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



↳ PROLOG
  ↳ PrologToPiTRSProof

ordered1({}0).
ordered1(.2(underscore, {}0)).
ordered1(.2(X, .2(Y, Xs))) :- less2(X, s1(Y)), ordered1(.2(Y, Xs)).
less2(00, s1(underscore1)).
less2(s1(X), s1(Y)) :- less2(X, Y).


With regard to the inferred argument filtering the predicates were used in the following modes:
ordered1: (b)
less2: (b,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


ordered_1_in_g1([]_0) -> ordered_1_out_g1([]_0)
ordered_1_in_g1(._22(underscore, []_0)) -> ordered_1_out_g1(._22(underscore, []_0))
ordered_1_in_g1(._22(X, ._22(Y, Xs))) -> if_ordered_1_in_1_g4(X, Y, Xs, less_2_in_gg2(X, s_11(Y)))
less_2_in_gg2(0_0, s_11(underscore1)) -> less_2_out_gg2(0_0, s_11(underscore1))
less_2_in_gg2(s_11(X), s_11(Y)) -> if_less_2_in_1_gg3(X, Y, less_2_in_gg2(X, Y))
if_less_2_in_1_gg3(X, Y, less_2_out_gg2(X, Y)) -> less_2_out_gg2(s_11(X), s_11(Y))
if_ordered_1_in_1_g4(X, Y, Xs, less_2_out_gg2(X, s_11(Y))) -> if_ordered_1_in_2_g4(X, Y, Xs, ordered_1_in_g1(._22(Y, Xs)))
if_ordered_1_in_2_g4(X, Y, Xs, ordered_1_out_g1(._22(Y, Xs))) -> ordered_1_out_g1(._22(X, ._22(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_1_in_g1(x1)  =  ordered_1_in_g1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
ordered_1_out_g1(x1)  =  ordered_1_out_g
if_ordered_1_in_1_g4(x1, x2, x3, x4)  =  if_ordered_1_in_1_g3(x2, x3, x4)
less_2_in_gg2(x1, x2)  =  less_2_in_gg2(x1, x2)
less_2_out_gg2(x1, x2)  =  less_2_out_gg
if_less_2_in_1_gg3(x1, x2, x3)  =  if_less_2_in_1_gg1(x3)
if_ordered_1_in_2_g4(x1, x2, x3, x4)  =  if_ordered_1_in_2_g1(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_1_in_g1([]_0) -> ordered_1_out_g1([]_0)
ordered_1_in_g1(._22(underscore, []_0)) -> ordered_1_out_g1(._22(underscore, []_0))
ordered_1_in_g1(._22(X, ._22(Y, Xs))) -> if_ordered_1_in_1_g4(X, Y, Xs, less_2_in_gg2(X, s_11(Y)))
less_2_in_gg2(0_0, s_11(underscore1)) -> less_2_out_gg2(0_0, s_11(underscore1))
less_2_in_gg2(s_11(X), s_11(Y)) -> if_less_2_in_1_gg3(X, Y, less_2_in_gg2(X, Y))
if_less_2_in_1_gg3(X, Y, less_2_out_gg2(X, Y)) -> less_2_out_gg2(s_11(X), s_11(Y))
if_ordered_1_in_1_g4(X, Y, Xs, less_2_out_gg2(X, s_11(Y))) -> if_ordered_1_in_2_g4(X, Y, Xs, ordered_1_in_g1(._22(Y, Xs)))
if_ordered_1_in_2_g4(X, Y, Xs, ordered_1_out_g1(._22(Y, Xs))) -> ordered_1_out_g1(._22(X, ._22(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_1_in_g1(x1)  =  ordered_1_in_g1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
ordered_1_out_g1(x1)  =  ordered_1_out_g
if_ordered_1_in_1_g4(x1, x2, x3, x4)  =  if_ordered_1_in_1_g3(x2, x3, x4)
less_2_in_gg2(x1, x2)  =  less_2_in_gg2(x1, x2)
less_2_out_gg2(x1, x2)  =  less_2_out_gg
if_less_2_in_1_gg3(x1, x2, x3)  =  if_less_2_in_1_gg1(x3)
if_ordered_1_in_2_g4(x1, x2, x3, x4)  =  if_ordered_1_in_2_g1(x4)


