𝑦 / 𝑥⦌𝐵))
7	3, 6	sbiev 2347	1 ⊢ ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 = ⦋𝑦 / 𝑥⦌𝐵)

Syntax hints:   ↔ wb 208   = wceq 1561  [wsb 2091  ⦋csb 3853

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1816  ax-4 1830  ax-5 1931  ax-6 1988  ax-7 2029  ax-8 2145  ax-9 2153  ax-10 2176  ax-11 2192  ax-12 2213  ax-ext 2735

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1564  df-ex 1801  df-nf 1805  df-sb 2092  df-clab 2742  df-cleq 2755  df-clel 2838  df-nfc 2912  df-sbc 3746  df-csb 3854

This theorem is referenced by: (None)