We introduce the anti-rectangle refining property for forcing notions and investigate fragments of Martin's axiom for ℵ 1 dense sets related to the anti-rectangle refining property, which is close to some fragment of Martin's axiom for ℵ 1 dense sets related to the rectangle refining property, and prove that they are really weaker fragments. Definition 2.1. • (Knaster, [8]) A forcing notion P has property K if for every I ∈ [P] ℵ1 , there is a refinement I ∈ [I] ℵ1 such that p and q are compatible in P (i.e. there exists r ∈ P such that both r ≤ P p and r ≤ P q hold, denoted by p ⊥ P q) for every p and q in I. • A forcing notion P has the anti-rectangle refining property if it is uncountable and for every I and J in [P] ℵ1 , there are refinements I ∈ [I] ℵ1 and J ∈ [J] ℵ1 such that p and q are incompatible in P (denoted by p ⊥ P q) for every p ∈ I and q ∈ J. For a forcing notion P, we define another forcing notion a(P) which consists of finite antichains in P, ordered by reverse inclusion. If a(P) has the ccc, then by the genericity, a(P) can add an uncountable antichain in P. So it follows from the next lemma that if P has the ccc and the anti-rectangle refining property, then a(P) can destroy the ccc property of P. Proposition 2.2. If a forcing notion P has the anti-rectangle refining property, then for every I and J in the set [a(P)] ℵ1 with I ∪ J pairwise disjoint, i.e. σ ∩ τ = ∅ for any distinct σ and τ in I ∪ J, there are refinements I ∈ [I] ℵ1 and J ∈ [J] ℵ1 such that for any σ ∈ I and τ ∈ J , σ ⊥ a(P) τ , i.e. σ ∪ τ ∈ a(P). This property is very similar to the rectangle refining property. Definition 2.3 (Larson-Todorčević, [11, Definition 4.1.]). A partition [ω 1 ] 2 = K 0 ∪ K 1 satisfies the rectangle refining property if for every I and J in the set [ω 1 ] ℵ1 , there are refinements I ∈ [I] ℵ1 and J ∈ [J] ℵ1 such that {{α, β} ; α ∈ I & β ∈ J & α < β} ⊆ K 0. This is the reason why I call the anti-rectangle refining property. We note that if a partition [ω 1 ] 2 = K 0 ∪ K 1 satisfies the rectangle refining property, then for every I and J in [ω 1 ] ℵ1 , there are refinements I ∈ [I] ℵ1 and J ∈ [J] ℵ1 such that (I and J are disjoint and) {{α, β} ; α ∈ I & β ∈ J } ⊆ K 0. Moreover we should note that this property is stronger than the ccc, so if P has the anti-rectangle refining property, then a(P) has the ccc. Proof of Proposition 2.2. Assume that P has the anti-rectangle refining property and let I and J be uncountable subsets of a(P) such that I ∪ J is pairwise disjoint. By shrinking I and J respectively if need, we may assume that each element of I has size m and each element of J has size n. Applying the anti-rectangle refining property mn times, we can find uncountable subsets I and J of I and J respectively such that for every σ ∈ I and τ ∈ J , the i-th condition in σ and the j-th condition