Left Termination of the query pattern ordered_in_1(g) w.r.t. the given Prolog program could successfully be proven:



Prolog
  ↳ PrologToPiTRSProof

Clauses:

ordered([]).
ordered(.(X, [])).
ordered(.(X, .(Y, Xs))) :- ','(less(X, s(Y)), ordered(.(Y, Xs))).
less(0, s(X)).
less(s(X), s(Y)) :- less(X, Y).

Queries:

ordered(g).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
ordered_in: (b)
less_in: (b,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, less_in_gg(X, s(Y)))
less_in_gg(0, s(X)) → less_out_gg(0, s(X))
less_in_gg(s(X), s(Y)) → U3_gg(X, Y, less_in_gg(X, Y))
U3_gg(X, Y, less_out_gg(X, Y)) → less_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, less_out_gg(X, s(Y))) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_in_g(x1)  =  ordered_in_g(x1)
[]  =  []
ordered_out_g(x1)  =  ordered_out_g
.(x1, x2)  =  .(x1, x2)
U1_g(x1, x2, x3, x4)  =  U1_g(x2, x3, x4)
less_in_gg(x1, x2)  =  less_in_gg(x1, x2)
0  =  0
s(x1)  =  s(x1)
less_out_gg(x1, x2)  =  less_out_gg
U3_gg(x1, x2, x3)  =  U3_gg(x3)
U2_g(x1, x2, x3, x4)  =  U2_g(x4)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
PiTRS
      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, less_in_gg(X, s(Y)))
less_in_gg(0, s(X)) → less_out_gg(0, s(X))
less_in_gg(s(X), s(Y)) → U3_gg(X, Y, less_in_gg(X, Y))
U3_gg(X, Y, less_out_gg(X, Y)) → less_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, less_out_gg(X, s(Y))) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_in_g(x1)  =  ordered_in_g(x1)
[]  =  []
ordered_out_g(x1)  =  ordered_out_g
.(x1, x2)  =  .(x1, x2)
U1_g(x1, x2, x3, x4)  =  U1_g(x2, x3, x4)
less_in_gg(x1, x2)  =  less_in_gg(x1, x2)
0  =  0
s(x1)  =  s(x1)
less_out_gg(x1, x2)  =  less_out_gg
U3_gg(x1, x2, x3)  =  U3_gg(x3)
U2_g(x1, x2, x3, x4)  =  U2_g(x4)


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, less_in_gg(X, s(Y)))
ORDERED_IN_G(.(X, .(Y, Xs))) → LESS_IN_GG(X, s(Y))
LESS_IN_GG(s(X), s(Y)) → U3_GG(X, Y, less_in_gg(X, Y))
LESS_IN_GG(s(X), s(Y)) → LESS_IN_GG(X, Y)
U1_G(X, Y, Xs, less_out_gg(X, s(Y))) → U2_G(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U1_G(X, Y, Xs, less_out_gg(X, s(Y))) → ORDERED_IN_G(.(Y, Xs))

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, less_in_gg(X, s(Y)))
less_in_gg(0, s(X)) → less_out_gg(0, s(X))
less_in_gg(s(X), s(Y)) → U3_gg(X, Y, less_in_gg(X, Y))
U3_gg(X, Y, less_out_gg(X, Y)) → less_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, less_out_gg(X, s(Y))) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_in_g(x1)  =  ordered_in_g(x1)
[]  =  []
ordered_out_g(x1)  =  ordered_out_g
.(x1, x2)  =  .(x1, x2)
U1_g(x1, x2, x3, x4)  =  U1_g(x2, x3, x4)
less_in_gg(x1, x2)  =  less_in_gg(x1, x2)
0  =  0
s(x1)  =  s(x1)
less_out_gg(x1, x2)  =  less_out_gg
U3_gg(x1, x2, x3)  =  U3_gg(x3)
U2_g(x1, x2, x3, x4)  =  U2_g(x4)
LESS_IN_GG(x1, x2)  =  LESS_IN_GG(x1, x2)
ORDERED_IN_G(x1)  =  ORDERED_IN_G(x1)
U1_G(x1, x2, x3, x4)  =  U1_G(x2, x3, x4)
U2_G(x1, x2, x3, x4)  =  U2_G(x4)
U3_GG(x1, x2, x3)  =  U3_GG(x3)

