Fair Division of Indivisible Items: Envy-Freeness vs. Efficiency Revisited

Abstract

We study conflicts between envy-based fairness and efficiency for allocating indivisible items under additive utilities. We formalize several small, transparent instances showing that standard envy-freeness (EF) or its relaxations EFX and EFX₀—i.e., envy-freeness up to any item, where EFX restricts attention to positively valued items and EFX₀ allows removing zero-valued items as well—can conflict with Pareto-optimality (PO), maximin (MM), or maximum Nash welfare (MNW). Normatively, we argue that envy-freeness (even as EFX or EFX₀) is not a panacea for allocating indivisible items and should be weighed against efficiency and welfare criteria.

1. Introduction

For items (goods) with additive preferences, several desiderata compete: envy-freeness (EF), which precludes agents from preferring another agent’s bundle of items; Pareto-optimality (PO), which precludes wasteful allocations; maximin (MM), which protects the worst-off; and maximum Nash welfare (MNW), which balances efficiency and equity by maximizing the product of utilities. With divisible resources, these aims often align; with indivisibilities, they may clash sharply.

Our contribution is to make several of these conflicts explicit and checkable via formally stated examples1 under a fully specified, simple model (additivity, non-wastefulness, and an equal-sum normalization). In particular, we do the following:

(i)

Exhibit two-agent instances where no allocation is both EF and PO.

(ii)

Give multi-agent examples where EF, PO, and MM cannot be simultaneously attained.

(iii)

Analyze envy-freeness up to any item (EFX₀) and show that while it can co-exist with PO/MM in some cases, it neither guarantees PO nor fully resolves fairness–efficiency tensions, especially with zero-valued items.

(iv)

Relate these constructions to broader results around MNW, goods and chores, mixed divisible and indivisible resources, and cake or land division of divisible items.

2. Related Work

For indivisible goods and additive valuations, the maximum Nash welfare (MNW) rule provides a central link between fairness and efficiency. Caragiannis et al. (2019) showed that MNW allocations are both envy-free up to one good (EF1) and Pareto-optimal (PO), and introduced the relaxation of envy-freeness up to any positively valued item (EFX) that we adopt here. This result established MNW as a key reference point for reconciling fairness and efficiency in the allocation of indivisible items.

We next situate our contributions within the literature by highlighting four key strands:

2.1. Maximum Nash Welfare and Fairness

Amanatidis et al. (2021) proved that MNW allocations are EFX₀ when item values take at most two distinct levels, though this guarantee fails for richer domains. For general additive valuations, they show that an MNW allocation does not guarantee any non-trivial approximation of EFX. Feldman et al. (2024) further quantified the efficiency loss required to achieve varying degrees of EFX, reinforcing MNW’s role as the benchmark for balanced outcomes.

2.2. Existence and Computation

The general existence of EFX and EFX₀ allocations remains open, but several special cases are known. Chaudhury et al. (2020) established this existence of EFX allocations for three agents with additive valuations. Garg and Murhekar (2023) showed that EFX and PO allocations can be efficiently found for binary instances but may not exist for richer domains, where the decision problem becomes NP-hard.

2.3. Beyond Pure Goods: Chores, Mixed Goods, and Money

Fair division with both goods and chores introduces non-monotone preferences. Aziz et al. (2022) designed EF1 procedures, achieving PO in several such environments. Bei et al. (2021) extended fair division to combined divisible and indivisible goods, defining envy-freeness for mixed goods (EFM) and documenting its conflict with PO. Nishimura and Sumita (2025) proposed EFXM, a generalization of EFX to mixed domains, and proved that MNW allocations satisfy it for binary-linear valuations. With money, Meertens et al. (2002) identified sufficient conditions under which EF and PO can coexist, linking discrete fairness to market equilibria.

2.4. Approximate and Geometric Fairness

Barman and Suzuki (2024) showed that for subadditive valuations, there always exists an allocation that is EFX and achieves at least half of the optimal Nash welfare, connecting fairness with approximate efficiency. In the geometric domain, Segal-Halevi and Sziklai (2018) examined the division of piecewise homogeneous goods in connected cake-cutting and demonstrated that resource- and population-monotonicity may conflict with PO. More broadly, a recent survey synthesizes progress on EF1, EFX (EFX₀), and MMS guarantees and outlines important open problems (Amanatidis et al., 2023).

