Theorem brres2 37184

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 5989	. . 3 ⊢ (𝐶 ∈ ran (𝑅 ↾ 𝐴) → (𝐵(𝑅 ↾ 𝐴)𝐶 ↔ (𝐵 ∈ 𝐴 ∧ 𝐵𝑅𝐶)))
2	1	pm5.32i 576	. 2 ⊢ ((𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ 𝐵(𝑅 ↾ 𝐴)𝐶) ↔ (𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ (𝐵 ∈ 𝐴 ∧ 𝐵𝑅𝐶)))
3		relres 6011	. . . 4 ⊢ Rel (𝑅 ↾ 𝐴)
4	3	relelrni 5949	. . 3 ⊢ (𝐵(𝑅 ↾ 𝐴)𝐶 → 𝐶 ∈ ran (𝑅 ↾ 𝐴))
5	4	pm4.71ri 562	. 2 ⊢ (𝐵(𝑅 ↾ 𝐴)𝐶 ↔ (𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ 𝐵(𝑅 ↾ 𝐴)𝐶))
6		brinxp2 5754	. . 3 ⊢ (𝐵(𝑅 ∩ (𝐴 × ran (𝑅 ↾ 𝐴)))𝐶 ↔ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ran (𝑅 ↾ 𝐴)) ∧ 𝐵𝑅𝐶))
7		df-3an 1090	. . 3 ⊢ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ 𝐵𝑅𝐶) ↔ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ran (𝑅 ↾ 𝐴)) ∧ 𝐵𝑅𝐶))
8		3anan12 1097	. . 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 205   ∧ wa 397   ∧ w3a 1088   ∈ wcel 2107   ∩ cin 3948   class class class wbr 5149   × cxp 5675  ran crn 5678   ↾ cres 5679

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704  ax-sep 5300  ax-nul 5307  ax-pr 5428

This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-3an 1090  df-tru 1545  df-fal 1555  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-ral 3063  df-rex 3072  df-rab 3434  df-v 3477  df-dif 3952  df-un 3954  df-in 3956  df-ss 3966  df-nul 4324  df-if 4530  df-sn 4630  df-pr 4632  df-op 4636  df-br 5150  df-opab 5212  df-xp 5683  df-rel 5684  df-cnv 5685  df-dm 5687  df-rn 5688  df-res 5689

This theorem is referenced by:  brinxprnres  37208