Theorem ref5 38988

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 2048	. . . . . 6 ⊢ (𝑦 = 𝑥 ↔ 𝑥 = 𝑦)
2	1	imbi1i 352	. . . . 5 ⊢ ((𝑦 = 𝑥 → 𝑥𝑅𝑦) ↔ (𝑥 = 𝑦 → 𝑥𝑅𝑦))
3	2	ralbii 3111	. . . 4 ⊢ (∀𝑦 ∈ 𝐵 (𝑦 = 𝑥 → 𝑥𝑅𝑦) ↔ ∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 → 𝑥𝑅𝑦))
4		breq2 5113	. . . . 5 ⊢ (𝑦 = 𝑥 → (𝑥𝑅𝑦 ↔ 𝑥𝑅𝑥))
5	4	ceqsralbv 3616	. . . 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 3111	. 2 ⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 → 𝑥𝑅𝑦) ↔ ∀𝑥 ∈ 𝐴 (𝑥 ∈ 𝐵 → 𝑥𝑅𝑥))
8		idinxpss 38987	. 2 ⊢ (( I ∩ (𝐴 × 𝐵)) ⊆ 𝑅 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 → 𝑥𝑅𝑦))
9		ralin 4202	. 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 1570   ∈ wcel 2143  ∀wral 3079   ∩ cin 3904   ⊆ wss 3905   class class class wbr 5109   I cid 5555   × cxp 5659

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735  ax-sep 5257  ax-pr 5404

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-ral 3080  df-rex 3090  df-rab 3417  df-v 3457  df-dif 3908  df-un 3910  df-in 3912  df-ss 3922  df-nul 4287  df-if 4488  df-sn 4590  df-pr 4592  df-op 4596  df-br 5110  df-opab 5174  df-id 5556  df-xp 5667  df-rel 5668

This theorem is referenced by:  dfrefrel5  39266