Overall, the literature agrees that exact envy-freeness (EF) is rarely achievable for indivisible goods, but MNW and its variants provide a principled compromise between fairness and efficiency. The persistent tension between EFX, EFX₀, and PO—along with complications from chores, mixed goods, and geometric resources—motivates the explicit constructions in this paper that make these conflicts transparent.

3. Model and Definitions

3.1. Environment

Let $N=\{1,\dots ,n\}$ be a finite set of agents, $n\ge 2$, and M a finite set of indivisible items, $\left|M\right|\ge 2$. An allocation is a partition $A=(A_1,\dots ,A_n)$ of M, where $A_i\subseteq M$ is the bundle assigned to agent i. Note that, $A_i\cap A_j=\varnothing$ for $i\ne j$ and ${\bigcup }_i{A}_i=M$.

We assume throughout:

• Additivity. Each agent i has an additive utility function $u_i:2^M\to {\mathbb{R}}_{\ge 0}$ of the form $u_i\left(S\right)={\sum }_{g\in S}{u}_i\left(g\right)$ with $u_i\left(g\right)\ge 0$ for all g.
• Non-wastefulness. All items are allocated.
• Equal-sum normalization. ${\sum }_{g\in M}{u}_i\left(g\right)$ is equal for all i. All agents have the same total value and the same entitlement.

Whenever we speak of “existence” or “non-existence” of an allocation satisfying a combination of properties, we mean within the assumptions of this model.

3.2. Fairness and Efficiency Notions

3.2.1. Envy-Freeness (EF)

An allocation A is envy-free if

$$u_i\left(A_i\right)\ge u_i\left(A_j\right)\mathrm{for}\mathrm{all}i,j\in N.$$

3.2.2. Pareto-Optimality (PO)

An allocation A is Pareto-optimal if there is no other allocation B with $u_i\left(B_i\right)\ge u_i\left(A_i\right)$ for all i, and $u_j\left(B_j\right)>u_j\left(A_j\right)$ for at least one j.

3.2.3. Maximin (MM)

An allocation A is maximin-optimal if it maximizes ${min}_{i\in N}{u}_i\left(A_i\right)$ over all allocations.

3.2.4. Maximum Nash Welfare (MNW)

An allocation A is a maximum Nash welfare allocation if it maximizes the product ${\prod }_{i\in N}{u}_i\left(A_i\right)$ over all allocations. As usual, we interpret this product as 0 whenever $u_i\left(A_i\right)=0$ for some i.2

3.3. Envy-Freeness up to Any (Positive) Item (EFX and EFX₀)

One relaxation of EF is to aim at allocations where the removal of any item x from an envied agent’s bundle eliminates another agent’s envy. We follow Caragiannis et al. (2019) in defining EFX only with respect to positively valued items, and Amanatidis et al. (2021) in using EFX₀ for the variant where any item may be removed, including items of value 0. Throughout the paper, EFX and EFX₀ are used in this precise sense. The definitions are now as follows:

3.3.1. Envy-Freeness up to Any Positive Item (EFX)—Caragiannis et al. (2019)

An allocation A is EFX if for every pair $i,j$ with $u_i\left(A_i\right)<u_i\left(A_j\right)$ and for every item $g\in A_j$ with $u_i\left(g\right)>0$,

$$u_i\left(A_i\right)\ge u_i\left(A_j\setminus \left\{g\right\}\right).$$

A slightly stronger version of EFX, called EFX₀, differs by allowing for items that are zero-valued by one agent.3

3.3.2. Envy-Freeness up to Any Item (EFX₀)—Amanatidis et al. (2021)

An allocation A is EFX₀ if for every pair $i,j$ with $u_i\left(A_i\right)<u_i\left(A_j\right)$ and for every item $g\in A_j$,

$$u_i\left(A_i\right)\ge u_i\left(A_j\setminus \left\{g\right\}\right).$$

4. Incompatibility 1: EF and PO for Two Agents

We begin with a two-agent instance showing EF and PO may be incompatible, even under additivity and normalization.

