Theorem ref5 38818

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 2038	. . . . . 6 ⊢ (𝑦 = 𝑥 ↔ 𝑥 = 𝑦)
2	1	imbi1i 351	. . . . 5 ⊢ ((𝑦 = 𝑥 → 𝑥𝑅𝑦) ↔ (𝑥 = 𝑦 → 𝑥𝑅𝑦))
3	2	ralbii 3108	. . . 4 ⊢ (∀𝑦 ∈ 𝐵 (𝑦 = 𝑥 → 𝑥𝑅𝑦) ↔ ∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 → 𝑥𝑅𝑦))
4		breq2 5104	. . . . 5 ⊢ (𝑦 = 𝑥 → (𝑥𝑅𝑦 ↔ 𝑥𝑅𝑥))
5	4	ceqsralbv 3616	. . . 4 ⊢ (∀𝑦 ∈ 𝐵 (𝑦 = 𝑥 → 𝑥𝑅𝑦) ↔ (𝑥 ∈ 𝐵 → 𝑥𝑅𝑥))
6	3, 5	bitr3i 279	. . 3 ⊢ (∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 → 𝑥𝑅𝑦) ↔ (𝑥 ∈ 𝐵 → 𝑥𝑅𝑥))
7	6	ralbii 3108	. 2 ⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 → 𝑥𝑅𝑦) ↔ ∀𝑥 ∈ 𝐴 (𝑥 ∈ 𝐵 → 𝑥𝑅𝑥))
8		idinxpss 38817	. 2 ⊢ (( I ∩ (𝐴 × 𝐵)) ⊆ 𝑅 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 → 𝑥𝑅𝑦))
9		ralin 4201	. 2 ⊢ (∀𝑥 ∈ (𝐴 ∩ 𝐵)𝑥𝑅𝑥 ↔ ∀𝑥 ∈ 𝐴 (𝑥 ∈ 𝐵 → 𝑥𝑅𝑥))
10	7, 8, 9	3bitr4i 305	1 ⊢ (( I ∩ (𝐴 × 𝐵)) ⊆ 𝑅 ↔ ∀𝑥 ∈ (𝐴 ∩ 𝐵)𝑥𝑅𝑥)

Syntax hints:   → wi 4   ↔ wb 208   = wceq 1560   ∈ wcel 2142  ∀wral 3076   ∩ cin 3903   ⊆ wss 3904   class class class wbr 5100   I cid 5541   × cxp 5645

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-8 2144  ax-9 2152  ax-ext 2734  ax-sep 5246  ax-pr 5390

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1100  df-tru 1563  df-fal 1573  df-ex 1800  df-sb 2091  df-clab 2741  df-cleq 2754  df-clel 2837  df-ral 3077  df-rex 3087  df-rab 3415  df-v 3456  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-nul 4286  df-if 4481  df-sn 4583  df-pr 4585  df-op 4589  df-br 5101  df-opab 5163  df-id 5542  df-xp 5653  df-rel 5654

This theorem is referenced by:  dfrefrel5  39096