Theorem sbcoreleleq 42528

Description: Substitution of a setvar variable for another setvar variable in a 3-conjunct formula. Derived automatically from sbcoreleleqVD 42852. (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		sbc3or 42525	. 2 ⊢ ([𝐴 / 𝑦](𝑥 ∈ 𝑦 ∨ 𝑦 ∈ 𝑥 ∨ 𝑥 = 𝑦) ↔ ([𝐴 / 𝑦]𝑥 ∈ 𝑦 ∨ [𝐴 / 𝑦]𝑦 ∈ 𝑥 ∨ [𝐴 / 𝑦]𝑥 = 𝑦))
2		sbcel2gv 3799	. . 3 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑦]𝑥 ∈ 𝑦 ↔ 𝑥 ∈ 𝐴))
3		sbcel1v 3798	. . . 4 ⊢ ([𝐴 / 𝑦]𝑦 ∈ 𝑥 ↔ 𝐴 ∈ 𝑥)
4	3	a1i 11	. . 3 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑦]𝑦 ∈ 𝑥 ↔ 𝐴 ∈ 𝑥))
5		eqsbc2 3796	. . 3 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑦]𝑥 = 𝑦 ↔ 𝑥 = 𝐴))
6		3orbi123 42504	. . . 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)))
7	6	3impexpbicomi 42473	. . 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)))))
8	2, 4, 5, 7	syl3c 66	. 2 ⊢ (𝐴 ∈ 𝑉 → ((𝑥 ∈ 𝐴 ∨ 𝐴 ∈ 𝑥 ∨ 𝑥 = 𝐴) ↔ ([𝐴 / 𝑦]𝑥 ∈ 𝑦 ∨ [𝐴 / 𝑦]𝑦 ∈ 𝑥 ∨ [𝐴 / 𝑦]𝑥 = 𝑦)))
9	1, 8	bitr4id 289	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑦](𝑥 ∈ 𝑦 ∨ 𝑦 ∈ 𝑥 ∨ 𝑥 = 𝑦) ↔ (𝑥 ∈ 𝐴 ∨ 𝐴 ∈ 𝑥 ∨ 𝑥 = 𝐴)))

Syntax hints:   → wi 4   ↔ wb 205   ∨ w3o 1085   = wceq 1540   ∈ wcel 2105  [wsbc 3727

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-10 2136  ax-12 2170  ax-ext 2707

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3or 1087  df-3an 1088  df-tru 1543  df-ex 1781  df-nf 1785  df-sb 2067  df-clab 2714  df-cleq 2728  df-clel 2814  df-v 3443  df-sbc 3728

This theorem is referenced by:  tratrb  42529  tratrbVD  42854