Left Termination of the query pattern insert_in_3(g, a, a) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

insert(X, void, tree(X, void, void)).
insert(X, tree(X, Left, Right), tree(X, Left, Right)).
insert(X, tree(Y, Left, Right), tree(Y, Left1, Right)) :- ','(less(X, Y), insert(X, Left, Left1)).
insert(X, tree(Y, Left, Right), tree(Y, Left, Right1)) :- ','(less(Y, X), insert(X, Right, Right1)).
less(0, s(X)).
less(s(X), s(Y)) :- less(X, Y).

Queries:

insert(g,a,a).

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

insert_in_gaa(X, void, tree(X, void, void)) → insert_out_gaa(X, void, tree(X, void, void))
insert_in_gaa(X, tree(X, Left, Right), tree(X, Left, Right)) → insert_out_gaa(X, tree(X, Left, Right), tree(X, Left, Right))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_gaa(X, Y, Left, Right, Left1, less_in_ga(X, Y))
less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U5_ga(X, Y, less_in_ga(X, Y))
U5_ga(X, Y, less_out_ga(X, Y)) → less_out_ga(s(X), s(Y))
U1_gaa(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → U2_gaa(X, Y, Left, Right, Left1, insert_in_gaa(X, Left, Left1))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_gaa(X, Y, Left, Right, Right1, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U5_ag(X, Y, less_in_ag(X, Y))
U5_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U3_gaa(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → U4_gaa(X, Y, Left, Right, Right1, insert_in_gaa(X, Right, Right1))
U4_gaa(X, Y, Left, Right, Right1, insert_out_gaa(X, Right, Right1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1))
U2_gaa(X, Y, Left, Right, Left1, insert_out_gaa(X, Left, Left1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right))

The argument filtering Pi contains the following mapping:
insert_in_gaa(x1, x2, x3)  =  insert_in_gaa(x1)
insert_out_gaa(x1, x2, x3)  =  insert_out_gaa
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x6)
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga
s(x1)  =  s(x1)
U5_ga(x1, x2, x3)  =  U5_ga(x3)
U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x6)
U3_gaa(x1, x2, x3, x4, x5, x6)  =  U3_gaa(x1, x6)
less_in_ag(x1, x2)  =  less_in_ag(x2)
less_out_ag(x1, x2)  =  less_out_ag(x1)
U5_ag(x1, x2, x3)  =  U5_ag(x3)
U4_gaa(x1, x2, x3, x4, x5, x6)  =  U4_gaa(x6)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
  ↳ PrologToPiTRSProof

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

insert_in_gaa(X, void, tree(X, void, void)) → insert_out_gaa(X, void, tree(X, void, void))
insert_in_gaa(X, tree(X, Left, Right), tree(X, Left, Right)) → insert_out_gaa(X, tree(X, Left, Right), tree(X, Left, Right))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_gaa(X, Y, Left, Right, Left1, less_in_ga(X, Y))
less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U5_ga(X, Y, less_in_ga(X, Y))
U5_ga(X, Y, less_out_ga(X, Y)) → less_out_ga(s(X), s(Y))
U1_gaa(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → U2_gaa(X, Y, Left, Right, Left1, insert_in_gaa(X, Left, Left1))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_gaa(X, Y, Left, Right, Right1, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U5_ag(X, Y, less_in_ag(X, Y))
U5_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U3_gaa(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → U4_gaa(X, Y, Left, Right, Right1, insert_in_gaa(X, Right, Right1))
U4_gaa(X, Y, Left, Right, Right1, insert_out_gaa(X, Right, Right1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1))
U2_gaa(X, Y, Left, Right, Left1, insert_out_gaa(X, Left, Left1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right))

The argument filtering Pi contains the following mapping:
insert_in_gaa(x1, x2, x3)  =  insert_in_gaa(x1)
insert_out_gaa(x1, x2, x3)  =  insert_out_gaa
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x6)
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga
s(x1)  =  s(x1)
U5_ga(x1, x2, x3)  =  U5_ga(x3)
U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x6)
U3_gaa(x1, x2, x3, x4, x5, x6)  =  U3_gaa(x1, x6)
less_in_ag(x1, x2)  =  less_in_ag(x2)
less_out_ag(x1, x2)  =  less_out_ag(x1)
U5_ag(x1, x2, x3)  =  U5_ag(x3)
U4_gaa(x1, x2, x3, x4, x5, x6)  =  U4_gaa(x6)


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

INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_GAA(X, Y, Left, Right, Left1, less_in_ga(X, Y))
INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → LESS_IN_GA(X, Y)
LESS_IN_GA(s(X), s(Y)) → U5_GA(X, Y, less_in_ga(X, Y))
LESS_IN_GA(s(X), s(Y)) → LESS_IN_GA(X, Y)
U1_GAA(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → U2_GAA(X, Y, Left, Right, Left1, insert_in_gaa(X, Left, Left1))
U1_GAA(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → INSERT_IN_GAA(X, Left, Left1)
INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_GAA(X, Y, Left, Right, Right1, less_in_ag(Y, X))
INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → LESS_IN_AG(Y, X)
LESS_IN_AG(s(X), s(Y)) → U5_AG(X, Y, less_in_ag(X, Y))
LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)
U3_GAA(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → U4_GAA(X, Y, Left, Right, Right1, insert_in_gaa(X, Right, Right1))
U3_GAA(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → INSERT_IN_GAA(X, Right, Right1)

The TRS R consists of the following rules:

insert_in_gaa(X, void, tree(X, void, void)) → insert_out_gaa(X, void, tree(X, void, void))
insert_in_gaa(X, tree(X, Left, Right), tree(X, Left, Right)) → insert_out_gaa(X, tree(X, Left, Right), tree(X, Left, Right))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_gaa(X, Y, Left, Right, Left1, less_in_ga(X, Y))
less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U5_ga(X, Y, less_in_ga(X, Y))
U5_ga(X, Y, less_out_ga(X, Y)) → less_out_ga(s(X), s(Y))
U1_gaa(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → U2_gaa(X, Y, Left, Right, Left1, insert_in_gaa(X, Left, Left1))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_gaa(X, Y, Left, Right, Right1, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U5_ag(X, Y, less_in_ag(X, Y))
U5_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U3_gaa(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → U4_gaa(X, Y, Left, Right, Right1, insert_in_gaa(X, Right, Right1))
U4_gaa(X, Y, Left, Right, Right1, insert_out_gaa(X, Right, Right1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1))
U2_gaa(X, Y, Left, Right, Left1, insert_out_gaa(X, Left, Left1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right))

