Proof: Suppose ⟦no⟧a,M(f)(g)=1 for some f,g∈D⟨e,t⟩. This means set(f)⋂set(g)=ø. Now take an arbitrary g'∈D⟨e,t⟩ such that set(g') ⊆ set(g). Since set(f)⋂set(g)=ø and set(g') ⊆ set(g), we have set(f)⋂set(g')=ø. Then, ⟦no⟧a,M(f)(g')=1. ▢
Proof: Suppose ⟦every⟧a,M(f)(g)=1 for some f,g∈D⟨e,t⟩. This means set(f) ⊆ set(g). Now take a funciton f'∈D⟨e,t⟩ such that set(f') ⊆ set(f). From set(f) ⊆ set(g) and set(f') ⊆ set(f), we have set(f') ⊆ set(g). Then, ⟦every⟧a,M(f')(g)=1. ▢