Example 1 

(Two agents, four items). Consider two agents, A and B, having the following utilities over a set of four items $M=\{a,b,c,d\}$:

$$\begin{array}{cccccc}& a& b& c& d& \mathit{Sum}\\ A:\hfill & 5& 8& 7& 1& 21\\ B:\hfill & 4& 7& 8& 2& 21\end{array}$$

The allocation $\left(\right\{a,b\},\{c,d\left\}\right)$, gives payoffs $(13,10)$ to $(A,B)$, which is PO and MM: There is no Pareto improvement possible, and it maximizes the minimum utility. However, B envies A since $u_B\left(\{a,b\}\right)=11>10=u_B\left(\{c,d\}\right)$, so it is not EF. In fact, there is no EF allocation in Example 1 because for EF both agents would need to get two items from $\{a,b,c\}$, which is not possible.4 This establishes a clear conflict between EF and efficiency, even in the simplest nontrivial case. A full enumeration summary of the example, checking for all relevant properties, is provided in

Table A1. All allocations for the 2-agent, 4-item instance (Example 1) and the properties they satisfy.

#	A	B	${\mathit{u}}_{\mathit{A}}$	${\mathit{u}}_{\mathit{B}}$	EF	EFX0	EFX	PO	MM	MNW
1	∅	$\{a,b,c,d\}$	0	21	×	×	×	✓	×	×
2	$\left\{a\right\}$	$\{b,c,d\}$	5	17	×	×	×	✓	×	×
3	$\left\{b\right\}$	$\{a,c,d\}$	8	14	×	×	×	✓	×	×
4	$\{a,b\}$	$\{c,d\}$	13	10	×	✓	✓	✓	✓	✓
5	$\left\{c\right\}$	$\{a,b,d\}$	7	13	×	×	×	×	×	×
6	$\{a,c\}$	$\{b,d\}$	12	9	×	✓	✓	×	×	×
7	$\{b,c\}$	$\{a,d\}$	15	6	×	×	×	✓	×	×
8	$\{a,b,c\}$	$\left\{d\right\}$	20	2	×	×	×	✓	×	×
9	$\left\{d\right\}$	$\{a,b,c\}$	1	19	×	×	×	✓	×	×
10	$\{a,d\}$	$\{b,c\}$	6	15	×	×	×	✓	×	×
11	$\{b,d\}$	$\{a,c\}$	9	12	×	✓	✓	✓	×	×
12	$\{a,b,d\}$	$\left\{c\right\}$	14	8	×	×	×	✓	×	×
13	$\{c,d\}$	$\{a,b\}$	8	11	×	✓	✓	×	×	×
14	$\{a,c,d\}$	$\left\{b\right\}$	13	7	×	×	×	×	×	×
15	$\{b,c,d\}$	$\left\{a\right\}$	16	4	×	×	×	✓	×	×
16	$\{a,b,c,d\}$	∅	21	0	×	×	×	✓	×	×

in the Appendix A.

Proposition 1. 

If $\left|M\right|\ge 4$, there exist allocation problems with two agents and $\left|M\right|$ items for which no allocation is simultaneously EF and PO.

Proof. 

Consider Example 1. It can be extended to any number of items $\left|M\right|\ge 4$. Add $\left|M\right|-4$ extra items, each worth the same very small value $ϵ$ to both agents with $\left(\right|M|-4)ϵ<<1$. In the extended instance, the allocation that gives A items $\{a,b\}$, and B items $\{c,d\}$ and all extra items, remains PO (and even MM if all the extra items are given to B), but B still envies A, so it is not EF. Moreover, as in Example 1, no EF allocation exists, because for EF to occur, both agents would need to receive two items from the set $\{a,b,c\}$ which is not possible. Hence, for any $\left|M\right|\ge 4$, there is no allocation that is both EF and PO. □

Fortunately, if there is an EF allocation, there is always an EF–PO–MM allocation when there are $n=2$ agents. It can be found by making Pareto improvements in an EF allocation (Brams et al., 2023).

5. Incompatibility 2: EF, PO, and Maximin with More than Two Agents

We now show that, with more agents, simultaneously imposing EF, PO, and MM may be impossible.