We have to consider all (P,R,Pi)-chains

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
PiDP
          ↳ DependencyGraphProof

Pi DP problem:
The TRS P consists of the following rules:

ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, less_in_gg(X, s(Y)))
ORDERED_IN_G(.(X, .(Y, Xs))) → LESS_IN_GG(X, s(Y))
LESS_IN_GG(s(X), s(Y)) → U3_GG(X, Y, less_in_gg(X, Y))
LESS_IN_GG(s(X), s(Y)) → LESS_IN_GG(X, Y)
U1_G(X, Y, Xs, less_out_gg(X, s(Y))) → U2_G(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U1_G(X, Y, Xs, less_out_gg(X, s(Y))) → ORDERED_IN_G(.(Y, Xs))

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, less_in_gg(X, s(Y)))
less_in_gg(0, s(X)) → less_out_gg(0, s(X))
less_in_gg(s(X), s(Y)) → U3_gg(X, Y, less_in_gg(X, Y))
U3_gg(X, Y, less_out_gg(X, Y)) → less_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, less_out_gg(X, s(Y))) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_in_g(x1)  =  ordered_in_g(x1)
[]  =  []
ordered_out_g(x1)  =  ordered_out_g
.(x1, x2)  =  .(x1, x2)
U1_g(x1, x2, x3, x4)  =  U1_g(x2, x3, x4)
less_in_gg(x1, x2)  =  less_in_gg(x1, x2)
0  =  0
s(x1)  =  s(x1)
less_out_gg(x1, x2)  =  less_out_gg
U3_gg(x1, x2, x3)  =  U3_gg(x3)
U2_g(x1, x2, x3, x4)  =  U2_g(x4)
LESS_IN_GG(x1, x2)  =  LESS_IN_GG(x1, x2)
ORDERED_IN_G(x1)  =  ORDERED_IN_G(x1)
U1_G(x1, x2, x3, x4)  =  U1_G(x2, x3, x4)
U2_G(x1, x2, x3, x4)  =  U2_G(x4)
U3_GG(x1, x2, x3)  =  U3_GG(x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 3 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
PiDP
                ↳ UsableRulesProof
              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

LESS_IN_GG(s(X), s(Y)) → LESS_IN_GG(X, Y)

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, less_in_gg(X, s(Y)))
less_in_gg(0, s(X)) → less_out_gg(0, s(X))
less_in_gg(s(X), s(Y)) → U3_gg(X, Y, less_in_gg(X, Y))
U3_gg(X, Y, less_out_gg(X, Y)) → less_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, less_out_gg(X, s(Y))) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_in_g(x1)  =  ordered_in_g(x1)
[]  =  []
ordered_out_g(x1)  =  ordered_out_g
.(x1, x2)  =  .(x1, x2)
U1_g(x1, x2, x3, x4)  =  U1_g(x2, x3, x4)
less_in_gg(x1, x2)  =  less_in_gg(x1, x2)
0  =  0
s(x1)  =  s(x1)
less_out_gg(x1, x2)  =  less_out_gg
U3_gg(x1, x2, x3)  =  U3_gg(x3)
U2_g(x1, x2, x3, x4)  =  U2_g(x4)
LESS_IN_GG(x1, x2)  =  LESS_IN_GG(x1, x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

LESS_IN_GG(s(X), s(Y)) → LESS_IN_GG(X, Y)

R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP

Q DP problem:
The TRS P consists of the following rules:

LESS_IN_GG(s(X), s(Y)) → LESS_IN_GG(X, Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
PiDP
                ↳ UsableRulesProof

Pi DP problem:
The TRS P consists of the following rules:

U1_G(X, Y, Xs, less_out_gg(X, s(Y))) → ORDERED_IN_G(.(Y, Xs))
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, less_in_gg(X, s(Y)))