The argument filtering Pi contains the following mapping:
insert_in_gaa(x1, x2, x3)  =  insert_in_gaa(x1)
insert_out_gaa(x1, x2, x3)  =  insert_out_gaa
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x6)
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga
s(x1)  =  s(x1)
U5_ga(x1, x2, x3)  =  U5_ga(x3)
U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x6)
U3_gaa(x1, x2, x3, x4, x5, x6)  =  U3_gaa(x1, x6)
less_in_ag(x1, x2)  =  less_in_ag(x2)
less_out_ag(x1, x2)  =  less_out_ag(x1)
U5_ag(x1, x2, x3)  =  U5_ag(x3)
U4_gaa(x1, x2, x3, x4, x5, x6)  =  U4_gaa(x6)
INSERT_IN_GAA(x1, x2, x3)  =  INSERT_IN_GAA(x1)
U5_GA(x1, x2, x3)  =  U5_GA(x3)
U5_AG(x1, x2, x3)  =  U5_AG(x3)
U2_GAA(x1, x2, x3, x4, x5, x6)  =  U2_GAA(x6)
U4_GAA(x1, x2, x3, x4, x5, x6)  =  U4_GAA(x6)
U3_GAA(x1, x2, x3, x4, x5, x6)  =  U3_GAA(x1, x6)
LESS_IN_GA(x1, x2)  =  LESS_IN_GA(x1)
LESS_IN_AG(x1, x2)  =  LESS_IN_AG(x2)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x6)

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

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
  ↳ PrologToPiTRSProof

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

INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_GAA(X, Y, Left, Right, Left1, less_in_ga(X, Y))
INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → LESS_IN_GA(X, Y)
LESS_IN_GA(s(X), s(Y)) → U5_GA(X, Y, less_in_ga(X, Y))
LESS_IN_GA(s(X), s(Y)) → LESS_IN_GA(X, Y)
U1_GAA(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → U2_GAA(X, Y, Left, Right, Left1, insert_in_gaa(X, Left, Left1))
U1_GAA(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → INSERT_IN_GAA(X, Left, Left1)
INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_GAA(X, Y, Left, Right, Right1, less_in_ag(Y, X))
INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → LESS_IN_AG(Y, X)
LESS_IN_AG(s(X), s(Y)) → U5_AG(X, Y, less_in_ag(X, Y))
LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)
U3_GAA(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → U4_GAA(X, Y, Left, Right, Right1, insert_in_gaa(X, Right, Right1))
U3_GAA(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → INSERT_IN_GAA(X, Right, Right1)

The TRS R consists of the following rules:

insert_in_gaa(X, void, tree(X, void, void)) → insert_out_gaa(X, void, tree(X, void, void))
insert_in_gaa(X, tree(X, Left, Right), tree(X, Left, Right)) → insert_out_gaa(X, tree(X, Left, Right), tree(X, Left, Right))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_gaa(X, Y, Left, Right, Left1, less_in_ga(X, Y))
less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U5_ga(X, Y, less_in_ga(X, Y))
U5_ga(X, Y, less_out_ga(X, Y)) → less_out_ga(s(X), s(Y))
U1_gaa(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → U2_gaa(X, Y, Left, Right, Left1, insert_in_gaa(X, Left, Left1))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_gaa(X, Y, Left, Right, Right1, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U5_ag(X, Y, less_in_ag(X, Y))
U5_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U3_gaa(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → U4_gaa(X, Y, Left, Right, Right1, insert_in_gaa(X, Right, Right1))
U4_gaa(X, Y, Left, Right, Right1, insert_out_gaa(X, Right, Right1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1))
U2_gaa(X, Y, Left, Right, Left1, insert_out_gaa(X, Left, Left1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right))

The argument filtering Pi contains the following mapping:
insert_in_gaa(x1, x2, x3)  =  insert_in_gaa(x1)
insert_out_gaa(x1, x2, x3)  =  insert_out_gaa
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x6)
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga
s(x1)  =  s(x1)
U5_ga(x1, x2, x3)  =  U5_ga(x3)
U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x6)
U3_gaa(x1, x2, x3, x4, x5, x6)  =  U3_gaa(x1, x6)
less_in_ag(x1, x2)  =  less_in_ag(x2)
less_out_ag(x1, x2)  =  less_out_ag(x1)
U5_ag(x1, x2, x3)  =  U5_ag(x3)
U4_gaa(x1, x2, x3, x4, x5, x6)  =  U4_gaa(x6)
INSERT_IN_GAA(x1, x2, x3)  =  INSERT_IN_GAA(x1)
U5_GA(x1, x2, x3)  =  U5_GA(x3)
U5_AG(x1, x2, x3)  =  U5_AG(x3)
U2_GAA(x1, x2, x3, x4, x5, x6)  =  U2_GAA(x6)
U4_GAA(x1, x2, x3, x4, x5, x6)  =  U4_GAA(x6)
U3_GAA(x1, x2, x3, x4, x5, x6)  =  U3_GAA(x1, x6)
LESS_IN_GA(x1, x2)  =  LESS_IN_GA(x1)
LESS_IN_AG(x1, x2)  =  LESS_IN_AG(x2)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x6)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 3 SCCs with 6 less nodes.

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

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

LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)

The TRS R consists of the following rules:

insert_in_gaa(X, void, tree(X, void, void)) → insert_out_gaa(X, void, tree(X, void, void))
insert_in_gaa(X, tree(X, Left, Right), tree(X, Left, Right)) → insert_out_gaa(X, tree(X, Left, Right), tree(X, Left, Right))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_gaa(X, Y, Left, Right, Left1, less_in_ga(X, Y))
less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U5_ga(X, Y, less_in_ga(X, Y))
U5_ga(X, Y, less_out_ga(X, Y)) → less_out_ga(s(X), s(Y))
U1_gaa(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → U2_gaa(X, Y, Left, Right, Left1, insert_in_gaa(X, Left, Left1))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_gaa(X, Y, Left, Right, Right1, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U5_ag(X, Y, less_in_ag(X, Y))
U5_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U3_gaa(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → U4_gaa(X, Y, Left, Right, Right1, insert_in_gaa(X, Right, Right1))
U4_gaa(X, Y, Left, Right, Right1, insert_out_gaa(X, Right, Right1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1))
U2_gaa(X, Y, Left, Right, Left1, insert_out_gaa(X, Left, Left1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right))

The argument filtering Pi contains the following mapping:
insert_in_gaa(x1, x2, x3)  =  insert_in_gaa(x1)
insert_out_gaa(x1, x2, x3)  =  insert_out_gaa
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x6)
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga
s(x1)  =  s(x1)
U5_ga(x1, x2, x3)  =  U5_ga(x3)
U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x6)
U3_gaa(x1, x2, x3, x4, x5, x6)  =  U3_gaa(x1, x6)
less_in_ag(x1, x2)  =  less_in_ag(x2)
less_out_ag(x1, x2)  =  less_out_ag(x1)
U5_ag(x1, x2, x3)  =  U5_ag(x3)
U4_gaa(x1, x2, x3, x4, x5, x6)  =  U4_gaa(x6)
LESS_IN_AG(x1, x2)  =  LESS_IN_AG(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
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
LESS_IN_AG(x1, x2)  =  LESS_IN_AG(x2)

