We consider the system Applicative_05__BTreeMember.

  Alphabet:

    0 : [] --> a 
    eq : [a * a] --> c 
    false : [] --> c 
    fork : [b * a * b] --> b 
    if : [c * c * c] --> c 
    lt : [a * a] --> c 
    member : [a * b] --> c 
    null : [] --> b 
    s : [a] --> a 
    true : [] --> c 

  Rules:

    lt(s(x), s(y)) => lt(x, y) 
    lt(0, s(x)) => true 
    lt(x, 0) => false 
    eq(x, x) => true 
    eq(s(x), 0) => false 
    eq(0, s(x)) => false 
    member(x, null) => false 
    member(x, fork(y, z, u)) => if(lt(x, z), member(x, y), if(eq(x, z), true, member(x, u))) 

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:

  lt(s(X), s(Y)) => lt(X, Y) 
  lt(0, s(X)) => true 
  lt(X, 0) => false 
  eq(X, X) => true 
  eq(s(X), 0) => false 
  eq(0, s(X)) => false 
  member(X, null) => false 
  member(X, fork(Y, Z, U)) => if(lt(X, Z), member(X, Y), if(eq(X, Z), true, member(X, U))) 

Moreover, the system is finitely branching.  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 Ce-terminating when seen as a many-sorted first-order TRS.

According to the external first-order termination prover, this system is indeed Ce-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) RisEmptyProof [EQUIVALENT]
 || (4) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    lt(s(%X), s(%Y)) -> lt(%X, %Y)
 ||    lt(0, s(%X)) -> true
 ||    lt(%X, 0) -> false
 ||    eq(%X, %X) -> true
 ||    eq(s(%X), 0) -> false
 ||    eq(0, s(%X)) -> false
 ||    member(%X, null) -> false
 ||    member(%X, fork(%Y, %Z, %U)) -> if(lt(%X, %Z), member(%X, %Y), if(eq(%X, %Z), true, member(%X, %U)))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Quasi precedence:
 || s_1 > [lt_2, true, eq_2] > [false, null]
 || member_2 > [lt_2, true, eq_2] > [false, null]
 || member_2 > if_3
 || 
 || 
 || Status:
 || lt_2: multiset status
 || s_1: multiset status
 || 0: multiset status
 || true: multiset status
 || false: multiset status
 || eq_2: [2,1]
 || member_2: multiset status
 || null: multiset status
 || fork_3: multiset status
 || if_3: multiset status
 || ~PAIR_2: multiset status
 || 
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    lt(s(%X), s(%Y)) -> lt(%X, %Y)
 ||    lt(0, s(%X)) -> true
 ||    lt(%X, 0) -> false
 ||    eq(%X, %X) -> true
 ||    eq(s(%X), 0) -> false
 ||    eq(0, s(%X)) -> false
 ||    member(%X, null) -> false
 ||    member(%X, fork(%Y, %Z, %U)) -> if(lt(%X, %Z), member(%X, %Y), if(eq(%X, %Z), true, member(%X, %U)))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || R is empty.
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (3) RisEmptyProof (EQUIVALENT)
 || The TRS R is empty. Hence, termination is trivially proven.
 || ----------------------------------------
 || 
 || (4)
 || 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, all):

  Dependency Pairs P_0:


  Rules R_0:

    lt(s(X), s(Y)) => lt(X, Y) 
    lt(0, s(X)) => true 
    lt(X, 0) => false 
    eq(X, X) => true 
    eq(s(X), 0) => false 
    eq(0, s(X)) => false 
    member(X, null) => false 
    member(X, fork(Y, Z, U)) => if(lt(X, Z), member(X, Y), if(eq(X, Z), true, member(X, U))) 

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

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

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.