Theorem eqsbc2 3827

Description: Substitution for the right-hand side in an equality. (Contributed by Alan Sare, 24-Oct-2011.) (Proof shortened by JJ, 7-Jul-2021.)

Assertion

Ref	Expression
eqsbc2	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝑥 ↔ 𝐵 = 𝐴))

Distinct variable group:   𝑥,𝐵

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

Proof of Theorem eqsbc2

Step	Hyp	Ref	Expression
1		eqsbc1 3810	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝑥 = 𝐵 ↔ 𝐴 = 𝐵))
2		eqcom 2741	. . 3 ⊢ (𝐵 = 𝑥 ↔ 𝑥 = 𝐵)
3	2	sbcbii 3820	. 2 ⊢ ([𝐴 / 𝑥]𝐵 = 𝑥 ↔ [𝐴 / 𝑥]𝑥 = 𝐵)
4		eqcom 2741	. 2 ⊢ (𝐵 = 𝐴 ↔ 𝐴 = 𝐵)
5	1, 3, 4	3bitr4g 314	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝑥 ↔ 𝐵 = 𝐴))

Syntax hints:   → wi 4   ↔ wb 206   = wceq 1539   ∈ wcel 2107  [wsbc 3763

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1542  df-ex 1779  df-sb 2064  df-clab 2713  df-cleq 2726  df-clel 2808  df-sbc 3764

This theorem is referenced by:  sbcoreleleq  44486  sbcoreleleqVD  44810