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import FormalConjecturesUtilErdős Problem 599
[AhBe09] Aharoni, Ron and Berger, Eli,
open SimpleGraph
namespace Erdos599
Erdős Problem 599 (the Erdős–Menger conjecture).
Let $G$ be a (possibly infinite) graph and let $A, B$ be disjoint independent sets of vertices. Must there exist a family $P$ of pairwise vertex-disjoint paths from $A$ to $B$, and a set $S$ of vertices containing exactly one vertex from each path in $P$, such that every path from $A$ to $B$ contains at least one vertex of $S$?
For finite $G$ this is equivalent to Menger's theorem. The answer is yes, proved by Aharoni and Berger [AhBe09].
@[category research solved, AMS 5]
theorem erdos_599 : answer(True) ↔
∀ (V : Type) (G : SimpleGraph V) (A B : Set V),
Disjoint A B → G.IsIndepSet A → G.IsIndepSet B →
∃ (ι : Type) (a b : ι → V) (p : ∀ i, G.Walk (a i) (b i)) (S : Set V),
(∀ i, a i ∈ A) ∧ (∀ i, b i ∈ B) ∧ (∀ i, (p i).IsPath) ∧
(Pairwise fun i j => Disjoint {v | v ∈ (p i).support} {v | v ∈ (p j).support}) ∧
S ⊆ {v | ∃ i, v ∈ (p i).support} ∧
(∀ i, ∃! v, v ∈ S ∧ v ∈ (p i).support) ∧
(∀ a' ∈ A, ∀ b' ∈ B, ∀ q : G.Walk a' b', q.IsPath → ∃ v ∈ q.support, v ∈ S) := ⊢ True ↔
∀ (V : Type) (G : SimpleGraph V) (A B : Set V),
Disjoint A B →
G.IsIndepSet A →
G.IsIndepSet B →
∃ ι a b p S,
(∀ (i : ι), a i ∈ A) ∧
(∀ (i : ι), b i ∈ B) ∧
(∀ (i : ι), (p i).IsPath) ∧
(Pairwise fun i j => Disjoint {v | v ∈ (p i).support} {v | v ∈ (p j).support}) ∧
S ⊆ {v | ∃ i, v ∈ (p i).support} ∧
(∀ (i : ι), ∃! v, v ∈ S ∧ v ∈ (p i).support) ∧
∀ a' ∈ A, ∀ b' ∈ B, ∀ (q : G.Walk a' b'), q.IsPath → ∃ v ∈ q.support, v ∈ S
All goals completed! 🐙
Menger's theorem for infinite graphs (Aharoni–Berger [AhBe09]).
The theorem actually proved by Aharoni and Berger holds for arbitrary vertex sets $A$
and $B$: in any (possibly infinite) graph $G$ there is a family $P$ of pairwise
vertex-disjoint $A$--$B$ paths together with an $A$--$B$ separator $S$ consisting of the
choice of exactly one vertex from each path in $P$. The disjointness and independence
hypotheses of erdos_599 are not needed.
@[category research solved, AMS 5]
theorem erdos_599.variants.aharoni_berger :
∀ (V : Type) (G : SimpleGraph V) (A B : Set V),
∃ (ι : Type) (a b : ι → V) (p : ∀ i, G.Walk (a i) (b i)) (S : Set V),
(∀ i, a i ∈ A) ∧ (∀ i, b i ∈ B) ∧ (∀ i, (p i).IsPath) ∧
(Pairwise fun i j => Disjoint {v | v ∈ (p i).support} {v | v ∈ (p j).support}) ∧
S ⊆ {v | ∃ i, v ∈ (p i).support} ∧
(∀ i, ∃! v, v ∈ S ∧ v ∈ (p i).support) ∧
(∀ a' ∈ A, ∀ b' ∈ B, ∀ q : G.Walk a' b', q.IsPath → ∃ v ∈ q.support, v ∈ S) := ⊢ ∀ (V : Type) (G : SimpleGraph V) (A B : Set V),
∃ ι a b p S,
(∀ (i : ι), a i ∈ A) ∧
(∀ (i : ι), b i ∈ B) ∧
(∀ (i : ι), (p i).IsPath) ∧
(Pairwise fun i j => Disjoint {v | v ∈ (p i).support} {v | v ∈ (p j).support}) ∧
S ⊆ {v | ∃ i, v ∈ (p i).support} ∧
(∀ (i : ι), ∃! v, v ∈ S ∧ v ∈ (p i).support) ∧
∀ a' ∈ A, ∀ b' ∈ B, ∀ (q : G.Walk a' b'), q.IsPath → ∃ v ∈ q.support, v ∈ S
All goals completed! 🐙
Sanity check: when $A = \varnothing$ the conclusion of erdos_599 holds trivially, with
the empty family of paths and $S = \varnothing$ (the covering condition is vacuous since
there is no path starting in $\varnothing$).
@[category test, AMS 5]
theorem erdos_599.test.empty_A (V : Type) (G : SimpleGraph V) (B : Set V) :
∃ (ι : Type) (a b : ι → V) (p : ∀ i, G.Walk (a i) (b i)) (S : Set V),
(∀ i, a i ∈ (∅ : Set V)) ∧ (∀ i, b i ∈ B) ∧ (∀ i, (p i).IsPath) ∧
(Pairwise fun i j => Disjoint {v | v ∈ (p i).support} {v | v ∈ (p j).support}) ∧
S ⊆ {v | ∃ i, v ∈ (p i).support} ∧
(∀ i, ∃! v, v ∈ S ∧ v ∈ (p i).support) ∧
(∀ a' ∈ (∅ : Set V), ∀ b' ∈ B, ∀ q : G.Walk a' b', q.IsPath →
∃ v ∈ q.support, v ∈ S) := V:TypeG:SimpleGraph VB:Set V⊢ ∃ ι a b p S,
(∀ (i : ι), a i ∈ ∅) ∧
(∀ (i : ι), b i ∈ B) ∧
(∀ (i : ι), (p i).IsPath) ∧
(Pairwise fun i j => Disjoint {v | v ∈ (p i).support} {v | v ∈ (p j).support}) ∧
S ⊆ {v | ∃ i, v ∈ (p i).support} ∧
(∀ (i : ι), ∃! v, v ∈ S ∧ v ∈ (p i).support) ∧
∀ a' ∈ ∅, ∀ b' ∈ B, ∀ (q : G.Walk a' b'), q.IsPath → ∃ v ∈ q.support, v ∈ S
All goals completed! 🐙
end Erdos599