Given a commutative algebra $A$ smooth over a field $k$ of characteristic zero, the module of K"ahler differentials $\Omega^{1}$ is projective of finite rank and so the sum of all wedge powers $\Omega^{\bullet}=\oplus_{p} \Omega^{p}$ is again projective of finite rank and is a commutative differential graded algebra with respect to the de Rham differential $d_{dR}$. Given a derivation $X$ of $A$, we can define a `contraction' operator on $\Omega^{\bullet}$ by requiring $\iota_{X}a=0$ for $a \in A$, $\iota_{X}(d_{dR}(a)$, and extending $\iota_{X}$ to all of $\Omega^{\bullet}$ by requiring it to be a derivation of degree $-1$.

Now if $\dots \rightarrow A_{1} \rightarrow A_{0}=A$ is a cdga with differential $d$ of degree $-1$, then the K"ahler differentials $\Omega^{1}$ are a dg-module over $A$ with the (internal) differential also denoted $d$. Given suitable finiteness hypotheses on $A$, $\Omega^{1}$ is projective of finite rank. Then $\Omega^{\bullet}$ is also a dg-module over $A$, with the differential $d$ of degree $-1$ with respect to the internal grading and of degree $0$ with respect to the grading by $p$. Again, it has a de Rham differential $d_{dR}$ of degree $1$ with respect to $p$ and degree $0$ with respect to the internal grading.

Question:

Given a derivation $X$ of $A$ of some degree $|X|$, I would like to consider the contraction $\iota_{X}$ acting on $\Omega^{\bullet}$. It should have the same definition on $a \in A$ and $d_{dR}a \in \Omega^{1}$, and then I would like it to be uniquely extending by making it a derivation of some (bi?)degree. I can imagine the degree to be various things. Roughly, the formula should be something like

$$\iota_{X}(\alpha \wedge \beta)=\iota_{X}(\alpha) \wedge \beta + (-1)^{|X||\alpha|}\alpha \wedge \iota_{X}(\beta),$$

except I don't know if $|\alpha|$ should take into account only the $p$ in $\alpha \in \Omega^{p}$ or also the internal grading $i$ in the dg-module $\Omega^{p}$.

What is the natural thing to do? Probably Theo Johnson-Freyd will give some nice monoidal answer, but I'll also appreciate something very concrete.