Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-01
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Equivalent characterisations of a nonempty tree by unique paths, edge count, minimal connectivity and maximal acyclicity

Statement

For a finite nonempty graph GG, the following are equivalent:

  1. GG is a tree.
  2. Every two vertices are joined by a unique path.
  3. GG is connected and E(G)=V(G)1|E(G)|=|V(G)|-1.
  4. GG is acyclic and E(G)=V(G)1|E(G)|=|V(G)|-1.
  5. GG is minimally connected: it is connected, but deleting any edge disconnects it.
  6. GG is maximally acyclic: it is acyclic, but adding any missing edge creates a cycle.

Facts & Assumptions

Given: A finite nonempty graph GG.

[L1]

Forests satisfy V=E+c|V|=|E|+c, and trees satisfy E=V1|E|=|V|-1 (For every forest, V=E+c|V|=|E|+c, where cc is the number of connected components, A tree on n1n\ge1 vertices has n1n-1 edges).

[L2]
[F1]

A tree is connected and acyclic (Trees, forests, leaves and isolated vertices).

Proof

technique · direct cycle of implications
1.1

If GG is a tree, connectedness supplies a path between each two vertices, and two distinct paths would contain a cycle. Thus condition 1 implies condition 2.

F1F2
1.2

Under condition 2, GG is connected. Deleting any edge uvuv leaves no uu-vv path, because such a path together with uvuv would give a second path in GG. Thus condition 2 implies condition 5.

F2
1.3

Suppose condition 5 holds. If GG contained a cycle, deleting one edge of that cycle would leave its endpoints joined around the rest of the cycle and would not disconnect GG. Hence GG is acyclic and therefore is a tree. Thus condition 5 implies condition 1.

F2F1
1.4

Condition 1 implies condition 3 by the tree edge count.

L1
1.5

Under condition 3, suppose GG had a cycle. Repeatedly delete an edge from a cycle; each deletion preserves connectedness, and finiteness makes the process stop at a connected acyclic spanning graph TT. At least one edge was deleted, so E(T)<V(G)1|E(T)|<|V(G)|-1, while the forest identity with one component gives E(T)=V(G)1|E(T)|=|V(G)|-1, a contradiction. Hence condition 3 implies condition 1.

L1F2
1.6

Under condition 4, the forest identity gives V=(V1)+c|V|=(|V|-1)+c, so c=1c=1 and GG is a tree. Conversely condition 1 implies condition 4 by definition and the tree edge count.

L1F1
1.7

Condition 1 implies condition 6 by edge addition.

L2
1.8

Under condition 6, if GG were disconnected, vertices in different components would be nonadjacent and adding an edge between them could create no cycle. Hence GG is connected, and its assumed acyclicity makes it a tree.

F1F2
2.1

All six conditions are therefore equivalent.

step 1.1step 1.2step 1.3step 1.4step 1.5step 1.6step 1.7step 1.8

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 50 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources