Theorem brres2 39125

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 5973	. . 3 ⊢ (𝐶 ∈ ran (𝑅 ↾ 𝐴) → (𝐵(𝑅 ↾ 𝐴)𝐶 ↔ (𝐵 ∈ 𝐴 ∧ 𝐵𝑅𝐶)))
2	1	pm5.32i 585	. 2 ⊢ ((𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ 𝐵(𝑅 ↾ 𝐴)𝐶) ↔ (𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ (𝐵 ∈ 𝐴 ∧ 𝐵𝑅𝐶)))
3		relres 5992	. . . 4 ⊢ Rel (𝑅 ↾ 𝐴)
4	3	relelrni 5927	. . 3 ⊢ (𝐵(𝑅 ↾ 𝐴)𝐶 → 𝐶 ∈ ran (𝑅 ↾ 𝐴))
5	4	pm4.71ri 570	. 2 ⊢ (𝐵(𝑅 ↾ 𝐴)𝐶 ↔ (𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ 𝐵(𝑅 ↾ 𝐴)𝐶))
6		brinxp2 5725	. . 3 ⊢ (𝐵(𝑅 ∩ (𝐴 × ran (𝑅 ↾ 𝐴)))𝐶 ↔ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ran (𝑅 ↾ 𝐴)) ∧ 𝐵𝑅𝐶))
7		df-3an 1105	. . 3 ⊢ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ 𝐵𝑅𝐶) ↔ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ran (𝑅 ↾ 𝐴)) ∧ 𝐵𝑅𝐶))
8		3anan12 1112	. . 3 ⊢ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ 𝐵𝑅𝐶) ↔ (𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ (𝐵 ∈ 𝐴 ∧ 𝐵𝑅𝐶)))
9	6, 7, 8	3bitr2i 302	. 2 ⊢ (𝐵(𝑅 ∩ (𝐴 × ran (𝑅 ↾ 𝐴)))𝐶 ↔ (𝐶 ∈ ran (𝑅 ↾ 𝐴) ∧ (𝐵 ∈ 𝐴 ∧ 𝐵𝑅𝐶)))
10	2, 5, 9	3bitr4i 306	1 ⊢ (𝐵(𝑅 ↾ 𝐴)𝐶 ↔ 𝐵(𝑅 ∩ (𝐴 × ran (𝑅 ↾ 𝐴)))𝐶)

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   ∧ w3a 1103   ∈ wcel 2145   ∩ cin 3897   class class class wbr 5102   × cxp 5645  ran crn 5648   ↾ cres 5649

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 2147  ax-9 2155  ax-ext 2732  ax-sep 5248  ax-pr 5390

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 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-dif 3901  df-un 3903  df-in 3905  df-ss 3915  df-nul 4279  df-if 4482  df-sn 4584  df-pr 4586  df-op 4590  df-br 5103  df-opab 5167  df-xp 5653  df-rel 5654  df-cnv 5655  df-dm 5657  df-rn 5658  df-res 5659

This theorem is used by:  brinxprnres  39149