Metamath Proof Explorer


Theorem ref5

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)

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

Proof

Step Hyp Ref Expression
1 equcom ( 𝑦 = 𝑥𝑥 = 𝑦 )
2 1 imbi1i ( ( 𝑦 = 𝑥𝑥 𝑅 𝑦 ) ↔ ( 𝑥 = 𝑦𝑥 𝑅 𝑦 ) )
3 2 ralbii ( ∀ 𝑦𝐵 ( 𝑦 = 𝑥𝑥 𝑅 𝑦 ) ↔ ∀ 𝑦𝐵 ( 𝑥 = 𝑦𝑥 𝑅 𝑦 ) )
4 breq2 ( 𝑦 = 𝑥 → ( 𝑥 𝑅 𝑦𝑥 𝑅 𝑥 ) )
5 4 ceqsralbv ( ∀ 𝑦𝐵 ( 𝑦 = 𝑥𝑥 𝑅 𝑦 ) ↔ ( 𝑥𝐵𝑥 𝑅 𝑥 ) )
6 3 5 bitr3i ( ∀ 𝑦𝐵 ( 𝑥 = 𝑦𝑥 𝑅 𝑦 ) ↔ ( 𝑥𝐵𝑥 𝑅 𝑥 ) )
7 6 ralbii ( ∀ 𝑥𝐴𝑦𝐵 ( 𝑥 = 𝑦𝑥 𝑅 𝑦 ) ↔ ∀ 𝑥𝐴 ( 𝑥𝐵𝑥 𝑅 𝑥 ) )
8 idinxpss ( ( I ∩ ( 𝐴 × 𝐵 ) ) ⊆ 𝑅 ↔ ∀ 𝑥𝐴𝑦𝐵 ( 𝑥 = 𝑦𝑥 𝑅 𝑦 ) )
9 ralin ( ∀ 𝑥 ∈ ( 𝐴𝐵 ) 𝑥 𝑅 𝑥 ↔ ∀ 𝑥𝐴 ( 𝑥𝐵𝑥 𝑅 𝑥 ) )
10 7 8 9 3bitr4i ( ( I ∩ ( 𝐴 × 𝐵 ) ) ⊆ 𝑅 ↔ ∀ 𝑥 ∈ ( 𝐴𝐵 ) 𝑥 𝑅 𝑥 )