Theorem brabg2 37112

Description: Relation by a binary relation abstraction. (Contributed by Jeff Madsen, 2-Sep-2009.)

Hypotheses

Ref	Expression
brabg2.1	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
brabg2.2	⊢ (𝑦 = 𝐵 → (𝜓 ↔ 𝜒))
brabg2.3	⊢ 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
brabg2.4	⊢ (𝜒 → 𝐴 ∈ 𝐶)

Assertion

Ref	Expression
brabg2	⊢ (𝐵 ∈ 𝐷 → (𝐴𝑅𝐵 ↔ 𝜒))

Distinct variable groups:   𝑥,𝐴,𝑦   𝑥,𝐵,𝑦   𝜒,𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦)   𝜓(𝑥,𝑦)   𝐶(𝑥,𝑦)   𝐷(𝑥,𝑦)   𝑅(𝑥,𝑦)

Proof of Theorem brabg2

Step	Hyp	Ref	Expression
1		brabg2.3	. . . . 5 ⊢ 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
2	1	relopabiv 5816	. . . 4 ⊢ Rel 𝑅
3	2	brrelex1i 5728	. . 3 ⊢ (𝐴𝑅𝐵 → 𝐴 ∈ V)
4		brabg2.1	. . . . . . 7 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
5		brabg2.2	. . . . . . 7 ⊢ (𝑦 = 𝐵 → (𝜓 ↔ 𝜒))
6	4, 5, 1	brabg 5535	. . . . . 6 ⊢ ((𝐴 ∈ V ∧ 𝐵 ∈ 𝐷) → (𝐴𝑅𝐵 ↔ 𝜒))
7	6	biimpd 228	. . . . 5 ⊢ ((𝐴 ∈ V ∧ 𝐵 ∈ 𝐷) → (𝐴𝑅𝐵 → 𝜒))
8	7	ex 412	. . . 4 ⊢ (𝐴 ∈ V → (𝐵 ∈ 𝐷 → (𝐴𝑅𝐵 → 𝜒)))
9	8	com3l 89	. . 3 ⊢ (𝐵 ∈ 𝐷 → (𝐴𝑅𝐵 → (𝐴 ∈ V → 𝜒)))
10	3, 9	mpdi 45	. 2 ⊢ (𝐵 ∈ 𝐷 → (𝐴𝑅𝐵 → 𝜒))
11		brabg2.4	. . 3 ⊢ (𝜒 → 𝐴 ∈ 𝐶)
12	4, 5, 1	brabg 5535	. . . . 5 ⊢ ((𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷) → (𝐴𝑅𝐵 ↔ 𝜒))
13	12	exbiri 810	. . . 4 ⊢ (𝐴 ∈ 𝐶 → (𝐵 ∈ 𝐷 → (𝜒 → 𝐴𝑅𝐵)))
14	13	com3l 89	. . 3 ⊢ (𝐵 ∈ 𝐷 → (𝜒 → (𝐴 ∈ 𝐶 → 𝐴𝑅𝐵)))
15	11, 14	mpdi 45	. 2 ⊢ (𝐵 ∈ 𝐷 → (𝜒 → 𝐴𝑅𝐵))
16	10, 15	impbid 211	1 ⊢ (𝐵 ∈ 𝐷 → (𝐴𝑅𝐵 ↔ 𝜒))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395   = wceq 1534   ∈ wcel 2099  Vcvv 3469   class class class wbr 5142  {copab 5204

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-ext 2698  ax-sep 5293  ax-nul 5300  ax-pr 5423

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-3an 1087  df-tru 1537  df-fal 1547  df-ex 1775  df-sb 2061  df-clab 2705  df-cleq 2719  df-clel 2805  df-ral 3057  df-rex 3066  df-rab 3428  df-v 3471  df-dif 3947  df-un 3949  df-in 3951  df-ss 3961  df-nul 4319  df-if 4525  df-sn 4625  df-pr 4627  df-op 4631  df-br 5143  df-opab 5205  df-xp 5678  df-rel 5679

This theorem is referenced by: (None)