Example 2 

(Three agents, six items). Let $N=\{A,B,C\}$ and $M=\{a,b,c,d,e,f\}$, and the utilities be as follows (with totals normalized to 21):5

$$\begin{array}{cccccccc}& a& b& c& d& e& f& Sum\\ A:\hfill & 6& 5& 4& 3& 2& 1& 21\\ B:\hfill & 3& 4& 5& 6& 2& 1& 21\\ C:\hfill & 4& 6& 1& 1& 5& 4& 21\end{array}$$

One EF allocation, which is not PO, is $\left(\right\{a,c\},\{b,d\},\{e,f\left\}\right)$ with utilities $(10,10,9)$. A distinct PO allocation, which is not EF, is $\left(\right\{a,b\},\{c,d\},\{e,f\left\}\right)$ with utilities $(11,11,9)$. Both are MM, since the minimum utility is 9 in each case.

The EF allocation in the above example is MM but not PO, being (weakly) Pareto-dominated by the PO allocation. The PO allocation is MM but not EF, because C envies A. In addition, the allocation $\left(\right\{a,c\},\{d,e\},\{b,f\left\}\right)$, with utilities $(10,8,10)$ is EF, PO, but not MM.6

We now show that there is no allocation that is EF, PO, and MM. Above, we saw that there exists an allocation with minimum value 9. Hence, the minimum value assigned to an agent in an EF, PO, and MM allocation must be 9. This requires each player to obtain at least two items, implying that each player receives exactly two items. The pairs of items that have a value of at least 9 for the players are as follows:

$$\begin{array}{cc}& \mathit{subsets}\mathit{with}\mathit{value}\mathit{at}\mathit{least}9\\ A:\hfill & \{a,b\},\{a,c\},\{a,d\},\{b,c\}\\ B:\hfill & \{a,d\},\{b,c\},\{b,d\},\{c,d\}\\ C:\hfill & \{a,b\},\{a,e\},\{b,e\},\{b,f\},\{e,f\}\end{array}$$

Because neither A nor B can receive any of the items e and f, agent C must receive $\{e,f\}$ for the allocation to be MM. Therefore, the following four MM allocations remain:

$$\begin{array}{ccc}& \mathit{allocation}& \mathit{properties}\\ 1:\hfill & \left(\right\{a,b\},\{c,d\},\{e,f\left\}\right)& MM,PO,\mathit{not}EF\\ 2:\hfill & \left(\right\{a,c\},\{b,d\},\{e,f\left\}\right)& MM,EF,\mathit{not}PO\\ 3:\hfill & \left(\right\{a,d\},\{b,c\},\{e,f\left\}\right)& MM,EF,\mathit{not}PO\\ 4:\hfill & \left(\right\{b,c\},\{a,d\},\{e,f\left\}\right)& MM,EF,\mathit{not}PO\end{array}$$

Hence, in the example, there is no allocation that simultaneously satisfies EF, PO, and MM.

Proposition 2. 

If $\left|M\right|\ge 6$, there exist allocation problems with three agents and $\left|M\right|$ items for which no allocation is simultaneously EF, PO, and MM.

Proof. 

Consider Example 2, which can be extended to any number of items $\left|M\right|\ge 6$ by adding $\left|M\right|-6$ items of very small value $ϵ$ for each agent, with $\left(\right|M|-6)ϵ<<1$. Denote the set of all extra items by $Z\subseteq M$. In any MM allocation of this extended instance, all $ϵ$-items are assigned to the worst-off agent (C in the first two allocations of Example 2). The first allocation in Example 2 now becomes $\left(\right\{a,b\},\{c,d\},\{e,f\}\cup Z)$ and remains EF and MM but not PO, the second allocation becomes $\left(\right\{a,c\},\{b,d\},\{e,f\}\cup Z)$ and remains PO and MM but not EF. The third allocation turns into $\left(\right\{a,c\},\{d,e\}\cup Z,\{b,f\left\}\right)$ and remains EF and PO but not MM, even if all the extra items are assigned to the worst-off agent B. □

Thus, even when MM can be achieved together with either EF or PO separately, the full triple EF, PO, and MM may be unattainable.

