Theorem bj-sbeqALT 34341

Description: Substitution in an equality (use the more general version bj-sbeq 34342 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 3852	. . 3 ⊢ Ⅎ𝑥⦋𝑦 / 𝑥⦌𝐴
2		nfcsb1v 3852	. . 3 ⊢ Ⅎ𝑥⦋𝑦 / 𝑥⦌𝐵
3	1, 2	nfeq 2968	. 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 2814	. 2 ⊢ (𝑥 = 𝑦 → (𝐴 = 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 = ⦋𝑦 / 𝑥⦌𝐵))
7	3, 6	sbiev 2322	1 ⊢ ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 = ⦋𝑦 / 𝑥⦌𝐵)

Syntax hints:   ↔ wb 209   = wceq 1538  [wsb 2069  ⦋csb 3828

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-10 2142  ax-11 2158  ax-12 2175  ax-ext 2770

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-ex 1782  df-nf 1786  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-nfc 2938  df-sbc 3721  df-csb 3829

This theorem is referenced by: (None)