On page 4 of this paper by H. Abbaspour, the author defines the **two-sided bar construction**
$$B(A,A,A):=A\otimes T(s\bar{A})\otimes A$$
of a differential graded algebra $(A,d_A)$ (over a field).
The definition of the differential $d=d_0+d_1$ on $B(A,A,A)$ is unclear to me. While $d_1$ seems to lower the *wordlength* on $T(s\bar{A})$ by $1$, $d_0$ seems to raise the *degree* (as a tensor product) of an element in $A\otimes T(s\bar{A})\otimes A$ by $1$.
As far as I understand, $(B(A,A,A),d)$ should be a chain complex, in fact it should give a free resolution of $(A,d_A)$ as an $(A\otimes A^{op},d_A\otimes 1+1\otimes d_A)$-module.

What is the grading on $B(A,A,A)$? Why does $d$ lower the degree by $1$?

Thanks to anyone who can shed some light on the two-sided bar construction.

chaincomplex giving a free resolution of $(A,d_A)$ as an $A\otimes A^{op}$-module. Or is this not true and what is really going on is that $(B(A,A,A),d_1)$ is a chain complex (graded bywordlengthon $T(s\bar{A})$) giving a free resolution of $(A,d_A)$ as an $A\otimes A^{op}$-module? $\endgroup$