We consider the system 05height.

  Alphabet:

    cons : [t * f] --> f 
    heightf : [f] --> N 
    heightt : [t] --> N 
    leaf : [] --> t 
    max : [N * N] --> N 
    nil : [] --> f 
    node : [f] --> t 
    s : [N] --> N 
    z : [] --> N 

  Rules:

    heightf(nil) => z 
    heightf(cons(x, y)) => max(heightt(x), heightf(y)) 
    heightt(leaf) => z 
    heightt(node(x)) => s(heightf(x)) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

We observe that the rules contain a first-order subset:

  heightf(nil) => z 
  heightf(cons(X, Y)) => max(heightt(X), heightf(Y)) 
  heightt(leaf) => z 
  heightt(node(X)) => s(heightf(X)) 

Moreover, the system is orthogonal.  Thus, by [Kop12, Thm. 7.55], we may omit all first-order dependency pairs from the dependency pair problem (DP(R), R) if this first-order part is terminating when seen as a many-sorted first-order TRS.

According to the external first-order termination prover, this system is indeed terminating:

 || proof of resources/system.trs
 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
 || 
 || 
 || Termination w.r.t. Q of the given QTRS could be proven:
 || 
 || (0) QTRS
 || (1) QTRSRRRProof [EQUIVALENT]
 || (2) QTRS
 || (3) QTRSRRRProof [EQUIVALENT]
 || (4) QTRS
 || (5) RisEmptyProof [EQUIVALENT]
 || (6) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    heightf(nil) -> z
 ||    heightf(cons(%X, %Y)) -> max(heightt(%X), heightf(%Y))
 ||    heightt(leaf) -> z
 ||    heightt(node(%X)) -> s(heightf(%X))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(cons(x_1, x_2)) = x_1 + 2*x_2
 ||    POL(heightf(x_1)) = 2*x_1
 ||    POL(heightt(x_1)) = 2*x_1
 ||    POL(leaf) = 2
 ||    POL(max(x_1, x_2)) = x_1 + x_2
 ||    POL(nil) = 2
 ||    POL(node(x_1)) = 2 + 2*x_1
 ||    POL(s(x_1)) = 1 + 2*x_1
 ||    POL(z) = 2
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    heightf(nil) -> z
 ||    heightt(leaf) -> z
 ||    heightt(node(%X)) -> s(heightf(%X))
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    heightf(cons(%X, %Y)) -> max(heightt(%X), heightf(%Y))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (3) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Knuth-Bendix order [KBO] with precedence:cons_2 > heightf_1 > heightt_1 > max_2
 || 
 || and weight map:
 || 
 ||    heightf_1=1
 ||    heightt_1=1
 ||    cons_2=2
 ||    max_2=0
 || 
 || The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    heightf(cons(%X, %Y)) -> max(heightt(%X), heightf(%Y))
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (4)
 || Obligation:
 || Q restricted rewrite system:
 || R is empty.
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (5) RisEmptyProof (EQUIVALENT)
 || The TRS R is empty. Hence, termination is trivially proven.
 || ----------------------------------------
 || 
 || (6)
 || YES
 || 
We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs.

We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:


  Rules R_0:

    heightf(nil) => z 
    heightf(cons(X, Y)) => max(heightt(X), heightf(Y)) 
    heightt(leaf) => z 
    heightt(node(X)) => s(heightf(X)) 

Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_0, minimal, formative).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:


This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.