Theorem ref5 37780

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 2014	. . . . . 6 ⊢ (𝑦 = 𝑥 ↔ 𝑥 = 𝑦)
2	1	imbi1i 349	. . . . 5 ⊢ ((𝑦 = 𝑥 → 𝑥𝑅𝑦) ↔ (𝑥 = 𝑦 → 𝑥𝑅𝑦))
3	2	ralbii 3089	. . . 4 ⊢ (∀𝑦 ∈ 𝐵 (𝑦 = 𝑥 → 𝑥𝑅𝑦) ↔ ∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 → 𝑥𝑅𝑦))
4		breq2 5147	. . . . 5 ⊢ (𝑦 = 𝑥 → (𝑥𝑅𝑦 ↔ 𝑥𝑅𝑥))
5	4	ceqsralbv 3642	. . . 4 ⊢ (∀𝑦 ∈ 𝐵 (𝑦 = 𝑥 → 𝑥𝑅𝑦) ↔ (𝑥 ∈ 𝐵 → 𝑥𝑅𝑥))
6	3, 5	bitr3i 277	. . 3 ⊢ (∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 → 𝑥𝑅𝑦) ↔ (𝑥 ∈ 𝐵 → 𝑥𝑅𝑥))
7	6	ralbii 3089	. 2 ⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 → 𝑥𝑅𝑦) ↔ ∀𝑥 ∈ 𝐴 (𝑥 ∈ 𝐵 → 𝑥𝑅𝑥))
8		idinxpss 37779	. 2 ⊢ (( I ∩ (𝐴 × 𝐵)) ⊆ 𝑅 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 → 𝑥𝑅𝑦))
9		ralin 37714	. 2 ⊢ (∀𝑥 ∈ (𝐴 ∩ 𝐵)𝑥𝑅𝑥 ↔ ∀𝑥 ∈ 𝐴 (𝑥 ∈ 𝐵 → 𝑥𝑅𝑥))
10	7, 8, 9	3bitr4i 303	1 ⊢ (( I ∩ (𝐴 × 𝐵)) ⊆ 𝑅 ↔ ∀𝑥 ∈ (𝐴 ∩ 𝐵)𝑥𝑅𝑥)

Syntax hints:   → wi 4   ↔ wb 205   = wceq 1534   ∈ wcel 2099  ∀wral 3057   ∩ cin 3944   ⊆ wss 3945   class class class wbr 5143   I cid 5570   × cxp 5671

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-ext 2699  ax-sep 5294  ax-nul 5301  ax-pr 5424

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-3an 1087  df-tru 1537  df-fal 1547  df-ex 1775  df-sb 2061  df-clab 2706  df-cleq 2720  df-clel 2806  df-ral 3058  df-rex 3067  df-rab 3429  df-v 3472  df-dif 3948  df-un 3950  df-in 3952  df-ss 3962  df-nul 4320  df-if 4526  df-sn 4626  df-pr 4628  df-op 4632  df-br 5144  df-opab 5206  df-id 5571  df-xp 5679  df-rel 5680

This theorem is referenced by:  dfrefrel5  37984