Theorem ref5 38858

Description: Two ways to say that an intersection of the identity relation with a Cartesian product is a subclass. (Contributed by Peter Mazsa, 12-Dec-2023.)

Assertion

Ref	Expression
ref5	⊢ (( I ∩ (𝐴 × 𝐵)) ⊆ 𝑅 ↔ ∀𝑥 ∈ (𝐴 ∩ 𝐵)𝑥𝑅𝑥)

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝑥,𝑅

Proof of Theorem ref5

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		equcom 2045	. . . . . 6 ⊢ (𝑦 = 𝑥 ↔ 𝑥 = 𝑦)
2	1	imbi1i 352	. . . . 5 ⊢ ((𝑦 = 𝑥 → 𝑥𝑅𝑦) ↔ (𝑥 = 𝑦 → 𝑥𝑅𝑦))
3	2	ralbii 3117	. . . 4 ⊢ (∀𝑦 ∈ 𝐵 (𝑦 = 𝑥 → 𝑥𝑅𝑦) ↔ ∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 → 𝑥𝑅𝑦))
4		breq2 5117	. . . . 5 ⊢ (𝑦 = 𝑥 → (𝑥𝑅𝑦 ↔ 𝑥𝑅𝑥))
5	4	ceqsralbv 3625	. . . 4 ⊢ (∀𝑦 ∈ 𝐵 (𝑦 = 𝑥 → 𝑥𝑅𝑦) ↔ (𝑥 ∈ 𝐵 → 𝑥𝑅𝑥))
6	3, 5	bitr3i 280	. . 3 ⊢ (∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 → 𝑥𝑅𝑦) ↔ (𝑥 ∈ 𝐵 → 𝑥𝑅𝑥))
7	6	ralbii 3117	. 2 ⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 → 𝑥𝑅𝑦) ↔ ∀𝑥 ∈ 𝐴 (𝑥 ∈ 𝐵 → 𝑥𝑅𝑥))
8		idinxpss 38857	. 2 ⊢ (( I ∩ (𝐴 × 𝐵)) ⊆ 𝑅 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 → 𝑥𝑅𝑦))
9		ralin 4210	. 2 ⊢ (∀𝑥 ∈ (𝐴 ∩ 𝐵)𝑥𝑅𝑥 ↔ ∀𝑥 ∈ 𝐴 (𝑥 ∈ 𝐵 → 𝑥𝑅𝑥))
10	7, 8, 9	3bitr4i 306	1 ⊢ (( I ∩ (𝐴 × 𝐵)) ⊆ 𝑅 ↔ ∀𝑥 ∈ (𝐴 ∩ 𝐵)𝑥𝑅𝑥)

Syntax hints:   → wi 4   ↔ wb 209   = wceq 1567   ∈ wcel 2149  ∀wral 3085   ∩ cin 3912   ⊆ wss 3913   class class class wbr 5113   I cid 5556   × cxp 5660

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741  ax-sep 5261  ax-pr 5405

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1103  df-tru 1570  df-fal 1580  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-ral 3086  df-rex 3096  df-rab 3424  df-v 3465  df-dif 3916  df-un 3918  df-in 3920  df-ss 3930  df-nul 4295  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-br 5114  df-opab 5178  df-id 5557  df-xp 5668  df-rel 5669

This theorem is referenced by:  dfrefrel5  39136