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
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

LESS_IN_AG(s(Y)) → LESS_IN_AG(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
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

LESS_IN_GA(s(X), s(Y)) → LESS_IN_GA(X, Y)

The TRS R consists of the following rules:

insert_in_gaa(X, void, tree(X, void, void)) → insert_out_gaa(X, void, tree(X, void, void))
insert_in_gaa(X, tree(X, Left, Right), tree(X, Left, Right)) → insert_out_gaa(X, tree(X, Left, Right), tree(X, Left, Right))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_gaa(X, Y, Left, Right, Left1, less_in_ga(X, Y))
less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U5_ga(X, Y, less_in_ga(X, Y))
U5_ga(X, Y, less_out_ga(X, Y)) → less_out_ga(s(X), s(Y))
U1_gaa(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → U2_gaa(X, Y, Left, Right, Left1, insert_in_gaa(X, Left, Left1))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_gaa(X, Y, Left, Right, Right1, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U5_ag(X, Y, less_in_ag(X, Y))
U5_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U3_gaa(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → U4_gaa(X, Y, Left, Right, Right1, insert_in_gaa(X, Right, Right1))
U4_gaa(X, Y, Left, Right, Right1, insert_out_gaa(X, Right, Right1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1))
U2_gaa(X, Y, Left, Right, Left1, insert_out_gaa(X, Left, Left1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right))

The argument filtering Pi contains the following mapping:
insert_in_gaa(x1, x2, x3)  =  insert_in_gaa(x1)
insert_out_gaa(x1, x2, x3)  =  insert_out_gaa
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x6)
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga
s(x1)  =  s(x1)
U5_ga(x1, x2, x3)  =  U5_ga(x3)
U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x6)
U3_gaa(x1, x2, x3, x4, x5, x6)  =  U3_gaa(x1, x6)
less_in_ag(x1, x2)  =  less_in_ag(x2)
less_out_ag(x1, x2)  =  less_out_ag(x1)
U5_ag(x1, x2, x3)  =  U5_ag(x3)
U4_gaa(x1, x2, x3, x4, x5, x6)  =  U4_gaa(x6)
LESS_IN_GA(x1, x2)  =  LESS_IN_GA(x1)

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
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

LESS_IN_GA(s(X), s(Y)) → LESS_IN_GA(X, Y)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
LESS_IN_GA(x1, x2)  =  LESS_IN_GA(x1)

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
                        ↳ QDPSizeChangeProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

LESS_IN_GA(s(X)) → LESS_IN_GA(X)

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
              ↳ PiDP
                ↳ UsableRulesProof
  ↳ PrologToPiTRSProof

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

INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_GAA(X, Y, Left, Right, Left1, less_in_ga(X, Y))
U1_GAA(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → INSERT_IN_GAA(X, Left, Left1)
U3_GAA(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → INSERT_IN_GAA(X, Right, Right1)
INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_GAA(X, Y, Left, Right, Right1, less_in_ag(Y, X))

The TRS R consists of the following rules:

insert_in_gaa(X, void, tree(X, void, void)) → insert_out_gaa(X, void, tree(X, void, void))
insert_in_gaa(X, tree(X, Left, Right), tree(X, Left, Right)) → insert_out_gaa(X, tree(X, Left, Right), tree(X, Left, Right))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_gaa(X, Y, Left, Right, Left1, less_in_ga(X, Y))
less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U5_ga(X, Y, less_in_ga(X, Y))
U5_ga(X, Y, less_out_ga(X, Y)) → less_out_ga(s(X), s(Y))
U1_gaa(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → U2_gaa(X, Y, Left, Right, Left1, insert_in_gaa(X, Left, Left1))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_gaa(X, Y, Left, Right, Right1, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U5_ag(X, Y, less_in_ag(X, Y))
U5_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U3_gaa(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → U4_gaa(X, Y, Left, Right, Right1, insert_in_gaa(X, Right, Right1))
U4_gaa(X, Y, Left, Right, Right1, insert_out_gaa(X, Right, Right1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1))
U2_gaa(X, Y, Left, Right, Left1, insert_out_gaa(X, Left, Left1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right))

The argument filtering Pi contains the following mapping:
insert_in_gaa(x1, x2, x3)  =  insert_in_gaa(x1)
insert_out_gaa(x1, x2, x3)  =  insert_out_gaa
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x6)
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga
s(x1)  =  s(x1)
U5_ga(x1, x2, x3)  =  U5_ga(x3)
U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x6)
U3_gaa(x1, x2, x3, x4, x5, x6)  =  U3_gaa(x1, x6)
less_in_ag(x1, x2)  =  less_in_ag(x2)
less_out_ag(x1, x2)  =  less_out_ag(x1)
U5_ag(x1, x2, x3)  =  U5_ag(x3)
U4_gaa(x1, x2, x3, x4, x5, x6)  =  U4_gaa(x6)
INSERT_IN_GAA(x1, x2, x3)  =  INSERT_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4, x5, x6)  =  U3_GAA(x1, x6)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x6)

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
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

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

INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_GAA(X, Y, Left, Right, Left1, less_in_ga(X, Y))
U1_GAA(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → INSERT_IN_GAA(X, Left, Left1)
U3_GAA(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → INSERT_IN_GAA(X, Right, Right1)
INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_GAA(X, Y, Left, Right, Right1, less_in_ag(Y, X))

The TRS R consists of the following rules:

less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U5_ga(X, Y, less_in_ga(X, Y))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U5_ag(X, Y, less_in_ag(X, Y))
U5_ga(X, Y, less_out_ga(X, Y)) → less_out_ga(s(X), s(Y))
U5_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))

The argument filtering Pi contains the following mapping:
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga
s(x1)  =  s(x1)
U5_ga(x1, x2, x3)  =  U5_ga(x3)
less_in_ag(x1, x2)  =  less_in_ag(x2)
less_out_ag(x1, x2)  =  less_out_ag(x1)
U5_ag(x1, x2, x3)  =  U5_ag(x3)
INSERT_IN_GAA(x1, x2, x3)  =  INSERT_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4, x5, x6)  =  U3_GAA(x1, x6)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x6)

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
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
  ↳ PrologToPiTRSProof

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

INSERT_IN_GAA(X) → U3_GAA(X, less_in_ag(X))
U3_GAA(X, less_out_ag(Y)) → INSERT_IN_GAA(X)
INSERT_IN_GAA(X) → U1_GAA(X, less_in_ga(X))
U1_GAA(X, less_out_ga) → INSERT_IN_GAA(X)

