SPFA by finite conditions
نویسندگان
چکیده
Proof. Clearly (A ∩ B)[G] ⊆ A[G] ∩ B[G]. Let us show the opposite inclusion. It is enough to deal with ordinals. Note that B[G]∩ V = B, since Q ⊆ B. Let δ ∈ A[G]∩B[G]∩On = A[G]∩B ∩On. Pick a canonical name δ ∼ ∈ A such that δ ∼ = δ. It is of the form ⟨⟨qi, τ̌i⟩ | i < ρ ≤ |Q|⟩, where τi’s are in B and ⟨qi | i < ρ⟩ is a maximal antichain. Now, δ ∼ ∈ B, since |Q|B ⊆ B. So, δ ∼ ∈ A∩B. Hence, δ = δ ∼ ∈ (A∩B)[G] and we are done.
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عنوان ژورنال:
- Arch. Math. Log.
دوره 55 شماره
صفحات -
تاریخ انتشار 2016