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].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

ordered_in(.(X, .(Y, Xs))) → U1(X, Y, Xs, less_in(X, s(Y)))
less_in(s(X), s(Y)) → U3(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U3(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U1(X, Y, Xs, less_out(X, s(Y))) → U2(X, Y, Xs, ordered_in(.(Y, Xs)))
ordered_in(.(X, [])) → ordered_out(.(X, []))
ordered_in([]) → ordered_out([])
U2(X, Y, Xs, ordered_out(.(Y, Xs))) → ordered_out(.(X, .(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_in(x1)  =  ordered_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x2, x3, x4)
less_in(x1, x2)  =  less_in(x1, x2)
s(x1)  =  s(x1)
U3(x1, x2, x3)  =  U3(x3)
0  =  0
less_out(x1, x2)  =  less_out
U2(x1, x2, x3, x4)  =  U2(x4)
[]  =  []
ordered_out(x1)  =  ordered_out

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(.(X, .(Y, Xs))) → U1(X, Y, Xs, less_in(X, s(Y)))
less_in(s(X), s(Y)) → U3(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U3(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U1(X, Y, Xs, less_out(X, s(Y))) → U2(X, Y, Xs, ordered_in(.(Y, Xs)))
ordered_in(.(X, [])) → ordered_out(.(X, []))
ordered_in([]) → ordered_out([])
U2(X, Y, Xs, ordered_out(.(Y, Xs))) → ordered_out(.(X, .(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_in(x1)  =  ordered_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x2, x3, x4)
less_in(x1, x2)  =  less_in(x1, x2)
s(x1)  =  s(x1)
U3(x1, x2, x3)  =  U3(x3)
0  =  0
less_out(x1, x2)  =  less_out
U2(x1, x2, x3, x4)  =  U2(x4)
[]  =  []
ordered_out(x1)  =  ordered_out


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(.(X, .(Y, Xs))) → U11(X, Y, Xs, less_in(X, s(Y)))
ORDERED_IN(.(X, .(Y, Xs))) → LESS_IN(X, s(Y))
LESS_IN(s(X), s(Y)) → U31(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U11(X, Y, Xs, less_out(X, s(Y))) → U21(X, Y, Xs, ordered_in(.(Y, Xs)))
U11(X, Y, Xs, less_out(X, s(Y))) → ORDERED_IN(.(Y, Xs))

The TRS R consists of the following rules:

ordered_in(.(X, .(Y, Xs))) → U1(X, Y, Xs, less_in(X, s(Y)))
less_in(s(X), s(Y)) → U3(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U3(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U1(X, Y, Xs, less_out(X, s(Y))) → U2(X, Y, Xs, ordered_in(.(Y, Xs)))
ordered_in(.(X, [])) → ordered_out(.(X, []))
ordered_in([]) → ordered_out([])
U2(X, Y, Xs, ordered_out(.(Y, Xs))) → ordered_out(.(X, .(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_in(x1)  =  ordered_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x2, x3, x4)
less_in(x1, x2)  =  less_in(x1, x2)
s(x1)  =  s(x1)
U3(x1, x2, x3)  =  U3(x3)
0  =  0
less_out(x1, x2)  =  less_out
U2(x1, x2, x3, x4)  =  U2(x4)
[]  =  []
ordered_out(x1)  =  ordered_out
LESS_IN(x1, x2)  =  LESS_IN(x1, x2)
U31(x1, x2, x3)  =  U31(x3)
ORDERED_IN(x1)  =  ORDERED_IN(x1)
U21(x1, x2, x3, x4)  =  U21(x4)
U11(x1, x2, x3, x4)  =  U11(x2, x3, x4)

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(.(X, .(Y, Xs))) → U11(X, Y, Xs, less_in(X, s(Y)))
ORDERED_IN(.(X, .(Y, Xs))) → LESS_IN(X, s(Y))
LESS_IN(s(X), s(Y)) → U31(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U11(X, Y, Xs, less_out(X, s(Y))) → U21(X, Y, Xs, ordered_in(.(Y, Xs)))
U11(X, Y, Xs, less_out(X, s(Y))) → ORDERED_IN(.(Y, Xs))

The TRS R consists of the following rules:

ordered_in(.(X, .(Y, Xs))) → U1(X, Y, Xs, less_in(X, s(Y)))
less_in(s(X), s(Y)) → U3(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U3(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U1(X, Y, Xs, less_out(X, s(Y))) → U2(X, Y, Xs, ordered_in(.(Y, Xs)))
ordered_in(.(X, [])) → ordered_out(.(X, []))
ordered_in([]) → ordered_out([])
U2(X, Y, Xs, ordered_out(.(Y, Xs))) → ordered_out(.(X, .(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_in(x1)  =  ordered_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x2, x3, x4)
less_in(x1, x2)  =  less_in(x1, x2)
s(x1)  =  s(x1)
U3(x1, x2, x3)  =  U3(x3)
0  =  0
less_out(x1, x2)  =  less_out
U2(x1, x2, x3, x4)  =  U2(x4)
[]  =  []
ordered_out(x1)  =  ordered_out
LESS_IN(x1, x2)  =  LESS_IN(x1, x2)
U31(x1, x2, x3)  =  U31(x3)
ORDERED_IN(x1)  =  ORDERED_IN(x1)
U21(x1, x2, x3, x4)  =  U21(x4)
U11(x1, x2, x3, x4)  =  U11(x2, x3, x4)

