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 ∈ ( 𝐴 ∩ 𝐵 ) 𝑥 𝑅 𝑥 )

Proof

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