Theorem brres2 38224

Description: Binary relation on a restriction. (Contributed by Peter Mazsa, 2-Jan-2019.) (Revised by Peter Mazsa, 16-Dec-2021.)

Assertion

Ref	Expression
brres2	⊢ (𝐵(𝑅 ↾ 𝐴)𝐶 ↔ 𝐵(𝑅 ∩ (𝐴 × ran (𝑅 ↾ 𝐴)))𝐶)

Proof of Theorem brres2

Step	Hyp	Ref	Expression
1		brres 6016	. . 3 ⊢ (𝐶 ∈ ran (𝑅 ↾ 𝐴) → (𝐵(𝑅 ↾ 𝐴)𝐶 ↔ (𝐵 ∈ 𝐴 ∧ 𝐵𝑅𝐶)))
2	1	pm5.32i 574	. 2 ⊢ ((𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ 𝐵(𝑅 ↾ 𝐴)𝐶) ↔ (𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ (𝐵 ∈ 𝐴 ∧ 𝐵𝑅𝐶)))
3		relres 6035	. . . 4 ⊢ Rel (𝑅 ↾ 𝐴)
4	3	relelrni 5974	. . 3 ⊢ (𝐵(𝑅 ↾ 𝐴)𝐶 → 𝐶 ∈ ran (𝑅 ↾ 𝐴))
5	4	pm4.71ri 560	. 2 ⊢ (𝐵(𝑅 ↾ 𝐴)𝐶 ↔ (𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ 𝐵(𝑅 ↾ 𝐴)𝐶))
6		brinxp2 5777	. . 3 ⊢ (𝐵(𝑅 ∩ (𝐴 × ran (𝑅 ↾ 𝐴)))𝐶 ↔ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ran (𝑅 ↾ 𝐴)) ∧ 𝐵𝑅𝐶))
7		df-3an 1089	. . 3 ⊢ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ 𝐵𝑅𝐶) ↔ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ran (𝑅 ↾ 𝐴)) ∧ 𝐵𝑅𝐶))
8		3anan12 1096	. . 3 ⊢ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ 𝐵𝑅𝐶) ↔ (𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ (𝐵 ∈ 𝐴 ∧ 𝐵𝑅𝐶)))
9	6, 7, 8	3bitr2i 299	. 2 ⊢ (𝐵(𝑅 ∩ (𝐴 × ran (𝑅 ↾ 𝐴)))𝐶 ↔ (𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ (𝐵 ∈ 𝐴 ∧ 𝐵𝑅𝐶)))
10	2, 5, 9	3bitr4i 303	1 ⊢ (𝐵(𝑅 ↾ 𝐴)𝐶 ↔ 𝐵(𝑅 ∩ (𝐴 × ran (𝑅 ↾ 𝐴)))𝐶)

Syntax hints:   ↔ wb 206   ∧ wa 395   ∧ w3a 1087   ∈ wcel 2108   ∩ cin 3975   class class class wbr 5166   × cxp 5698  ran crn 5701   ↾ cres 5702

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711  ax-sep 5317  ax-nul 5324  ax-pr 5447

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-3an 1089  df-tru 1540  df-fal 1550  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-ral 3068  df-rex 3077  df-rab 3444  df-v 3490  df-dif 3979  df-un 3981  df-in 3983  df-ss 3993  df-nul 4353  df-if 4549  df-sn 4649  df-pr 4651  df-op 4655  df-br 5167  df-opab 5229  df-xp 5706  df-rel 5707  df-cnv 5708  df-dm 5710  df-rn 5711  df-res 5712

This theorem is referenced by:  brinxprnres  38247