In the following example, we provide an extension to $n=4$ agents:

Example 3 

(Four agents, eight items). $N=\{A,B,C,D\}$, $M=\{a,b,c,d,e,f,g,h\}$:

$$\begin{array}{cccccccccc}& a& b& c& d& e& f& g& h& \mathit{Sum}\\ A:\hfill & 5& 8& 7& 1& 1& 1& 1& 1& 25\\ B:\hfill & 4& 7& 8& 2& 1& 1& 1& 1& 25\\ C:\hfill & 1& 1& 1& 1& 5& 8& 7& 1& 25\\ D:\hfill & 1& 1& 1& 1& 4& 7& 8& 2& 25\end{array}$$

One of several EF allocations is $\left(\right\{b,e\},\{c,h\},\{a,f\},\{d,g\left\}\right)$, all of which are worth $(9,9,9,9)$ to $(A,B,C,D)$. These are all strongly dominated by the PO-MM allocation $\left(\right\{a,b\},\{c,d\},$ $\{e,f\},\{g,h\})$, which is worth $(13,10,13,10)$ to $(A,B,C,D)$. The PO allocation is not EF, because B, whose utility for $\{c,d\}$ is 10, envies A for receiving $\{a,b\}$, which would have utility 11 for B. Also, D envies C for the same reason. That there is no EF, PO and MM allocation can be seen by the fact that the MM allocation is unique. To see why $\left(\right\{a,b\},\{c,d\},\{e,f\},\{g,h\left\}\right)$ is an MM allocation, consider that there was another allocation with a minimum utility of 11. Observe that no agent values a single item 9 or more. Hence, each agent needs to get at least two items, which implies that, for eight items, each agent receives exactly two items. The pairs of items that have a value of at least 11 for A and B would be any pair from the set $\{a,b,c\}$. Given that each of A and B needs to receive two items from that set, such an allocation is impossible. Hence, the minimal value in an MM allocation must be 10. Now, the pairs with value at least 10 for the agents are as follows:

$$\begin{array}{cc}& \mathit{subsets}\mathit{with}\mathit{value}\mathit{at}\mathit{least}10\\ A:\hfill & \{a,b\},\{a,c\},\{b,c\}\\ B:\hfill & \{a,b\},\{a,c\},\{b,c\},\{c,d\}\\ C:\hfill & \{e,f\},\{e,g\},\{f,g\}\\ D:\hfill & \{e,f\},\{e,g\},\{f,g\},\{g,h\}\end{array}$$

As one can see easily, there is only one allocation that gives each agent at least a value of 10, which is allocation $\left(\right\{a,b\},\{c,d\},\{e,f\},\{g,h\left\}\right)$. Hence, there is no allocation in the example that is EF, PO, and MM.

Proposition 3. 

If $\left|M\right|\ge 8$, there exist allocation problems with four agents and $\left|M\right|$ items for which no allocation is simultaneously EF, PO, and MM.

Proof. 

Consider the following extension of Example 3: For any $\left|M\right|>8$, add $\left(\right|M|-8)$ extra items, each of which has a very small value $ϵ$ for every agent, with $\left(\right|M|-8)ϵ<<1$. Denote the set of extra items by Z and its cardinality by z. In the stated EF allocation in Example 3, all agents strictly prefer their own bundle to any other by at least a value of 1. Hence, EF is preserved for any distribution of the extra items, because $zϵ<1$. In the unique PO and MM allocation of Example 3, the minimal value is 10. Therefore, in any other allocation in Example 3, the minimum level of an agent is at most 9. Hence, even if all the extra items are given to one player, the minimum value of a player in any extension of another allocation of Example 3 will be smaller than 10. Such an allocation cannot become MM.

Because the worst-off agents in the MM allocation of Example 3 are B and D, in any MM allocation of the extended instance, all extra items must be assigned to B and D and split as evenly as possible. For z even, this would lead to a split $Z_B\cup Z_D=Z$ with $Z_B\cap Z_D=\varnothing$ and $|Z_B|=|Z_D|$. For odd z, the split $(Z_B,Z_D)$ must be such that $|Z_B|={\displaystyle \frac{z-1}{2}}$ and $|Z_D|={\displaystyle \frac{z+1}{2}}$, or vice versa. The allocation would then be $(\{a,b\},\{c,d\}\cup Z_B,\{e,f\},\{g,h\}\cup Z_D)$. However, this (those) PO and MM allocation(s) would still not be EF, because it is still the case that B envies A and D envies C. □

6. Incompatibility 3: EFX, EFX₀, and Pareto-Optimality

We now turn to relaxations of EF and the role of zero-valued items. From the previous definitions of EF, EFX, and EFX₀ in Section 3.3, we see that EF ⇒ EFX₀ ⇒ EFX. Hence, it will be important to be precise about the definition used.

