Theorem sbcoreleleq 37766

Description: Substitution of a setvar variable for another setvar variable in a 3-conjunct formula. Derived automatically from sbcoreleleqVD 38117. (Contributed by Alan Sare, 31-Dec-2011.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
sbcoreleleq	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑦](𝑥 ∈ 𝑦 ∨ 𝑦 ∈ 𝑥 ∨ 𝑥 = 𝑦) ↔ (𝑥 ∈ 𝐴 ∨ 𝐴 ∈ 𝑥 ∨ 𝑥 = 𝐴)))

Distinct variable group:   𝑥,𝑦

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

Proof of Theorem sbcoreleleq

Step	Hyp	Ref	Expression
1		sbcel2gv 3463	. . 3 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑦]𝑥 ∈ 𝑦 ↔ 𝑥 ∈ 𝐴))
2		sbcel1v 3462	. . . 4 ⊢ ([𝐴 / 𝑦]𝑦 ∈ 𝑥 ↔ 𝐴 ∈ 𝑥)
3	2	a1i 11	. . 3 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑦]𝑦 ∈ 𝑥 ↔ 𝐴 ∈ 𝑥))
4		eqsbc3r 3459	. . 3 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑦]𝑥 = 𝑦 ↔ 𝑥 = 𝐴))
5		3orbi123 37738	. . . 4 ⊢ ((([𝐴 / 𝑦]𝑥 ∈ 𝑦 ↔ 𝑥 ∈ 𝐴) ∧ ([𝐴 / 𝑦]𝑦 ∈ 𝑥 ↔ 𝐴 ∈ 𝑥) ∧ ([𝐴 / 𝑦]𝑥 = 𝑦 ↔ 𝑥 = 𝐴)) → (([𝐴 / 𝑦]𝑥 ∈ 𝑦 ∨ [𝐴 / 𝑦]𝑦 ∈ 𝑥 ∨ [𝐴 / 𝑦]𝑥 = 𝑦) ↔ (𝑥 ∈ 𝐴 ∨ 𝐴 ∈ 𝑥 ∨ 𝑥 = 𝐴)))
6	5	3impexpbicomi 37707	. . 3 ⊢ (([𝐴 / 𝑦]𝑥 ∈ 𝑦 ↔ 𝑥 ∈ 𝐴) → (([𝐴 / 𝑦]𝑦 ∈ 𝑥 ↔ 𝐴 ∈ 𝑥) → (([𝐴 / 𝑦]𝑥 = 𝑦 ↔ 𝑥 = 𝐴) → ((𝑥 ∈ 𝐴 ∨ 𝐴 ∈ 𝑥 ∨ 𝑥 = 𝐴) ↔ ([𝐴 / 𝑦]𝑥 ∈ 𝑦 ∨ [𝐴 / 𝑦]𝑦 ∈ 𝑥 ∨ [𝐴 / 𝑦]𝑥 = 𝑦)))))
7	1, 3, 4, 6	syl3c 64	. 2 ⊢ (𝐴 ∈ 𝑉 → ((𝑥 ∈ 𝐴 ∨ 𝐴 ∈ 𝑥 ∨ 𝑥 = 𝐴) ↔ ([𝐴 / 𝑦]𝑥 ∈ 𝑦 ∨ [𝐴 / 𝑦]𝑦 ∈ 𝑥 ∨ [𝐴 / 𝑦]𝑥 = 𝑦)))
8		sbc3or 37759	. 2 ⊢ ([𝐴 / 𝑦](𝑥 ∈ 𝑦 ∨ 𝑦 ∈ 𝑥 ∨ 𝑥 = 𝑦) ↔ ([𝐴 / 𝑦]𝑥 ∈ 𝑦 ∨ [𝐴 / 𝑦]𝑦 ∈ 𝑥 ∨ [𝐴 / 𝑦]𝑥 = 𝑦))
9	7, 8	syl6rbbr 278	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑦](𝑥 ∈ 𝑦 ∨ 𝑦 ∈ 𝑥 ∨ 𝑥 = 𝑦) ↔ (𝑥 ∈ 𝐴 ∨ 𝐴 ∈ 𝑥 ∨ 𝑥 = 𝐴)))

Syntax hints:   → wi 4   ↔ wb 195   ∨ w3o 1030   = wceq 1475   ∈ wcel 1977  [wsbc 3402

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-3or 1032  df-3an 1033  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-v 3175  df-sbc 3403

This theorem is referenced by:  tratrb  37767  tratrbVD  38119