The TRS R consists of the following rules:

less_in_ga(0) → less_out_ga
less_in_ga(s(X)) → U5_ga(less_in_ga(X))
less_in_ag(s(X)) → less_out_ag(0)
less_in_ag(s(Y)) → U5_ag(less_in_ag(Y))
U5_ga(less_out_ga) → less_out_ga
U5_ag(less_out_ag(X)) → less_out_ag(s(X))

The set Q consists of the following terms:

less_in_ga(x0)
less_in_ag(x0)
U5_ga(x0)
U5_ag(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule INSERT_IN_GAA(X) → U1_GAA(X, less_in_ga(X)) at position [1] we obtained the following new rules:

INSERT_IN_GAA(s(x0)) → U1_GAA(s(x0), U5_ga(less_in_ga(x0)))
INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga)



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

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

INSERT_IN_GAA(X) → U3_GAA(X, less_in_ag(X))
U3_GAA(X, less_out_ag(Y)) → INSERT_IN_GAA(X)
U1_GAA(X, less_out_ga) → INSERT_IN_GAA(X)
INSERT_IN_GAA(s(x0)) → U1_GAA(s(x0), U5_ga(less_in_ga(x0)))
INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga)

The TRS R consists of the following rules:

less_in_ga(0) → less_out_ga
less_in_ga(s(X)) → U5_ga(less_in_ga(X))
less_in_ag(s(X)) → less_out_ag(0)
less_in_ag(s(Y)) → U5_ag(less_in_ag(Y))
U5_ga(less_out_ga) → less_out_ga
U5_ag(less_out_ag(X)) → less_out_ag(s(X))

The set Q consists of the following terms:

less_in_ga(x0)
less_in_ag(x0)
U5_ga(x0)
U5_ag(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule INSERT_IN_GAA(X) → U3_GAA(X, less_in_ag(X)) at position [1] we obtained the following new rules:

INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), U5_ag(less_in_ag(x0)))
INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), less_out_ag(0))



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ Instantiation
  ↳ PrologToPiTRSProof

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

INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), less_out_ag(0))
INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), U5_ag(less_in_ag(x0)))
U3_GAA(X, less_out_ag(Y)) → INSERT_IN_GAA(X)
U1_GAA(X, less_out_ga) → INSERT_IN_GAA(X)
INSERT_IN_GAA(s(x0)) → U1_GAA(s(x0), U5_ga(less_in_ga(x0)))
INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga)

The TRS R consists of the following rules:

less_in_ga(0) → less_out_ga
less_in_ga(s(X)) → U5_ga(less_in_ga(X))
less_in_ag(s(X)) → less_out_ag(0)
less_in_ag(s(Y)) → U5_ag(less_in_ag(Y))
U5_ga(less_out_ga) → less_out_ga
U5_ag(less_out_ag(X)) → less_out_ag(s(X))

The set Q consists of the following terms:

less_in_ga(x0)
less_in_ag(x0)
U5_ga(x0)
U5_ag(x0)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U1_GAA(X, less_out_ga) → INSERT_IN_GAA(X) we obtained the following new rules:

U1_GAA(0, less_out_ga) → INSERT_IN_GAA(0)
U1_GAA(s(z0), less_out_ga) → INSERT_IN_GAA(s(z0))



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
  ↳ PrologToPiTRSProof

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

INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), less_out_ag(0))
U1_GAA(0, less_out_ga) → INSERT_IN_GAA(0)
INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), U5_ag(less_in_ag(x0)))
U3_GAA(X, less_out_ag(Y)) → INSERT_IN_GAA(X)
U1_GAA(s(z0), less_out_ga) → INSERT_IN_GAA(s(z0))
INSERT_IN_GAA(s(x0)) → U1_GAA(s(x0), U5_ga(less_in_ga(x0)))
INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga)

The TRS R consists of the following rules:

less_in_ga(0) → less_out_ga
less_in_ga(s(X)) → U5_ga(less_in_ga(X))
less_in_ag(s(X)) → less_out_ag(0)
less_in_ag(s(Y)) → U5_ag(less_in_ag(Y))
U5_ga(less_out_ga) → less_out_ga
U5_ag(less_out_ag(X)) → less_out_ag(s(X))

The set Q consists of the following terms:

less_in_ga(x0)
less_in_ag(x0)
U5_ga(x0)
U5_ag(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ AND
                                        ↳ QDP
                                          ↳ UsableRulesProof
                                        ↳ QDP
  ↳ PrologToPiTRSProof

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

U1_GAA(0, less_out_ga) → INSERT_IN_GAA(0)
INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga)

The TRS R consists of the following rules:

less_in_ga(0) → less_out_ga
less_in_ga(s(X)) → U5_ga(less_in_ga(X))
less_in_ag(s(X)) → less_out_ag(0)
less_in_ag(s(Y)) → U5_ag(less_in_ag(Y))
U5_ga(less_out_ga) → less_out_ga
U5_ag(less_out_ag(X)) → less_out_ag(s(X))

The set Q consists of the following terms:

less_in_ga(x0)
less_in_ag(x0)
U5_ga(x0)
U5_ag(x0)

We have to consider all (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ AND
                                        ↳ QDP
                                          ↳ UsableRulesProof
                                            ↳ QDP
                                              ↳ QReductionProof
                                        ↳ QDP
  ↳ PrologToPiTRSProof

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

U1_GAA(0, less_out_ga) → INSERT_IN_GAA(0)
INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga)

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

less_in_ga(x0)
less_in_ag(x0)
U5_ga(x0)
U5_ag(x0)

We have to consider all (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

less_in_ga(x0)
less_in_ag(x0)
U5_ga(x0)
U5_ag(x0)



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ AND
                                        ↳ QDP
                                          ↳ UsableRulesProof
                                            ↳ QDP
                                              ↳ QReductionProof
                                                ↳ QDP
                                                  ↳ NonTerminationProof
                                        ↳ QDP
  ↳ PrologToPiTRSProof

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

U1_GAA(0, less_out_ga) → INSERT_IN_GAA(0)
INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

U1_GAA(0, less_out_ga) → INSERT_IN_GAA(0)
INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga)

The TRS R consists of the following rules:none


s = INSERT_IN_GAA(0) evaluates to t =INSERT_IN_GAA(0)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga)
with rule INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga) at position [] and matcher [ ]

U1_GAA(0, less_out_ga) → INSERT_IN_GAA(0)
with rule U1_GAA(0, less_out_ga) → INSERT_IN_GAA(0)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.





↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ AND
                                        ↳ QDP
                                        ↳ QDP
                                          ↳ Instantiation
  ↳ PrologToPiTRSProof

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

INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), less_out_ag(0))
INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), U5_ag(less_in_ag(x0)))
U3_GAA(X, less_out_ag(Y)) → INSERT_IN_GAA(X)
U1_GAA(s(z0), less_out_ga) → INSERT_IN_GAA(s(z0))
INSERT_IN_GAA(s(x0)) → U1_GAA(s(x0), U5_ga(less_in_ga(x0)))

The TRS R consists of the following rules:

less_in_ga(0) → less_out_ga
less_in_ga(s(X)) → U5_ga(less_in_ga(X))
less_in_ag(s(X)) → less_out_ag(0)
less_in_ag(s(Y)) → U5_ag(less_in_ag(Y))
U5_ga(less_out_ga) → less_out_ga
U5_ag(less_out_ag(X)) → less_out_ag(s(X))

The set Q consists of the following terms:

less_in_ga(x0)
less_in_ag(x0)
U5_ga(x0)
U5_ag(x0)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U3_GAA(X, less_out_ag(Y)) → INSERT_IN_GAA(X) we obtained the following new rules:

U3_GAA(s(z0), less_out_ag(0)) → INSERT_IN_GAA(s(z0))
U3_GAA(s(z0), less_out_ag(x1)) → INSERT_IN_GAA(s(z0))



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ AND
                                        ↳ QDP
                                        ↳ QDP
                                          ↳ Instantiation
                                            ↳ QDP
                                              ↳ NonTerminationProof
  ↳ PrologToPiTRSProof

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

INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), less_out_ag(0))
U3_GAA(s(z0), less_out_ag(0)) → INSERT_IN_GAA(s(z0))
INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), U5_ag(less_in_ag(x0)))
U3_GAA(s(z0), less_out_ag(x1)) → INSERT_IN_GAA(s(z0))
U1_GAA(s(z0), less_out_ga) → INSERT_IN_GAA(s(z0))
INSERT_IN_GAA(s(x0)) → U1_GAA(s(x0), U5_ga(less_in_ga(x0)))

The TRS R consists of the following rules:

less_in_ga(0) → less_out_ga
less_in_ga(s(X)) → U5_ga(less_in_ga(X))
less_in_ag(s(X)) → less_out_ag(0)
less_in_ag(s(Y)) → U5_ag(less_in_ag(Y))
U5_ga(less_out_ga) → less_out_ga
U5_ag(less_out_ag(X)) → less_out_ag(s(X))

The set Q consists of the following terms:

less_in_ga(x0)
less_in_ag(x0)
U5_ga(x0)
U5_ag(x0)

We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), less_out_ag(0))
U3_GAA(s(z0), less_out_ag(0)) → INSERT_IN_GAA(s(z0))
INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), U5_ag(less_in_ag(x0)))
U3_GAA(s(z0), less_out_ag(x1)) → INSERT_IN_GAA(s(z0))
U1_GAA(s(z0), less_out_ga) → INSERT_IN_GAA(s(z0))
INSERT_IN_GAA(s(x0)) → U1_GAA(s(x0), U5_ga(less_in_ga(x0)))

The TRS R consists of the following rules:

less_in_ga(0) → less_out_ga
less_in_ga(s(X)) → U5_ga(less_in_ga(X))
less_in_ag(s(X)) → less_out_ag(0)
less_in_ag(s(Y)) → U5_ag(less_in_ag(Y))
U5_ga(less_out_ga) → less_out_ga
U5_ag(less_out_ag(X)) → less_out_ag(s(X))


s = U3_GAA(s(z0), less_out_ag(0)) evaluates to t =U3_GAA(s(z0), less_out_ag(0))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

U3_GAA(s(z0), less_out_ag(0)) → INSERT_IN_GAA(s(z0))
with rule U3_GAA(s(z0'), less_out_ag(0)) → INSERT_IN_GAA(s(z0')) at position [] and matcher [z0' / z0]

INSERT_IN_GAA(s(z0)) → U3_GAA(s(z0), less_out_ag(0))
with rule INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), less_out_ag(0))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.




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

insert_in_gaa(X, void, tree(X, void, void)) → insert_out_gaa(X, void, tree(X, void, void))
insert_in_gaa(X, tree(X, Left, Right), tree(X, Left, Right)) → insert_out_gaa(X, tree(X, Left, Right), tree(X, Left, Right))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_gaa(X, Y, Left, Right, Left1, less_in_ga(X, Y))
less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U5_ga(X, Y, less_in_ga(X, Y))
U5_ga(X, Y, less_out_ga(X, Y)) → less_out_ga(s(X), s(Y))
U1_gaa(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → U2_gaa(X, Y, Left, Right, Left1, insert_in_gaa(X, Left, Left1))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_gaa(X, Y, Left, Right, Right1, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U5_ag(X, Y, less_in_ag(X, Y))
U5_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U3_gaa(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → U4_gaa(X, Y, Left, Right, Right1, insert_in_gaa(X, Right, Right1))
U4_gaa(X, Y, Left, Right, Right1, insert_out_gaa(X, Right, Right1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1))
U2_gaa(X, Y, Left, Right, Left1, insert_out_gaa(X, Left, Left1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right))

The argument filtering Pi contains the following mapping:
insert_in_gaa(x1, x2, x3)  =  insert_in_gaa(x1)
insert_out_gaa(x1, x2, x3)  =  insert_out_gaa(x1)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x6)
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga(x1)
s(x1)  =  s(x1)
U5_ga(x1, x2, x3)  =  U5_ga(x1, x3)
U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x1, x6)
U3_gaa(x1, x2, x3, x4, x5, x6)  =  U3_gaa(x1, x6)
less_in_ag(x1, x2)  =  less_in_ag(x2)
less_out_ag(x1, x2)  =  less_out_ag(x1, x2)
U5_ag(x1, x2, x3)  =  U5_ag(x2, x3)
U4_gaa(x1, x2, x3, x4, x5, x6)  =  U4_gaa(x1, x6)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

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

insert_in_gaa(X, void, tree(X, void, void)) → insert_out_gaa(X, void, tree(X, void, void))
insert_in_gaa(X, tree(X, Left, Right), tree(X, Left, Right)) → insert_out_gaa(X, tree(X, Left, Right), tree(X, Left, Right))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_gaa(X, Y, Left, Right, Left1, less_in_ga(X, Y))
less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U5_ga(X, Y, less_in_ga(X, Y))
U5_ga(X, Y, less_out_ga(X, Y)) → less_out_ga(s(X), s(Y))
U1_gaa(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → U2_gaa(X, Y, Left, Right, Left1, insert_in_gaa(X, Left, Left1))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_gaa(X, Y, Left, Right, Right1, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U5_ag(X, Y, less_in_ag(X, Y))
U5_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U3_gaa(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → U4_gaa(X, Y, Left, Right, Right1, insert_in_gaa(X, Right, Right1))
U4_gaa(X, Y, Left, Right, Right1, insert_out_gaa(X, Right, Right1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1))
U2_gaa(X, Y, Left, Right, Left1, insert_out_gaa(X, Left, Left1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right))