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

ORDERED_1_IN_G1(._22(X, ._22(Y, Xs))) -> IF_ORDERED_1_IN_1_G4(X, Y, Xs, less_2_in_gg2(X, s_11(Y)))
ORDERED_1_IN_G1(._22(X, ._22(Y, Xs))) -> LESS_2_IN_GG2(X, s_11(Y))
LESS_2_IN_GG2(s_11(X), s_11(Y)) -> IF_LESS_2_IN_1_GG3(X, Y, less_2_in_gg2(X, Y))
LESS_2_IN_GG2(s_11(X), s_11(Y)) -> LESS_2_IN_GG2(X, Y)
IF_ORDERED_1_IN_1_G4(X, Y, Xs, less_2_out_gg2(X, s_11(Y))) -> IF_ORDERED_1_IN_2_G4(X, Y, Xs, ordered_1_in_g1(._22(Y, Xs)))
IF_ORDERED_1_IN_1_G4(X, Y, Xs, less_2_out_gg2(X, s_11(Y))) -> ORDERED_1_IN_G1(._22(Y, Xs))

The TRS R consists of the following rules:

ordered_1_in_g1([]_0) -> ordered_1_out_g1([]_0)
ordered_1_in_g1(._22(underscore, []_0)) -> ordered_1_out_g1(._22(underscore, []_0))
ordered_1_in_g1(._22(X, ._22(Y, Xs))) -> if_ordered_1_in_1_g4(X, Y, Xs, less_2_in_gg2(X, s_11(Y)))
less_2_in_gg2(0_0, s_11(underscore1)) -> less_2_out_gg2(0_0, s_11(underscore1))
less_2_in_gg2(s_11(X), s_11(Y)) -> if_less_2_in_1_gg3(X, Y, less_2_in_gg2(X, Y))
if_less_2_in_1_gg3(X, Y, less_2_out_gg2(X, Y)) -> less_2_out_gg2(s_11(X), s_11(Y))
if_ordered_1_in_1_g4(X, Y, Xs, less_2_out_gg2(X, s_11(Y))) -> if_ordered_1_in_2_g4(X, Y, Xs, ordered_1_in_g1(._22(Y, Xs)))
if_ordered_1_in_2_g4(X, Y, Xs, ordered_1_out_g1(._22(Y, Xs))) -> ordered_1_out_g1(._22(X, ._22(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_1_in_g1(x1)  =  ordered_1_in_g1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
ordered_1_out_g1(x1)  =  ordered_1_out_g
if_ordered_1_in_1_g4(x1, x2, x3, x4)  =  if_ordered_1_in_1_g3(x2, x3, x4)
less_2_in_gg2(x1, x2)  =  less_2_in_gg2(x1, x2)
less_2_out_gg2(x1, x2)  =  less_2_out_gg
if_less_2_in_1_gg3(x1, x2, x3)  =  if_less_2_in_1_gg1(x3)
if_ordered_1_in_2_g4(x1, x2, x3, x4)  =  if_ordered_1_in_2_g1(x4)
IF_ORDERED_1_IN_2_G4(x1, x2, x3, x4)  =  IF_ORDERED_1_IN_2_G1(x4)
IF_ORDERED_1_IN_1_G4(x1, x2, x3, x4)  =  IF_ORDERED_1_IN_1_G3(x2, x3, x4)
IF_LESS_2_IN_1_GG3(x1, x2, x3)  =  IF_LESS_2_IN_1_GG1(x3)
ORDERED_1_IN_G1(x1)  =  ORDERED_1_IN_G1(x1)
LESS_2_IN_GG2(x1, x2)  =  LESS_2_IN_GG2(x1, x2)

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_1_IN_G1(._22(X, ._22(Y, Xs))) -> IF_ORDERED_1_IN_1_G4(X, Y, Xs, less_2_in_gg2(X, s_11(Y)))
ORDERED_1_IN_G1(._22(X, ._22(Y, Xs))) -> LESS_2_IN_GG2(X, s_11(Y))
LESS_2_IN_GG2(s_11(X), s_11(Y)) -> IF_LESS_2_IN_1_GG3(X, Y, less_2_in_gg2(X, Y))
LESS_2_IN_GG2(s_11(X), s_11(Y)) -> LESS_2_IN_GG2(X, Y)
IF_ORDERED_1_IN_1_G4(X, Y, Xs, less_2_out_gg2(X, s_11(Y))) -> IF_ORDERED_1_IN_2_G4(X, Y, Xs, ordered_1_in_g1(._22(Y, Xs)))
IF_ORDERED_1_IN_1_G4(X, Y, Xs, less_2_out_gg2(X, s_11(Y))) -> ORDERED_1_IN_G1(._22(Y, Xs))

