Metamath Proof Explorer


Theorem ssrel3

Description: Subclass relation in another form when the subclass is a relation. (Contributed by Peter Mazsa, 16-Feb-2019)

Ref Expression
Assertion ssrel3 ( Rel 𝐴 → ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝐴 𝑦 → 𝑥 𝐵 𝑦 ) ) )

Proof

Step Hyp Ref Expression
1 ssrel ⊢ ( Rel 𝐴 → ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ∀ 𝑦 ( ⟨ 𝑥 , 𝑦 ⟩ ∈ 𝐴 → ⟨ 𝑥 , 𝑦 ⟩ ∈ 𝐵 ) ) )
2 df-br ⊢ ( 𝑥 𝐴 𝑦 ↔ ⟨ 𝑥 , 𝑦 ⟩ ∈ 𝐴 )
3 df-br ⊢ ( 𝑥 𝐵 𝑦 ↔ ⟨ 𝑥 , 𝑦 ⟩ ∈ 𝐵 )
4 2 3 imbi12i ⊢ ( ( 𝑥 𝐴 𝑦 → 𝑥 𝐵 𝑦 ) ↔ ( ⟨ 𝑥 , 𝑦 ⟩ ∈ 𝐴 → ⟨ 𝑥 , 𝑦 ⟩ ∈ 𝐵 ) )
5 4 2albii ⊢ ( ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝐴 𝑦 → 𝑥 𝐵 𝑦 ) ↔ ∀ 𝑥 ∀ 𝑦 ( ⟨ 𝑥 , 𝑦 ⟩ ∈ 𝐴 → ⟨ 𝑥 , 𝑦 ⟩ ∈ 𝐵 ) )
6 1 5 bitr4di ⊢ ( Rel 𝐴 → ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝐴 𝑦 → 𝑥 𝐵 𝑦 ) ) )