Theorem eqsbc3r 2921

Description: eqsbc3 2900 with setvar variable on right side of equals sign. (Contributed by Alan Sare, 24-Oct-2011.) (Proof shortened by JJ, 7-Jul-2021.)

Assertion

Ref	Expression
eqsbc3r	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝑥 ↔ 𝐵 = 𝐴))

Distinct variable group:   𝑥,𝐵

Allowed substitution hints:   𝐴(𝑥)   𝑉(𝑥)

Proof of Theorem eqsbc3r

Step	Hyp	Ref	Expression
1		eqsbc3 2900	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝑥 = 𝐵 ↔ 𝐴 = 𝐵))
2		eqcom 2102	. . 3 ⊢ (𝐵 = 𝑥 ↔ 𝑥 = 𝐵)
3	2	sbcbii 2920	. 2 ⊢ ([𝐴 / 𝑥]𝐵 = 𝑥 ↔ [𝐴 / 𝑥]𝑥 = 𝐵)
4		eqcom 2102	. 2 ⊢ (𝐵 = 𝐴 ↔ 𝐴 = 𝐵)
5	1, 3, 4	3bitr4g 222	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝑥 ↔ 𝐵 = 𝐴))

Syntax hints:   → wi 4   ↔ wb 104   = wceq 1299   ∈ wcel 1448  [wsbc 2862

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 671  ax-5 1391  ax-7 1392  ax-gen 1393  ax-ie1 1437  ax-ie2 1438  ax-8 1450  ax-10 1451  ax-11 1452  ax-i12 1453  ax-bndl 1454  ax-4 1455  ax-17 1474  ax-i9 1478  ax-ial 1482  ax-i5r 1483  ax-ext 2082

This theorem depends on definitions:  df-bi 116  df-tru 1302  df-nf 1405  df-sb 1704  df-clab 2087  df-cleq 2093  df-clel 2096  df-nfc 2229  df-v 2643  df-sbc 2863

This theorem is referenced by: (None)