Theorem sbcoreleleq 39521

Description: Substitution of a setvar variable for another setvar variable in a 3-conjunct formula. Derived automatically from sbcoreleleqVD 39855. (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 3693	. . 3 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑦]𝑥 ∈ 𝑦 ↔ 𝑥 ∈ 𝐴))
2		sbcel1v 3692	. . . 4 ⊢ ([𝐴 / 𝑦]𝑦 ∈ 𝑥 ↔ 𝐴 ∈ 𝑥)
3	2	a1i 11	. . 3 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑦]𝑦 ∈ 𝑥 ↔ 𝐴 ∈ 𝑥))
4		eqsbc3r 3690	. . 3 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑦]𝑥 = 𝑦 ↔ 𝑥 = 𝐴))
5		3orbi123 39497	. . . 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 39466	. . 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 66	. 2 ⊢ (𝐴 ∈ 𝑉 → ((𝑥 ∈ 𝐴 ∨ 𝐴 ∈ 𝑥 ∨ 𝑥 = 𝐴) ↔ ([𝐴 / 𝑦]𝑥 ∈ 𝑦 ∨ [𝐴 / 𝑦]𝑦 ∈ 𝑥 ∨ [𝐴 / 𝑦]𝑥 = 𝑦)))
8		sbc3or 39518	. 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 282	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑦](𝑥 ∈ 𝑦 ∨ 𝑦 ∈ 𝑥 ∨ 𝑥 = 𝑦) ↔ (𝑥 ∈ 𝐴 ∨ 𝐴 ∈ 𝑥 ∨ 𝑥 = 𝐴)))

Syntax hints:   → wi 4   ↔ wb 198   ∨ w3o 1107   = wceq 1653   ∈ wcel 2157  [wsbc 3633

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-9 2166  ax-10 2185  ax-11 2200  ax-12 2213  ax-13 2377  ax-ext 2777

This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-3or 1109  df-3an 1110  df-tru 1657  df-ex 1876  df-nf 1880  df-sb 2065  df-clab 2786  df-cleq 2792  df-clel 2795  df-v 3387  df-sbc 3634

This theorem is referenced by:  tratrb  39522  tratrbVD  39857