Theorem bj-sbeqALT 36944

Description: Substitution in an equality (use the more general version bj-sbeq 36945 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 3869	. . 3 ⊢ Ⅎ𝑥⦋𝑦 / 𝑥⦌𝐴
2		nfcsb1v 3869	. . 3 ⊢ Ⅎ𝑥⦋𝑦 / 𝑥⦌𝐵
3	1, 2	nfeq 2908	. 2 ⊢ Ⅎ𝑥⦋𝑦 / 𝑥⦌𝐴 = ⦋𝑦 / 𝑥⦌𝐵
4		csbeq1a 3859	. . 3 ⊢ (𝑥 = 𝑦 → 𝐴 = ⦋𝑦 / 𝑥⦌𝐴)
5		csbeq1a 3859	. . 3 ⊢ (𝑥 = 𝑦 → 𝐵 = ⦋𝑦 / 𝑥⦌𝐵)
6	4, 5	eqeq12d 2747	. 2 ⊢ (𝑥 = 𝑦 → (𝐴 = 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 = ⦋𝑦 / 𝑥⦌𝐵))
7	3, 6	sbiev 2315	1 ⊢ ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 = ⦋𝑦 / 𝑥⦌𝐵)

Syntax hints:   ↔ wb 206   = wceq 1541  [wsb 2067  ⦋csb 3845

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-10 2144  ax-11 2160  ax-12 2180  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-ex 1781  df-nf 1785  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-nfc 2881  df-sbc 3737  df-csb 3846

This theorem is referenced by: (None)