The argument filtering Pi contains the following mapping:
insert_in_gaa(x1, x2, x3)  =  insert_in_gaa(x1)
insert_out_gaa(x1, x2, x3)  =  insert_out_gaa(x1)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x6)
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga(x1)
s(x1)  =  s(x1)
U5_ga(x1, x2, x3)  =  U5_ga(x1, x3)
U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x1, x6)
U3_gaa(x1, x2, x3, x4, x5, x6)  =  U3_gaa(x1, x6)
less_in_ag(x1, x2)  =  less_in_ag(x2)
less_out_ag(x1, x2)  =  less_out_ag(x1, x2)
U5_ag(x1, x2, x3)  =  U5_ag(x2, x3)
U4_gaa(x1, x2, x3, x4, x5, x6)  =  U4_gaa(x1, x6)


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

INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_GAA(X, Y, Left, Right, Left1, less_in_ga(X, Y))
INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → LESS_IN_GA(X, Y)
LESS_IN_GA(s(X), s(Y)) → U5_GA(X, Y, less_in_ga(X, Y))
LESS_IN_GA(s(X), s(Y)) → LESS_IN_GA(X, Y)
U1_GAA(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → U2_GAA(X, Y, Left, Right, Left1, insert_in_gaa(X, Left, Left1))
U1_GAA(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → INSERT_IN_GAA(X, Left, Left1)
INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_GAA(X, Y, Left, Right, Right1, less_in_ag(Y, X))
INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → LESS_IN_AG(Y, X)
LESS_IN_AG(s(X), s(Y)) → U5_AG(X, Y, less_in_ag(X, Y))
LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)
U3_GAA(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → U4_GAA(X, Y, Left, Right, Right1, insert_in_gaa(X, Right, Right1))
U3_GAA(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → INSERT_IN_GAA(X, Right, Right1)

The TRS R consists of the following rules:

insert_in_gaa(X, void, tree(X, void, void)) → insert_out_gaa(X, void, tree(X, void, void))
insert_in_gaa(X, tree(X, Left, Right), tree(X, Left, Right)) → insert_out_gaa(X, tree(X, Left, Right), tree(X, Left, Right))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_gaa(X, Y, Left, Right, Left1, less_in_ga(X, Y))
less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U5_ga(X, Y, less_in_ga(X, Y))
U5_ga(X, Y, less_out_ga(X, Y)) → less_out_ga(s(X), s(Y))
U1_gaa(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → U2_gaa(X, Y, Left, Right, Left1, insert_in_gaa(X, Left, Left1))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_gaa(X, Y, Left, Right, Right1, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U5_ag(X, Y, less_in_ag(X, Y))
U5_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U3_gaa(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → U4_gaa(X, Y, Left, Right, Right1, insert_in_gaa(X, Right, Right1))
U4_gaa(X, Y, Left, Right, Right1, insert_out_gaa(X, Right, Right1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1))
U2_gaa(X, Y, Left, Right, Left1, insert_out_gaa(X, Left, Left1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right))

The argument filtering Pi contains the following mapping:
insert_in_gaa(x1, x2, x3)  =  insert_in_gaa(x1)
insert_out_gaa(x1, x2, x3)  =  insert_out_gaa(x1)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x6)
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga(x1)
s(x1)  =  s(x1)
U5_ga(x1, x2, x3)  =  U5_ga(x1, x3)
U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x1, x6)
U3_gaa(x1, x2, x3, x4, x5, x6)  =  U3_gaa(x1, x6)
less_in_ag(x1, x2)  =  less_in_ag(x2)
less_out_ag(x1, x2)  =  less_out_ag(x1, x2)
U5_ag(x1, x2, x3)  =  U5_ag(x2, x3)
U4_gaa(x1, x2, x3, x4, x5, x6)  =  U4_gaa(x1, x6)
INSERT_IN_GAA(x1, x2, x3)  =  INSERT_IN_GAA(x1)
U5_GA(x1, x2, x3)  =  U5_GA(x1, x3)
U5_AG(x1, x2, x3)  =  U5_AG(x2, x3)
U2_GAA(x1, x2, x3, x4, x5, x6)  =  U2_GAA(x1, x6)
U4_GAA(x1, x2, x3, x4, x5, x6)  =  U4_GAA(x1, x6)
U3_GAA(x1, x2, x3, x4, x5, x6)  =  U3_GAA(x1, x6)
LESS_IN_GA(x1, x2)  =  LESS_IN_GA(x1)
LESS_IN_AG(x1, x2)  =  LESS_IN_AG(x2)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x6)

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

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof

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

INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_GAA(X, Y, Left, Right, Left1, less_in_ga(X, Y))
INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → LESS_IN_GA(X, Y)
LESS_IN_GA(s(X), s(Y)) → U5_GA(X, Y, less_in_ga(X, Y))
LESS_IN_GA(s(X), s(Y)) → LESS_IN_GA(X, Y)
U1_GAA(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → U2_GAA(X, Y, Left, Right, Left1, insert_in_gaa(X, Left, Left1))
U1_GAA(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → INSERT_IN_GAA(X, Left, Left1)
INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_GAA(X, Y, Left, Right, Right1, less_in_ag(Y, X))
INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → LESS_IN_AG(Y, X)
LESS_IN_AG(s(X), s(Y)) → U5_AG(X, Y, less_in_ag(X, Y))
LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)
U3_GAA(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → U4_GAA(X, Y, Left, Right, Right1, insert_in_gaa(X, Right, Right1))
U3_GAA(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → INSERT_IN_GAA(X, Right, Right1)

The TRS R consists of the following rules:

insert_in_gaa(X, void, tree(X, void, void)) → insert_out_gaa(X, void, tree(X, void, void))
insert_in_gaa(X, tree(X, Left, Right), tree(X, Left, Right)) → insert_out_gaa(X, tree(X, Left, Right), tree(X, Left, Right))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_gaa(X, Y, Left, Right, Left1, less_in_ga(X, Y))
less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U5_ga(X, Y, less_in_ga(X, Y))
U5_ga(X, Y, less_out_ga(X, Y)) → less_out_ga(s(X), s(Y))
U1_gaa(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → U2_gaa(X, Y, Left, Right, Left1, insert_in_gaa(X, Left, Left1))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_gaa(X, Y, Left, Right, Right1, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U5_ag(X, Y, less_in_ag(X, Y))
U5_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U3_gaa(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → U4_gaa(X, Y, Left, Right, Right1, insert_in_gaa(X, Right, Right1))
U4_gaa(X, Y, Left, Right, Right1, insert_out_gaa(X, Right, Right1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1))
U2_gaa(X, Y, Left, Right, Left1, insert_out_gaa(X, Left, Left1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right))

