MathLabs
TheoremProved

Well-founded monovariant termination theorem

Statement

Suppose every legal move s→s′s \to s' of a process decreases an integer-valued function M:S→ZM: \mathcal{S} \to \mathbb{Z} by at least 11, i.e. M(s′)≤M(s)−1M(s') \le M(s) - 1, and M(s)≥0M(s) \ge 0 for all states s∈Ss \in \mathcal{S}. Then starting from any state s0s_0, the process must terminate in at most M(s0)M(s_0) steps.

Why is it true?

You cannot step down a staircase of height M(s0)M(s_0) more than M(s0)M(s_0) times if each step drops at least one stair and you can never go below the ground floor 00.

Proof sketch

**Step 1 (inductive bound after kk moves).** Let s0→s1→s2→⋯→sks_0 \to s_1 \to s_2 \to \cdots \to s_k be any sequence of kk valid moves. Applying the hypothesis M(si)≤M(si−1)−1M(s_{i}) \le M(s_{i-1}) - 1 for i=1,2,…,ki = 1, 2, \dots, k and summing telescopes the inequality to M(sk)≤M(s0)−kM(s_k) \le M(s_0) - k.

**Step 2 (upper bound on kk).** Since M(sk)≥0M(s_k) \ge 0 for every reachable state sk∈Ss_k \in \mathcal{S}, combining the two inequalities gives 0≤M(sk)≤M(s0)−k0 \le M(s_k) \le M(s_0) - k, which rearranges immediately to k≤M(s0)k \le M(s_0). Therefore no valid trajectory can have length k>M(s0)k > M(s_0), and the process must terminate in at most M(s0)M(s_0) steps.

Topics that use this theorem

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

  1. Arthur Engel (1998). Problem-Solving Strategies · DOI:10.1007/b97682
  2. Jessica Striker (2017). Dynamical Algebraic Combinatorics: Promotion, Rowmotion, and Resonance · DOI:10.1090/noti1539