Metamath Proof Explorer


Theorem reldmlan

Description: The domain of Lan is a relation. (Contributed by Zhi Wang, 3-Nov-2025)

Ref Expression
Assertion reldmlan Rel dom Lan

Proof

Step Hyp Ref Expression
1 df-lan Lan = ( 𝑝 ∈ ( V × V ) , 𝑒 ∈ V ↦ ( 1st𝑝 ) / 𝑐 ( 2nd𝑝 ) / 𝑑 ( 𝑓 ∈ ( 𝑐 Func 𝑑 ) , 𝑥 ∈ ( 𝑐 Func 𝑒 ) ↦ ( ( ⟨ 𝑑 , 𝑒 ⟩ −∘F 𝑓 ) ( ( 𝑑 FuncCat 𝑒 ) UP ( 𝑐 FuncCat 𝑒 ) ) 𝑥 ) ) )
2 1 reldmmpo Rel dom Lan