6.1. A Benchmark Observation

Recall Example 1. The PO-MM allocation $\left(\right\{a,b\},\{c,d\left\}\right)$, giving $(13,10)$ to $(A,B)$, is not EF, because B envies A, as we showed earlier. However, removing item a from A’s bundle suffices to eliminate envy from B’s perspective. This observation lies behind the EFX and EFX₀ definitions given earlier: An allocation is EFX₀ (EFX) if the removal of any (positively valued) item from an envied agent’s bundle eliminates another agent’s envy. Hence, $\left(\right\{a,b\},\{c,d\left\}\right)$ is not EF but EFX₀ (and therefore EFX from EFX₀ ⇒ EFX). Because this allocation is PO and MM, an EFX₀ allocation can coexist with PO and MM in this instance. However, the incompatibility can be re-established if we assume that each agent has a different zero-valued item (if they were the same, we could simply eliminate the item from the distribution process).

6.2. How Zero-Valued Items Affect EFX₀

Example 4. 

Consider $N=\{A,B\}$, $M=\{a,b,c\}$ with

$$\begin{array}{ccccc}& a& b& c& \mathit{Sum}\\ A:\hfill & 2& 1& 0& 3\\ B:\hfill & 2& 0& 1& 3\end{array}$$

Out of the eight possible allocations, of which a complete enumeration can be found in

Table A2. All allocations for the 2-agent, 3-item instance (Example 4) and the properties they satisfy.

#	A	B	${\mathit{u}}_{\mathit{A}}$	${\mathit{u}}_{\mathit{B}}$	EF	EFX0	EFX	PO	MM	MNW
1	$\{a,b,c\}$	∅	3	0	×	×	×	×	×	×
2	$\{a,b\}$	$\left\{c\right\}$	3	1	×	×	✓	✓	✓	✓
3	$\{a,c\}$	$\left\{b\right\}$	2	0	×	×	×	×	×	×
4	$\left\{a\right\}$	$\{b,c\}$	2	1	×	✓	✓	×	✓	×
5	$\{b,c\}$	$\left\{a\right\}$	1	2	×	✓	✓	×	✓	×
6	$\left\{b\right\}$	$\{a,c\}$	1	3	×	×	✓	✓	✓	✓
7	$\left\{c\right\}$	$\{a,b\}$	0	2	×	×	×	×	×	×
8	∅	$\{a,b,c\}$	0	3	×	×	×	×	×	×

in the Appendix A, there are two PO allocations that give a $(3,1)$ split: $\left(\right\{a,b\},\{c\left\}\right)$ and $\left(\right\{b\},\{a,c\left\}\right)$. Both are PO, MM, and EFX (because removing the positively valued item a suffices to eliminate envy), but not EFX₀: The envying agent’s envy can not be eliminated by removing the zero-valued items b in the first allocation and c in the second allocation. A would still be envied by B in the first allocation if we take away item b from A’s bundle, because B would still prefer a, which she values at 2 over c which she values at 1. At the same time, there are EFX₀ allocations such as $\left(\right\{a\},\{b,c\left\}\right)$ and $\left(\right\{b,c\},\{a\left\}\right)$, but those are strictly Pareto-dominated by the previous PO allocations, although they would still be MM.

Proposition 4. 

If $\left|M\right|\ge 3$, there exist allocation problems with two agents and $\left|M\right|\ge 3$ items for which no allocation is simultaneously EFX₀, PO, and MM.

Proof. 