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, less_in_gg(X, s(Y)))
less_in_gg(0, s(X)) → less_out_gg(0, s(X))
less_in_gg(s(X), s(Y)) → U3_gg(X, Y, less_in_gg(X, Y))
U3_gg(X, Y, less_out_gg(X, Y)) → less_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, less_out_gg(X, s(Y))) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_in_g(x1)  =  ordered_in_g(x1)
[]  =  []
ordered_out_g(x1)  =  ordered_out_g
.(x1, x2)  =  .(x1, x2)
U1_g(x1, x2, x3, x4)  =  U1_g(x2, x3, x4)
less_in_gg(x1, x2)  =  less_in_gg(x1, x2)
0  =  0
s(x1)  =  s(x1)
less_out_gg(x1, x2)  =  less_out_gg
U3_gg(x1, x2, x3)  =  U3_gg(x3)
U2_g(x1, x2, x3, x4)  =  U2_g(x4)
ORDERED_IN_G(x1)  =  ORDERED_IN_G(x1)
U1_G(x1, x2, x3, x4)  =  U1_G(x2, x3, x4)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
PiDP
                    ↳ PiDPToQDPProof

Pi DP problem:
The TRS P consists of the following rules:

U1_G(X, Y, Xs, less_out_gg(X, s(Y))) → ORDERED_IN_G(.(Y, Xs))
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, less_in_gg(X, s(Y)))

The TRS R consists of the following rules:

less_in_gg(0, s(X)) → less_out_gg(0, s(X))
less_in_gg(s(X), s(Y)) → U3_gg(X, Y, less_in_gg(X, Y))
U3_gg(X, Y, less_out_gg(X, Y)) → less_out_gg(s(X), s(Y))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
less_in_gg(x1, x2)  =  less_in_gg(x1, x2)
0  =  0
s(x1)  =  s(x1)
less_out_gg(x1, x2)  =  less_out_gg
U3_gg(x1, x2, x3)  =  U3_gg(x3)
ORDERED_IN_G(x1)  =  ORDERED_IN_G(x1)
U1_G(x1, x2, x3, x4)  =  U1_G(x2, x3, x4)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ UsableRulesReductionPairsProof

Q DP problem:
The TRS P consists of the following rules:

U1_G(Y, Xs, less_out_gg) → ORDERED_IN_G(.(Y, Xs))
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(Y, Xs, less_in_gg(X, s(Y)))

The TRS R consists of the following rules:

less_in_gg(0, s(X)) → less_out_gg
less_in_gg(s(X), s(Y)) → U3_gg(less_in_gg(X, Y))
U3_gg(less_out_gg) → less_out_gg

The set Q consists of the following terms:

less_in_gg(x0, x1)
U3_gg(x0)

We have to consider all (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

less_in_gg(0, s(X)) → less_out_gg
Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x1, x2)) = 2·x1 + 2·x2   
POL(0) = 2   
POL(ORDERED_IN_G(x1)) = 1 + x1   
POL(U1_G(x1, x2, x3)) = 1 + 2·x1 + 2·x2 + 2·x3   
POL(U3_gg(x1)) = x1   
POL(less_in_gg(x1, x2)) = x1 + x2   
POL(less_out_gg) = 0   
POL(s(x1)) = x1   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ UsableRulesReductionPairsProof
QDP
                            ↳ RuleRemovalProof

Q DP problem:
The TRS P consists of the following rules:

U1_G(Y, Xs, less_out_gg) → ORDERED_IN_G(.(Y, Xs))
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(Y, Xs, less_in_gg(X, s(Y)))

The TRS R consists of the following rules:

less_in_gg(s(X), s(Y)) → U3_gg(less_in_gg(X, Y))
U3_gg(less_out_gg) → less_out_gg

The set Q consists of the following terms:

less_in_gg(x0, x1)
U3_gg(x0)

We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

U1_G(Y, Xs, less_out_gg) → ORDERED_IN_G(.(Y, Xs))

Strictly oriented rules of the TRS R:

U3_gg(less_out_gg) → less_out_gg

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x1, x2)) = 2·x1 + 2·x2   
POL(ORDERED_IN_G(x1)) = 1 + x1   
POL(U1_G(x1, x2, x3)) = 2·x1 + 2·x2 + x3   
POL(U3_gg(x1)) = 2 + x1   
POL(less_in_gg(x1, x2)) = x1 + x2   
POL(less_out_gg) = 2   
POL(s(x1)) = 1 + x1   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ UsableRulesReductionPairsProof
                          ↳ QDP
                            ↳ RuleRemovalProof
QDP
                                ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(Y, Xs, less_in_gg(X, s(Y)))

The TRS R consists of the following rules:

less_in_gg(s(X), s(Y)) → U3_gg(less_in_gg(X, Y))

The set Q consists of the following terms:

less_in_gg(x0, x1)
U3_gg(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.