The TRS R consists of the following rules:

ordered_1_in_g1([]_0) -> ordered_1_out_g1([]_0)
ordered_1_in_g1(._22(underscore, []_0)) -> ordered_1_out_g1(._22(underscore, []_0))
ordered_1_in_g1(._22(X, ._22(Y, Xs))) -> if_ordered_1_in_1_g4(X, Y, Xs, less_2_in_gg2(X, s_11(Y)))
less_2_in_gg2(0_0, s_11(underscore1)) -> less_2_out_gg2(0_0, s_11(underscore1))
less_2_in_gg2(s_11(X), s_11(Y)) -> if_less_2_in_1_gg3(X, Y, less_2_in_gg2(X, Y))
if_less_2_in_1_gg3(X, Y, less_2_out_gg2(X, Y)) -> less_2_out_gg2(s_11(X), s_11(Y))
if_ordered_1_in_1_g4(X, Y, Xs, less_2_out_gg2(X, s_11(Y))) -> if_ordered_1_in_2_g4(X, Y, Xs, ordered_1_in_g1(._22(Y, Xs)))
if_ordered_1_in_2_g4(X, Y, Xs, ordered_1_out_g1(._22(Y, Xs))) -> ordered_1_out_g1(._22(X, ._22(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_1_in_g1(x1)  =  ordered_1_in_g1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
ordered_1_out_g1(x1)  =  ordered_1_out_g
if_ordered_1_in_1_g4(x1, x2, x3, x4)  =  if_ordered_1_in_1_g3(x2, x3, x4)
less_2_in_gg2(x1, x2)  =  less_2_in_gg2(x1, x2)
less_2_out_gg2(x1, x2)  =  less_2_out_gg
if_less_2_in_1_gg3(x1, x2, x3)  =  if_less_2_in_1_gg1(x3)
if_ordered_1_in_2_g4(x1, x2, x3, x4)  =  if_ordered_1_in_2_g1(x4)
IF_ORDERED_1_IN_2_G4(x1, x2, x3, x4)  =  IF_ORDERED_1_IN_2_G1(x4)
IF_ORDERED_1_IN_1_G4(x1, x2, x3, x4)  =  IF_ORDERED_1_IN_1_G3(x2, x3, x4)
IF_LESS_2_IN_1_GG3(x1, x2, x3)  =  IF_LESS_2_IN_1_GG1(x3)
ORDERED_1_IN_G1(x1)  =  ORDERED_1_IN_G1(x1)
LESS_2_IN_GG2(x1, x2)  =  LESS_2_IN_GG2(x1, x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph 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_2_IN_GG2(s_11(X), s_11(Y)) -> LESS_2_IN_GG2(X, Y)

The TRS R consists of the following rules:

ordered_1_in_g1([]_0) -> ordered_1_out_g1([]_0)
ordered_1_in_g1(._22(underscore, []_0)) -> ordered_1_out_g1(._22(underscore, []_0))
ordered_1_in_g1(._22(X, ._22(Y, Xs))) -> if_ordered_1_in_1_g4(X, Y, Xs, less_2_in_gg2(X, s_11(Y)))
less_2_in_gg2(0_0, s_11(underscore1)) -> less_2_out_gg2(0_0, s_11(underscore1))
less_2_in_gg2(s_11(X), s_11(Y)) -> if_less_2_in_1_gg3(X, Y, less_2_in_gg2(X, Y))
if_less_2_in_1_gg3(X, Y, less_2_out_gg2(X, Y)) -> less_2_out_gg2(s_11(X), s_11(Y))
if_ordered_1_in_1_g4(X, Y, Xs, less_2_out_gg2(X, s_11(Y))) -> if_ordered_1_in_2_g4(X, Y, Xs, ordered_1_in_g1(._22(Y, Xs)))
if_ordered_1_in_2_g4(X, Y, Xs, ordered_1_out_g1(._22(Y, Xs))) -> ordered_1_out_g1(._22(X, ._22(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_1_in_g1(x1)  =  ordered_1_in_g1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
ordered_1_out_g1(x1)  =  ordered_1_out_g
if_ordered_1_in_1_g4(x1, x2, x3, x4)  =  if_ordered_1_in_1_g3(x2, x3, x4)
less_2_in_gg2(x1, x2)  =  less_2_in_gg2(x1, x2)
less_2_out_gg2(x1, x2)  =  less_2_out_gg
if_less_2_in_1_gg3(x1, x2, x3)  =  if_less_2_in_1_gg1(x3)
if_ordered_1_in_2_g4(x1, x2, x3, x4)  =  if_ordered_1_in_2_g1(x4)
LESS_2_IN_GG2(x1, x2)  =  LESS_2_IN_GG2(x1, x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting 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_2_IN_GG2(s_11(X), s_11(Y)) -> LESS_2_IN_GG2(X, Y)

