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Given two weighted formula utility representations with binary rational numbers as weights, decide whether their values agree on every Boolean valuation. This problem is co-NP-complete.

For membership, a valuation witnessing unequal values certifies non-equivalence. Evaluate all propositional formulas and add their weights exactly in polynomial time. For hardness, compare the one-rule base $\{(\theta ,1)\}$ with the empty base. They are equivalent exactly when $\theta$ is an unsatisfiable Boolean formula. Thus even very small weighted rule bases express a universal equivalence question.

Every finite collection of finite propositional formulas mentions only finitely many Boolean variables. Two Boolean valuations agreeing on those variables satisfy exactly the same formulas. They therefore have the same value under any weighted formula utility representation using that collection.

This observation limits representation over infinite sets. On $\mathbb{N}$, the function indicating whether the selected subset is infinite cannot be represented by a finite rule base: if $F$ is the finite set of mentioned indices, $\varnothing$ and $\mathbb{N}\setminus F$ agree on all those indices but require values zero and one. Completeness claims for finite propositional rule bases must therefore specify a finite underlying set or restrict the functions represented.

For the intended finite representation, put $m=|S|<\infty$. For each $T\subseteq S$, form the mintermA Boolean valuation satisfies ${\chi }_T$ exactly when its selected subset is $T$. Therefore the weighted formula utility representationrepresents $u$ exactly: at the Boolean valuation for $T$, only that minterm contributes, with weight $u(T)$. This proves completeness and the bound of $2^{|S|}$ rules simultaneously. Zero-weight rules may be omitted. The bound counts rules; writing each minterm also requires $O(m)$ symbols in Big O notation.

For the computational question, assume rule weights are rational numbers encoded in binary. This makes exact equality of sums a finite computational task. With this conventional encoding, equivalence of weighted formula utility representations is co-NP-complete.

For membership, the complement has a short witness: a Boolean valuation on the variables occurring in either rule base. Evaluate each propositional formula, sum the weights of satisfied rules, and accept the witness if the sums differ. Propositional formula evaluation takes polynomial time. Exact sums of binary rational numbers also have polynomially bounded bit lengths: the product of their denominators has bit length at most the sum of the input denominator lengths. Thus non-equivalence lies in NP, and equivalence lies in co-NP.

For hardness, take a Boolean satisfiability problem instance $\theta$ and constructThe first utility function is one at satisfying Boolean valuations and zero elsewhere, while the second is always zero. ConsequentlyThe unsatisfiable Boolean formula problem is co-NP-complete, as the complement of the NP-complete Boolean satisfiability problem. The construction is a polynomial-time reduction, proving hardness even with the single weight $1$.

Two qualifications matter in the literal wording. First, unrestricted real weights have no specified finite encoding, so a standard bit-complexity classification needs the encoding assumption above. Second, the problem says only that $S$ is nonempty. If infinite $S$ is allowed while rule bases remain finite, its completeness claim is false. For $S=\mathbb{N}$, take $u(T)=1$ when $T$ is infinite and $u(T)=0$ when $T$ is finite. A finite rule base mentions only a finite subset $F$ of the variables. The Boolean valuations for $\varnothing$ and $\mathbb{N}\setminus F$ agree on all those variables but have different required values. No such rule base represents this $u$. Thus the construction answers the intended finite-set question and identifies the missing hypothesis in the unrestricted statement.