We extend Example 4 to any $\left|M\right|\ge 3$ by adding $\left|M\right|-3$ extra items, each worth a very small value $ϵ$ to both agents, with $\left(\right|M|-3)ϵ<<1$. Denote the set of extra items by Z with $Z_A\cup Z_B=Z$ and $Z_A\cap Z_B=\varnothing$. In the extended instance, the two original PO allocations from Example 4 remain PO for any allocation $(Z_A,Z_B)$ of the $\left|M\right|-3$ extra items and are now as follows: $(\{a,b\}\cup Z_A,\left\{c\right\}\cup Z_B)$ and $(\left\{b\right\}\cup Z_A,\{a,c\}\cup Z_B)$. Those allocations will also be MM if all of the extra items go to the agent with the lower total valuation, i.e., $Z_B=Z$ in the first, and $Z_A=Z$ in the second allocation. Because the total value of the extra items is strictly smaller than one for both agents, in any such PO allocation, the agent who holds item a continues to be envied by the other agent, even if one of the items is removed. So the two allocations $\left(\right\{a,b\},\{c\}\cup Z)$ and $\left(\right\{b\}\cup Z,\{a,c\left\}\right)$ are PO and MM, but not EFX₀. The only two EFX₀ allocations in the extended instance are $\left(\right\{a\},\{b,c\}\cup Z)$ and $\left(\right\{b,c\}\cup Z,\{a\left\}\right)$, i.e., those where one agent receives only item a and the other agent receives all the remaining items (including the extra ones). Those allocations are MM and EFX₀. However, $\left(\right\{a\},\{b,c\}\cup Z)$ is Pareto-dominated by $\left(\right\{a,b\},\{c\}\cup Z)$, and $\left(\right\{b,c\}\cup Z,\{a\left\}\right)$ is Pareto-dominated by $\left(\right\{b\}\cup Z,\{a,c\left\}\right)$. Hence, even after adding the extra $ϵ$-valued items, there is no allocation that satisfies EFX₀, PO, and MM simultaneously. □

This illustrates that removal-based envy relaxations can be satisfied in ways that divert attention from substantive gains. It is known that EFX and EFX₀ allocations exist for two or three agents, but it is not known for four or more agents. Even if the latter is true, it does not solve the envy problem: Envy cannot actually be dispelled by hypothetically eliminating an item from an agent’s bundle, as EFX and EFX₀ counterfactually assume (Chaudhury et al., 2020).

6.3. MNW as a Reference Point

The maximum Nash welfare (MNW) solution is known to satisfy strong fairness and efficiency guarantees for indivisible goods in many settings. Our examples reinforce a complementary perspective: In instances like the ones above, MNW allocations can outperform some EF, EFX, or EFX₀ allocations in both welfare and perceived fairness. For example, in Example 4, which has no EF allocation, the product is 2 for the EFX₀ allocations, whereas it is 3 for the PO-MM allocations.7 Consequently, an MNW allocation does not always seem “unreasonably fair,” as claimed in Caragiannis et al. (2019), at least if one seeks to dispel envy.8

7. Discussion and Normative Implications

Our constructions can be read in two ways.

First, they document structural conflict: Even with additive valuations and normalized total utilities, natural fairness and efficiency desiderata cannot, in general, be satisfied simultaneously. The conflicts do not depend on contrived pathologies but arise in simple, symmetric environments.

Second, they qualify the normative force of envy-based criteria.

• EF is demanding: Insisting on it may rule out PO or MM in otherwise attractive allocations, and, as our examples show, may leave no allocation that satisfies all desiderata at once.
• EFX and EFX₀ weaken EF, but their justification is subtle: They rely on comparisons after hypothetically removing items, and interact delicately with zero-valued items.
• MNW and MM, by focusing on the welfare and protection of the worst-off, sometimes produce allocations that appear, if anything, more compelling than envy-free alternatives.

Our takeaway is not that envy-based notions of fairness should be dismissed, but they should be evaluated alongside efficiency and welfare criteria, especially for indivisible items.

8. Conclusions

We provided simple but fully specified examples that (i) distinguish EF from PO for two agents, (ii) preclude the triple EF, PO, and MM with more agents, and (iii) show that EFX₀ neither ensures PO nor eliminates normative concerns in the presence of zero-valued items.

Potential future work could include the following: Characterize valuation domains where EF/EFX/EFX₀ can coexist with PO/MM; and connect explicit incompatibilities with implementable mechanisms in real allocation environments (allocation of courses, housing, tasks).