Given $n$ boolean variables $x_1,\ldots,x_n$ each of which is assigned a positive cost $c_1,\ldots,c_n\in\mathbb{Z}_{>0}$ and a boolean function $f$ on these variables given in the form $$f(x_1,\ldots,x_n)=\bigwedge_{i=1}^k\bigoplus_{j=1}^{l_i}x_{r_{ij}}$$ ($\oplus$ denoting XOR) with $k\in\mathbb{Z}_{>0}$, integers $1\leq l_i\leq n$ and $1\leq r_{i1}<\cdots<r_{il_i}\leq n$ for all $i=1,\ldots,k$, $j=1,\ldots,l_i$, the problem is to find an assignment of minimum cost for $x_1,\ldots,x_n$ that satisfies $f$, if such an assignment exists. The cost of an assignment is simply given by $$\sum_{\substack{i\in\{1,\ldots,n\}\\x_i\,\text{true}}}c_i.$$ Is this problem NP-hard, that is to say, is the accompanying decision problem "Is there a satisfying assignment of cost at most some value $K$" NP-hard?
Now, the standard XOR-SAT problem is in P, for it maps directly to the question of solvability of a system of linear equations over $\mathbb{F}_2$ (see, e. g., https://en.wikipedia.org/wiki/Boolean_satisfiability_problem#XOR-satisfiability). The result of this solution (if it exists) is an affine subspace of $\mathbb{F}_2^n$. The problem is thus reduced to pick the element corresponding with minimal cost from that subspace. Alas, that subspace may be quite large, and indeed, rewriting $f$ in binary $k\times n$-matrix form, with a $1$ for each $x_{r_{ij}}$ at the $i$-th row and the $r_{ij}$-th column, and zero otherwise, we get a cost minimization problem subject to $$Ax=1,$$ where $A$ is said matrix, $x$ is the column vector consisting of the $x_1,\ldots,x_n$ and $1$ is the all-1-vector. This is an instance of a binary linear programming problem, which are known to be NP-hard in general. So the question is, is it NP-hard in this particular instance as well?