The argument filtering Pi contains the following mapping:
insert_in_gaa(x1, x2, x3)  =  insert_in_gaa(x1)
insert_out_gaa(x1, x2, x3)  =  insert_out_gaa(x1)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x6)
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga(x1)
s(x1)  =  s(x1)
U5_ga(x1, x2, x3)  =  U5_ga(x1, x3)
U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x1, x6)
U3_gaa(x1, x2, x3, x4, x5, x6)  =  U3_gaa(x1, x6)
less_in_ag(x1, x2)  =  less_in_ag(x2)
less_out_ag(x1, x2)  =  less_out_ag(x1, x2)
U5_ag(x1, x2, x3)  =  U5_ag(x2, x3)
U4_gaa(x1, x2, x3, x4, x5, x6)  =  U4_gaa(x1, x6)
INSERT_IN_GAA(x1, x2, x3)  =  INSERT_IN_GAA(x1)
U5_GA(x1, x2, x3)  =  U5_GA(x1, x3)
U5_AG(x1, x2, x3)  =  U5_AG(x2, x3)
U2_GAA(x1, x2, x3, x4, x5, x6)  =  U2_GAA(x1, x6)
U4_GAA(x1, x2, x3, x4, x5, x6)  =  U4_GAA(x1, x6)
U3_GAA(x1, x2, x3, x4, x5, x6)  =  U3_GAA(x1, x6)
LESS_IN_GA(x1, x2)  =  LESS_IN_GA(x1)
LESS_IN_AG(x1, x2)  =  LESS_IN_AG(x2)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x6)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 3 SCCs with 6 less nodes.

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

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

LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)

The TRS R consists of the following rules:

insert_in_gaa(X, void, tree(X, void, void)) → insert_out_gaa(X, void, tree(X, void, void))
insert_in_gaa(X, tree(X, Left, Right), tree(X, Left, Right)) → insert_out_gaa(X, tree(X, Left, Right), tree(X, Left, Right))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_gaa(X, Y, Left, Right, Left1, less_in_ga(X, Y))
less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U5_ga(X, Y, less_in_ga(X, Y))
U5_ga(X, Y, less_out_ga(X, Y)) → less_out_ga(s(X), s(Y))
U1_gaa(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → U2_gaa(X, Y, Left, Right, Left1, insert_in_gaa(X, Left, Left1))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_gaa(X, Y, Left, Right, Right1, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U5_ag(X, Y, less_in_ag(X, Y))
U5_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U3_gaa(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → U4_gaa(X, Y, Left, Right, Right1, insert_in_gaa(X, Right, Right1))
U4_gaa(X, Y, Left, Right, Right1, insert_out_gaa(X, Right, Right1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1))
U2_gaa(X, Y, Left, Right, Left1, insert_out_gaa(X, Left, Left1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right))

The argument filtering Pi contains the following mapping:
insert_in_gaa(x1, x2, x3)  =  insert_in_gaa(x1)
insert_out_gaa(x1, x2, x3)  =  insert_out_gaa(x1)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x6)
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga(x1)
s(x1)  =  s(x1)
U5_ga(x1, x2, x3)  =  U5_ga(x1, x3)
U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x1, x6)
U3_gaa(x1, x2, x3, x4, x5, x6)  =  U3_gaa(x1, x6)
less_in_ag(x1, x2)  =  less_in_ag(x2)
less_out_ag(x1, x2)  =  less_out_ag(x1, x2)
U5_ag(x1, x2, x3)  =  U5_ag(x2, x3)
U4_gaa(x1, x2, x3, x4, x5, x6)  =  U4_gaa(x1, x6)
LESS_IN_AG(x1, x2)  =  LESS_IN_AG(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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
              ↳ PiDP

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

LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
LESS_IN_AG(x1, x2)  =  LESS_IN_AG(x2)

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

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

LESS_IN_AG(s(Y)) → LESS_IN_AG(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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP

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

LESS_IN_GA(s(X), s(Y)) → LESS_IN_GA(X, Y)

The TRS R consists of the following rules:

insert_in_gaa(X, void, tree(X, void, void)) → insert_out_gaa(X, void, tree(X, void, void))
insert_in_gaa(X, tree(X, Left, Right), tree(X, Left, Right)) → insert_out_gaa(X, tree(X, Left, Right), tree(X, Left, Right))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_gaa(X, Y, Left, Right, Left1, less_in_ga(X, Y))
less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U5_ga(X, Y, less_in_ga(X, Y))
U5_ga(X, Y, less_out_ga(X, Y)) → less_out_ga(s(X), s(Y))
U1_gaa(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → U2_gaa(X, Y, Left, Right, Left1, insert_in_gaa(X, Left, Left1))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_gaa(X, Y, Left, Right, Right1, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U5_ag(X, Y, less_in_ag(X, Y))
U5_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U3_gaa(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → U4_gaa(X, Y, Left, Right, Right1, insert_in_gaa(X, Right, Right1))
U4_gaa(X, Y, Left, Right, Right1, insert_out_gaa(X, Right, Right1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1))
U2_gaa(X, Y, Left, Right, Left1, insert_out_gaa(X, Left, Left1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right))

The argument filtering Pi contains the following mapping:
insert_in_gaa(x1, x2, x3)  =  insert_in_gaa(x1)
insert_out_gaa(x1, x2, x3)  =  insert_out_gaa(x1)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x6)
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga(x1)
s(x1)  =  s(x1)
U5_ga(x1, x2, x3)  =  U5_ga(x1, x3)
U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x1, x6)
U3_gaa(x1, x2, x3, x4, x5, x6)  =  U3_gaa(x1, x6)
less_in_ag(x1, x2)  =  less_in_ag(x2)
less_out_ag(x1, x2)  =  less_out_ag(x1, x2)
U5_ag(x1, x2, x3)  =  U5_ag(x2, x3)
U4_gaa(x1, x2, x3, x4, x5, x6)  =  U4_gaa(x1, x6)
LESS_IN_GA(x1, x2)  =  LESS_IN_GA(x1)

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

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

LESS_IN_GA(s(X), s(Y)) → LESS_IN_GA(X, Y)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
LESS_IN_GA(x1, x2)  =  LESS_IN_GA(x1)

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

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

LESS_IN_GA(s(X)) → LESS_IN_GA(X)

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

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

INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_GAA(X, Y, Left, Right, Left1, less_in_ga(X, Y))
U1_GAA(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → INSERT_IN_GAA(X, Left, Left1)
U3_GAA(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → INSERT_IN_GAA(X, Right, Right1)
INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_GAA(X, Y, Left, Right, Right1, less_in_ag(Y, X))

