Theorem brres2 38777

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 5974	. . 3 ⊢ (𝐶 ∈ ran (𝑅 ↾ 𝐴) → (𝐵(𝑅 ↾ 𝐴)𝐶 ↔ (𝐵 ∈ 𝐴 ∧ 𝐵𝑅𝐶)))
2	1	pm5.32i 582	. 2 ⊢ ((𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ 𝐵(𝑅 ↾ 𝐴)𝐶) ↔ (𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ (𝐵 ∈ 𝐴 ∧ 𝐵𝑅𝐶)))
3		relres 5993	. . . 4 ⊢ Rel (𝑅 ↾ 𝐴)
4	3	relelrni 5927	. . 3 ⊢ (𝐵(𝑅 ↾ 𝐴)𝐶 → 𝐶 ∈ ran (𝑅 ↾ 𝐴))
5	4	pm4.71ri 568	. 2 ⊢ (𝐵(𝑅 ↾ 𝐴)𝐶 ↔ (𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ 𝐵(𝑅 ↾ 𝐴)𝐶))
6		brinxp2 5727	. . 3 ⊢ (𝐵(𝑅 ∩ (𝐴 × ran (𝑅 ↾ 𝐴)))𝐶 ↔ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ran (𝑅 ↾ 𝐴)) ∧ 𝐵𝑅𝐶))
7		df-3an 1101	. . 3 ⊢ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ 𝐵𝑅𝐶) ↔ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ran (𝑅 ↾ 𝐴)) ∧ 𝐵𝑅𝐶))
8		3anan12 1108	. . 3 ⊢ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ 𝐵𝑅𝐶) ↔ (𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ (𝐵 ∈ 𝐴 ∧ 𝐵𝑅𝐶)))
9	6, 7, 8	3bitr2i 301	. 2 ⊢ (𝐵(𝑅 ∩ (𝐴 × ran (𝑅 ↾ 𝐴)))𝐶 ↔ (𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ (𝐵 ∈ 𝐴 ∧ 𝐵𝑅𝐶)))
10	2, 5, 9	3bitr4i 305	1 ⊢ (𝐵(𝑅 ↾ 𝐴)𝐶 ↔ 𝐵(𝑅 ∩ (𝐴 × ran (𝑅 ↾ 𝐴)))𝐶)

Syntax hints:   ↔ wb 208   ∧ wa 399   ∧ w3a 1099   ∈ wcel 2144   ∩ cin 3905   class class class wbr 5102   × cxp 5647  ran crn 5650   ↾ cres 5651

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1817  ax-4 1831  ax-5 1932  ax-6 1989  ax-7 2030  ax-8 2146  ax-9 2154  ax-ext 2736  ax-sep 5248  ax-pr 5392

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1101  df-tru 1565  df-fal 1575  df-ex 1802  df-sb 2093  df-clab 2743  df-cleq 2756  df-clel 2839  df-ral 3079  df-rex 3089  df-rab 3417  df-v 3458  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4288  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-br 5103  df-opab 5165  df-xp 5655  df-rel 5656  df-cnv 5657  df-dm 5659  df-rn 5660  df-res 5661

This theorem is referenced by:  brinxprnres  38801