Theorem eqsbc3r 3625

Description: eqsbc3 3608 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 3608	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝑥 = 𝐵 ↔ 𝐴 = 𝐵))
2		eqcom 2759	. . 3 ⊢ (𝐵 = 𝑥 ↔ 𝑥 = 𝐵)
3	2	sbcbii 3624	. 2 ⊢ ([𝐴 / 𝑥]𝐵 = 𝑥 ↔ [𝐴 / 𝑥]𝑥 = 𝐵)
4		eqcom 2759	. 2 ⊢ (𝐵 = 𝐴 ↔ 𝐴 = 𝐵)
5	1, 3, 4	3bitr4g 303	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝑥 ↔ 𝐵 = 𝐴))

Syntax hints:   → wi 4   ↔ wb 196   = wceq 1624   ∈ wcel 2131  [wsbc 3568

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1863  ax-4 1878  ax-5 1980  ax-6 2046  ax-7 2082  ax-9 2140  ax-10 2160  ax-11 2175  ax-12 2188  ax-13 2383  ax-ext 2732

This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1627  df-ex 1846  df-nf 1851  df-sb 2039  df-clab 2739  df-cleq 2745  df-clel 2748  df-v 3334  df-sbc 3569

This theorem is referenced by:  sbcoreleleq  39239  sbcoreleleqVD  39586