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(s(X), s(Y)) → LESS_IN(X, Y)

The TRS R consists of the following rules:

ordered_in(.(X, .(Y, Xs))) → U1(X, Y, Xs, less_in(X, s(Y)))
less_in(s(X), s(Y)) → U3(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U3(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U1(X, Y, Xs, less_out(X, s(Y))) → U2(X, Y, Xs, ordered_in(.(Y, Xs)))
ordered_in(.(X, [])) → ordered_out(.(X, []))
ordered_in([]) → ordered_out([])
U2(X, Y, Xs, ordered_out(.(Y, Xs))) → ordered_out(.(X, .(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_in(x1)  =  ordered_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x2, x3, x4)
less_in(x1, x2)  =  less_in(x1, x2)
s(x1)  =  s(x1)
U3(x1, x2, x3)  =  U3(x3)
0  =  0
less_out(x1, x2)  =  less_out
U2(x1, x2, x3, x4)  =  U2(x4)
[]  =  []
ordered_out(x1)  =  ordered_out
LESS_IN(x1, x2)  =  LESS_IN(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(s(X), s(Y)) → LESS_IN(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(s(X), s(Y)) → LESS_IN(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:

ORDERED_IN(.(X, .(Y, Xs))) → U11(X, Y, Xs, less_in(X, s(Y)))
U11(X, Y, Xs, less_out(X, s(Y))) → ORDERED_IN(.(Y, Xs))

The TRS R consists of the following rules:

ordered_in(.(X, .(Y, Xs))) → U1(X, Y, Xs, less_in(X, s(Y)))
less_in(s(X), s(Y)) → U3(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U3(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U1(X, Y, Xs, less_out(X, s(Y))) → U2(X, Y, Xs, ordered_in(.(Y, Xs)))
ordered_in(.(X, [])) → ordered_out(.(X, []))
ordered_in([]) → ordered_out([])
U2(X, Y, Xs, ordered_out(.(Y, Xs))) → ordered_out(.(X, .(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_in(x1)  =  ordered_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x2, x3, x4)
less_in(x1, x2)  =  less_in(x1, x2)
s(x1)  =  s(x1)
U3(x1, x2, x3)  =  U3(x3)
0  =  0
less_out(x1, x2)  =  less_out
U2(x1, x2, x3, x4)  =  U2(x4)
[]  =  []
ordered_out(x1)  =  ordered_out
ORDERED_IN(x1)  =  ORDERED_IN(x1)
U11(x1, x2, x3, x4)  =  U11(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:

ORDERED_IN(.(X, .(Y, Xs))) → U11(X, Y, Xs, less_in(X, s(Y)))
U11(X, Y, Xs, less_out(X, s(Y))) → ORDERED_IN(.(Y, Xs))

The TRS R consists of the following rules:

less_in(s(X), s(Y)) → U3(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U3(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
less_in(x1, x2)  =  less_in(x1, x2)
s(x1)  =  s(x1)
U3(x1, x2, x3)  =  U3(x3)
0  =  0
less_out(x1, x2)  =  less_out
ORDERED_IN(x1)  =  ORDERED_IN(x1)
U11(x1, x2, x3, x4)  =  U11(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:

ORDERED_IN(.(X, .(Y, Xs))) → U11(Y, Xs, less_in(X, s(Y)))
U11(Y, Xs, less_out) → ORDERED_IN(.(Y, Xs))

The TRS R consists of the following rules:

less_in(s(X), s(Y)) → U3(less_in(X, Y))
less_in(0, s(X)) → less_out
U3(less_out) → less_out

The set Q consists of the following terms:

less_in(x0, x1)
U3(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.

The following dependency pairs can be deleted:

ORDERED_IN(.(X, .(Y, Xs))) → U11(Y, Xs, less_in(X, s(Y)))
U11(Y, Xs, less_out) → ORDERED_IN(.(Y, Xs))
The following rules are removed from R:

less_in(s(X), s(Y)) → U3(less_in(X, Y))
less_in(0, s(X)) → less_out
U3(less_out) → less_out
Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(0) = 0   
POL(ORDERED_IN(x1)) = x1   
POL(U11(x1, x2, x3)) = 2 + 2·x1 + 2·x2 + x3   
POL(U3(x1)) = 2 + x1   
POL(less_in(x1, x2)) = x1 + x2   
POL(less_out) = 2   
POL(s(x1)) = 2 + x1   



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

Q DP problem:
P is empty.
R is empty.
The set Q consists of the following terms:

less_in(x0, x1)
U3(x0)

We have to consider all (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.