R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem 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_2_IN_GG2(s_11(X), s_11(Y)) -> LESS_2_IN_GG2(X, Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {LESS_2_IN_GG2}.
By using the subterm criterion together with the size-change analysis 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_1_IN_G1(._22(X, ._22(Y, Xs))) -> IF_ORDERED_1_IN_1_G4(X, Y, Xs, less_2_in_gg2(X, s_11(Y)))
IF_ORDERED_1_IN_1_G4(X, Y, Xs, less_2_out_gg2(X, s_11(Y))) -> ORDERED_1_IN_G1(._22(Y, Xs))

The TRS R consists of the following rules:

ordered_1_in_g1([]_0) -> ordered_1_out_g1([]_0)
ordered_1_in_g1(._22(underscore, []_0)) -> ordered_1_out_g1(._22(underscore, []_0))
ordered_1_in_g1(._22(X, ._22(Y, Xs))) -> if_ordered_1_in_1_g4(X, Y, Xs, less_2_in_gg2(X, s_11(Y)))
less_2_in_gg2(0_0, s_11(underscore1)) -> less_2_out_gg2(0_0, s_11(underscore1))
less_2_in_gg2(s_11(X), s_11(Y)) -> if_less_2_in_1_gg3(X, Y, less_2_in_gg2(X, Y))
if_less_2_in_1_gg3(X, Y, less_2_out_gg2(X, Y)) -> less_2_out_gg2(s_11(X), s_11(Y))
if_ordered_1_in_1_g4(X, Y, Xs, less_2_out_gg2(X, s_11(Y))) -> if_ordered_1_in_2_g4(X, Y, Xs, ordered_1_in_g1(._22(Y, Xs)))
if_ordered_1_in_2_g4(X, Y, Xs, ordered_1_out_g1(._22(Y, Xs))) -> ordered_1_out_g1(._22(X, ._22(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_1_in_g1(x1)  =  ordered_1_in_g1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
ordered_1_out_g1(x1)  =  ordered_1_out_g
if_ordered_1_in_1_g4(x1, x2, x3, x4)  =  if_ordered_1_in_1_g3(x2, x3, x4)
less_2_in_gg2(x1, x2)  =  less_2_in_gg2(x1, x2)
less_2_out_gg2(x1, x2)  =  less_2_out_gg
if_less_2_in_1_gg3(x1, x2, x3)  =  if_less_2_in_1_gg1(x3)
if_ordered_1_in_2_g4(x1, x2, x3, x4)  =  if_ordered_1_in_2_g1(x4)
IF_ORDERED_1_IN_1_G4(x1, x2, x3, x4)  =  IF_ORDERED_1_IN_1_G3(x2, x3, x4)
ORDERED_1_IN_G1(x1)  =  ORDERED_1_IN_G1(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting 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_1_IN_G1(._22(X, ._22(Y, Xs))) -> IF_ORDERED_1_IN_1_G4(X, Y, Xs, less_2_in_gg2(X, s_11(Y)))
IF_ORDERED_1_IN_1_G4(X, Y, Xs, less_2_out_gg2(X, s_11(Y))) -> ORDERED_1_IN_G1(._22(Y, Xs))

The TRS R consists of the following rules:

less_2_in_gg2(0_0, s_11(underscore1)) -> less_2_out_gg2(0_0, s_11(underscore1))
less_2_in_gg2(s_11(X), s_11(Y)) -> if_less_2_in_1_gg3(X, Y, less_2_in_gg2(X, Y))
if_less_2_in_1_gg3(X, Y, less_2_out_gg2(X, Y)) -> less_2_out_gg2(s_11(X), s_11(Y))

