Theorem ref5 38570

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 2020	. . . . . 6 ⊢ (𝑦 = 𝑥 ↔ 𝑥 = 𝑦)
2	1	imbi1i 349	. . . . 5 ⊢ ((𝑦 = 𝑥 → 𝑥𝑅𝑦) ↔ (𝑥 = 𝑦 → 𝑥𝑅𝑦))
3	2	ralbii 3084	. . . 4 ⊢ (∀𝑦 ∈ 𝐵 (𝑦 = 𝑥 → 𝑥𝑅𝑦) ↔ ∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 → 𝑥𝑅𝑦))
4		breq2 5104	. . . . 5 ⊢ (𝑦 = 𝑥 → (𝑥𝑅𝑦 ↔ 𝑥𝑅𝑥))
5	4	ceqsralbv 3613	. . . 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 3084	. 2 ⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 → 𝑥𝑅𝑦) ↔ ∀𝑥 ∈ 𝐴 (𝑥 ∈ 𝐵 → 𝑥𝑅𝑥))
8		idinxpss 38569	. 2 ⊢ (( I ∩ (𝐴 × 𝐵)) ⊆ 𝑅 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 → 𝑥𝑅𝑦))
9		ralin 4203	. 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 206   = wceq 1542   ∈ wcel 2114  ∀wral 3052   ∩ cin 3902   ⊆ wss 3903   class class class wbr 5100   I cid 5526   × cxp 5630

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709  ax-sep 5243  ax-pr 5379

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ral 3053  df-rex 3063  df-rab 3402  df-v 3444  df-dif 3906  df-un 3908  df-in 3910  df-ss 3920  df-nul 4288  df-if 4482  df-sn 4583  df-pr 4585  df-op 4589  df-br 5101  df-opab 5163  df-id 5527  df-xp 5638  df-rel 5639

This theorem is referenced by:  dfrefrel5  38848