Theorem brabg2 35801

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 5719	. . . 4 ⊢ Rel 𝑅
3	2	brrelex1i 5634	. . 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 5445	. . . . . 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 5445	. . . . 5 ⊢ ((𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷) → (𝐴𝑅𝐵 ↔ 𝜒))
13	12	exbiri 807	. . . 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 1539   ∈ wcel 2108  Vcvv 3422   class class class wbr 5070  {copab 5132

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709  ax-sep 5218  ax-nul 5225  ax-pr 5347

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-ral 3068  df-rex 3069  df-rab 3072  df-v 3424  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4254  df-if 4457  df-sn 4559  df-pr 4561  df-op 4565  df-br 5071  df-opab 5133  df-xp 5586  df-rel 5587

This theorem is referenced by: (None)