Theorem sbcoreleleq 45626

Description: Substitution of a setvar variable for another setvar variable in a 3-conjunct formula. Derived automatically from sbcoreleleqVD 45949. (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 45623	. 2 ⊢ ([𝐴 / 𝑦](𝑥 ∈ 𝑦 ∨ 𝑦 ∈ 𝑥 ∨ 𝑥 = 𝑦) ↔ ([𝐴 / 𝑦]𝑥 ∈ 𝑦 ∨ [𝐴 / 𝑦]𝑦 ∈ 𝑥 ∨ [𝐴 / 𝑦]𝑥 = 𝑦))
2		sbcel2gv 3804	. . 3 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑦]𝑥 ∈ 𝑦 ↔ 𝑥 ∈ 𝐴))
3		sbcel1v 3803	. . . 4 ⊢ ([𝐴 / 𝑦]𝑦 ∈ 𝑥 ↔ 𝐴 ∈ 𝑥)
4	3	a1i 11	. . 3 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑦]𝑦 ∈ 𝑥 ↔ 𝐴 ∈ 𝑥))
5		eqsbc2 3801	. . 3 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑦]𝑥 = 𝑦 ↔ 𝑥 = 𝐴))
6		3orbi123 45602	. . . 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 45572	. . 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 67	. 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 293	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑦](𝑥 ∈ 𝑦 ∨ 𝑦 ∈ 𝑥 ∨ 𝑥 = 𝑦) ↔ (𝑥 ∈ 𝐴 ∨ 𝐴 ∈ 𝑥 ∨ 𝑥 = 𝐴)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∨ w3o 1102   = wceq 1569   ∈ wcel 2144  [wsbc 3738

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3or 1104  df-3an 1105  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-sbc 3739

This theorem is used by:  tratrb  45627  tratrbVD  45951