EFX from No-Envy + Monotonicity
Theorem. For a monotone valuation $v$ on indivisible goods, envy-freeness ([[social_choice.fair_division.indivisible.envy_free]]) implies EFX ([[social_choice.fair_division.indivisible.efx]]).
In Lean: SocialChoice.FairDivision.Indivisible.IsEFX.of_noEnvy_mono.
Statement and proof
This is the same content as
[[social_choice.fair_division.indivisible.ef_implies_efx_mono]], stated
inside EFX.lean as a local sufficiency lemma for the EFX existence
proofs in the same file. The Lean signatures differ only in how the
hypothesis is bundled:
IsEnvyFree.isEFX_of_monoinFairness.lean— phrased as a method on the EF predicate.IsEFX.of_noEnvy_monoinEFX.lean— phrased as a constructor of the EFX predicate.
Both prove $T \subseteq S \Rightarrow v_i(T) \le v_i(S)$, combined with EF, yields EFX, by the same calculation: removing an item from the envied bundle (a subset operation) only decreases its value, so the EF inequality $v_i(A(j)) \le v_i(A(i))$ propagates to bundles missing one item.
Why duplicate
Having both forms lets EFX.lean use the constructor directly without
importing the symmetric Fairness-side lemma. It is purely an
organisational convenience; the proof is the same.
References
- [AGT Chapter 11] Nisan, Roughgarden, Tardos, and Vazirani, Algorithmic Game Theory. EF-EFX implication.