The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
less_2_in_gg2(x1, x2)  =  less_2_in_gg2(x1, x2)
less_2_out_gg2(x1, x2)  =  less_2_out_gg
if_less_2_in_1_gg3(x1, x2, x3)  =  if_less_2_in_1_gg1(x3)
IF_ORDERED_1_IN_1_G4(x1, x2, x3, x4)  =  IF_ORDERED_1_IN_1_G3(x2, x3, x4)
ORDERED_1_IN_G1(x1)  =  ORDERED_1_IN_G1(x1)

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

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof

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

ORDERED_1_IN_G1(._22(X, ._22(Y, Xs))) -> IF_ORDERED_1_IN_1_G3(Y, Xs, less_2_in_gg2(X, s_11(Y)))
IF_ORDERED_1_IN_1_G3(Y, Xs, less_2_out_gg) -> ORDERED_1_IN_G1(._22(Y, Xs))

The TRS R consists of the following rules:

less_2_in_gg2(0_0, s_11(underscore1)) -> less_2_out_gg
less_2_in_gg2(s_11(X), s_11(Y)) -> if_less_2_in_1_gg1(less_2_in_gg2(X, Y))
if_less_2_in_1_gg1(less_2_out_gg) -> less_2_out_gg

The set Q consists of the following terms:

less_2_in_gg2(x0, x1)
if_less_2_in_1_gg1(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_ORDERED_1_IN_1_G3, ORDERED_1_IN_G1}.
By using a polynomial ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

IF_ORDERED_1_IN_1_G3(Y, Xs, less_2_out_gg) -> ORDERED_1_IN_G1(._22(Y, Xs))

Strictly oriented rules of the TRS R:

less_2_in_gg2(0_0, s_11(underscore1)) -> less_2_out_gg

Used ordering: POLO with Polynomial interpretation:

POL(0_0) = 2   
POL(less_2_out_gg) = 1   
POL(._22(x1, x2)) = x1 + 2·x2   
POL(ORDERED_1_IN_G1(x1)) = x1   
POL(if_less_2_in_1_gg1(x1)) = x1   
POL(s_11(x1)) = x1   
POL(less_2_in_gg2(x1, x2)) = x1 + x2   
POL(IF_ORDERED_1_IN_1_G3(x1, x2, x3)) = x1 + 2·x2 + x3   



↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ RuleRemovalProof

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

ORDERED_1_IN_G1(._22(X, ._22(Y, Xs))) -> IF_ORDERED_1_IN_1_G3(Y, Xs, less_2_in_gg2(X, s_11(Y)))

The TRS R consists of the following rules:

less_2_in_gg2(s_11(X), s_11(Y)) -> if_less_2_in_1_gg1(less_2_in_gg2(X, Y))
if_less_2_in_1_gg1(less_2_out_gg) -> less_2_out_gg

The set Q consists of the following terms:

less_2_in_gg2(x0, x1)
if_less_2_in_1_gg1(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_ORDERED_1_IN_1_G3, ORDERED_1_IN_G1}.
By using a polynomial ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

less_2_in_gg2(s_11(X), s_11(Y)) -> if_less_2_in_1_gg1(less_2_in_gg2(X, Y))
if_less_2_in_1_gg1(less_2_out_gg) -> less_2_out_gg

Used ordering: POLO with Polynomial interpretation:

POL(less_2_out_gg) = 0   
POL(._22(x1, x2)) = 2·x1 + x2   
POL(ORDERED_1_IN_G1(x1)) = 2 + x1   
POL(if_less_2_in_1_gg1(x1)) = 1 + x1   
POL(s_11(x1)) = 1 + x1   
POL(less_2_in_gg2(x1, x2)) = 1 + x1 + x2   
POL(IF_ORDERED_1_IN_1_G3(x1, x2, x3)) = x1 + x2 + x3   



↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ DependencyGraphProof

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

ORDERED_1_IN_G1(._22(X, ._22(Y, Xs))) -> IF_ORDERED_1_IN_1_G3(Y, Xs, less_2_in_gg2(X, s_11(Y)))

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

less_2_in_gg2(x0, x1)
if_less_2_in_1_gg1(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_ORDERED_1_IN_1_G3, ORDERED_1_IN_G1}.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.