Theorem brabg2 38430

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 5809	. . . 4 ⊢ Rel 𝑅
3	2	brrelex1i 5719	. . 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 5526	. . . . . 6 ⊢ ((𝐴 ∈ V ∧ 𝐵 ∈ 𝐷) → (𝐴𝑅𝐵 ↔ 𝜒))
7	6	biimpd 232	. . . . 5 ⊢ ((𝐴 ∈ V ∧ 𝐵 ∈ 𝐷) → (𝐴𝑅𝐵 → 𝜒))
8	7	ex 418	. . . 4 ⊢ (𝐴 ∈ V → (𝐵 ∈ 𝐷 → (𝐴𝑅𝐵 → 𝜒)))
9	8	com3l 90	. . 3 ⊢ (𝐵 ∈ 𝐷 → (𝐴𝑅𝐵 → (𝐴 ∈ V → 𝜒)))
10	3, 9	mpdi 46	. 2 ⊢ (𝐵 ∈ 𝐷 → (𝐴𝑅𝐵 → 𝜒))
11		brabg2.4	. . 3 ⊢ (𝜒 → 𝐴 ∈ 𝐶)
12	4, 5, 1	brabg 5526	. . . . 5 ⊢ ((𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷) → (𝐴𝑅𝐵 ↔ 𝜒))
13	12	exbiri 823	. . . 4 ⊢ (𝐴 ∈ 𝐶 → (𝐵 ∈ 𝐷 → (𝜒 → 𝐴𝑅𝐵)))
14	13	com3l 90	. . 3 ⊢ (𝐵 ∈ 𝐷 → (𝜒 → (𝐴 ∈ 𝐶 → 𝐴𝑅𝐵)))
15	11, 14	mpdi 46	. 2 ⊢ (𝐵 ∈ 𝐷 → (𝜒 → 𝐴𝑅𝐵))
16	10, 15	impbid 215	1 ⊢ (𝐵 ∈ 𝐷 → (𝐴𝑅𝐵 ↔ 𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2146  Vcvv 3457   class class class wbr 5111  {copab 5175

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737  ax-sep 5259  ax-pr 5406

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-ral 3082  df-rex 3092  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-br 5112  df-opab 5176  df-xp 5669  df-rel 5670

This theorem is used by: (None)