Acyclic complex in Khovanov homology

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I am reading Dror Bar-Natan's paper Categorification. In section 3.5.1 (page 9), "Invariance under R1", it is claimed "It is easy to check that $\mathcal{C}'$ is subcomplex of $\mathcal{C}$, and as that $v_+$ is a unit for the product $m$, $\mathcal{C}'$ is acyclic".

The question is: Which result is used to deduce $\mathcal{C}'$ is acyclic?

As a side note, I am not familiarized with this use of the word "complex", the definition of (chain) complex I am most familiar with is an object of the form $ A_0 \rightarrow A_1 \rightarrow A_2 \rightarrow \cdots$, where $A_i$ are say $\mathbb{Z}$-modules, from which one can compute its homology.

I am also slightly confused, because it is said that in section 3.5.1 everything is "flattened", so I guess $\mathcal{C}$ and $\mathcal{C'}$ are basically two (two-term?) chain complexes like $A\rightarrow B$, but you need at least two boundary maps to compute homology.