The TRS R consists of the following rules:

insert_in_gaa(X, void, tree(X, void, void)) → insert_out_gaa(X, void, tree(X, void, void))
insert_in_gaa(X, tree(X, Left, Right), tree(X, Left, Right)) → insert_out_gaa(X, tree(X, Left, Right), tree(X, Left, Right))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_gaa(X, Y, Left, Right, Left1, less_in_ga(X, Y))
less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U5_ga(X, Y, less_in_ga(X, Y))
U5_ga(X, Y, less_out_ga(X, Y)) → less_out_ga(s(X), s(Y))
U1_gaa(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → U2_gaa(X, Y, Left, Right, Left1, insert_in_gaa(X, Left, Left1))
insert_in_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_gaa(X, Y, Left, Right, Right1, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U5_ag(X, Y, less_in_ag(X, Y))
U5_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U3_gaa(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → U4_gaa(X, Y, Left, Right, Right1, insert_in_gaa(X, Right, Right1))
U4_gaa(X, Y, Left, Right, Right1, insert_out_gaa(X, Right, Right1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left, Right1))
U2_gaa(X, Y, Left, Right, Left1, insert_out_gaa(X, Left, Left1)) → insert_out_gaa(X, tree(Y, Left, Right), tree(Y, Left1, Right))

The argument filtering Pi contains the following mapping:
insert_in_gaa(x1, x2, x3)  =  insert_in_gaa(x1)
insert_out_gaa(x1, x2, x3)  =  insert_out_gaa(x1)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x6)
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga(x1)
s(x1)  =  s(x1)
U5_ga(x1, x2, x3)  =  U5_ga(x1, x3)
U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x1, x6)
U3_gaa(x1, x2, x3, x4, x5, x6)  =  U3_gaa(x1, x6)
less_in_ag(x1, x2)  =  less_in_ag(x2)
less_out_ag(x1, x2)  =  less_out_ag(x1, x2)
U5_ag(x1, x2, x3)  =  U5_ag(x2, x3)
U4_gaa(x1, x2, x3, x4, x5, x6)  =  U4_gaa(x1, x6)
INSERT_IN_GAA(x1, x2, x3)  =  INSERT_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4, x5, x6)  =  U3_GAA(x1, x6)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x6)

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

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

INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U1_GAA(X, Y, Left, Right, Left1, less_in_ga(X, Y))
U1_GAA(X, Y, Left, Right, Left1, less_out_ga(X, Y)) → INSERT_IN_GAA(X, Left, Left1)
U3_GAA(X, Y, Left, Right, Right1, less_out_ag(Y, X)) → INSERT_IN_GAA(X, Right, Right1)
INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_GAA(X, Y, Left, Right, Right1, less_in_ag(Y, X))

The TRS R consists of the following rules:

less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U5_ga(X, Y, less_in_ga(X, Y))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U5_ag(X, Y, less_in_ag(X, Y))
U5_ga(X, Y, less_out_ga(X, Y)) → less_out_ga(s(X), s(Y))
U5_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))

The argument filtering Pi contains the following mapping:
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga(x1)
s(x1)  =  s(x1)
U5_ga(x1, x2, x3)  =  U5_ga(x1, x3)
less_in_ag(x1, x2)  =  less_in_ag(x2)
less_out_ag(x1, x2)  =  less_out_ag(x1, x2)
U5_ag(x1, x2, x3)  =  U5_ag(x2, x3)
INSERT_IN_GAA(x1, x2, x3)  =  INSERT_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4, x5, x6)  =  U3_GAA(x1, x6)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x6)

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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing

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

U3_GAA(X, less_out_ag(Y, X)) → INSERT_IN_GAA(X)
U1_GAA(X, less_out_ga(X)) → INSERT_IN_GAA(X)
INSERT_IN_GAA(X) → U3_GAA(X, less_in_ag(X))
INSERT_IN_GAA(X) → U1_GAA(X, less_in_ga(X))

The TRS R consists of the following rules:

less_in_ga(0) → less_out_ga(0)
less_in_ga(s(X)) → U5_ga(X, less_in_ga(X))
less_in_ag(s(X)) → less_out_ag(0, s(X))
less_in_ag(s(Y)) → U5_ag(Y, less_in_ag(Y))
U5_ga(X, less_out_ga(X)) → less_out_ga(s(X))
U5_ag(Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))

The set Q consists of the following terms:

less_in_ga(x0)
less_in_ag(x0)
U5_ga(x0, x1)
U5_ag(x0, x1)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule INSERT_IN_GAA(X) → U1_GAA(X, less_in_ga(X)) at position [1] we obtained the following new rules:

INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga(0))
INSERT_IN_GAA(s(x0)) → U1_GAA(s(x0), U5_ga(x0, less_in_ga(x0)))



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

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

U3_GAA(X, less_out_ag(Y, X)) → INSERT_IN_GAA(X)
INSERT_IN_GAA(s(x0)) → U1_GAA(s(x0), U5_ga(x0, less_in_ga(x0)))
U1_GAA(X, less_out_ga(X)) → INSERT_IN_GAA(X)
INSERT_IN_GAA(X) → U3_GAA(X, less_in_ag(X))
INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga(0))

The TRS R consists of the following rules:

less_in_ga(0) → less_out_ga(0)
less_in_ga(s(X)) → U5_ga(X, less_in_ga(X))
less_in_ag(s(X)) → less_out_ag(0, s(X))
less_in_ag(s(Y)) → U5_ag(Y, less_in_ag(Y))
U5_ga(X, less_out_ga(X)) → less_out_ga(s(X))
U5_ag(Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))

The set Q consists of the following terms:

less_in_ga(x0)
less_in_ag(x0)
U5_ga(x0, x1)
U5_ag(x0, x1)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule INSERT_IN_GAA(X) → U3_GAA(X, less_in_ag(X)) at position [1] we obtained the following new rules:

INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), U5_ag(x0, less_in_ag(x0)))
INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), less_out_ag(0, s(x0)))



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ Instantiation

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

INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), U5_ag(x0, less_in_ag(x0)))
INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), less_out_ag(0, s(x0)))
U3_GAA(X, less_out_ag(Y, X)) → INSERT_IN_GAA(X)
U1_GAA(X, less_out_ga(X)) → INSERT_IN_GAA(X)
INSERT_IN_GAA(s(x0)) → U1_GAA(s(x0), U5_ga(x0, less_in_ga(x0)))
INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga(0))

The TRS R consists of the following rules:

less_in_ga(0) → less_out_ga(0)
less_in_ga(s(X)) → U5_ga(X, less_in_ga(X))
less_in_ag(s(X)) → less_out_ag(0, s(X))
less_in_ag(s(Y)) → U5_ag(Y, less_in_ag(Y))
U5_ga(X, less_out_ga(X)) → less_out_ga(s(X))
U5_ag(Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))

The set Q consists of the following terms:

less_in_ga(x0)
less_in_ag(x0)
U5_ga(x0, x1)
U5_ag(x0, x1)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U1_GAA(X, less_out_ga(X)) → INSERT_IN_GAA(X) we obtained the following new rules:

U1_GAA(0, less_out_ga(0)) → INSERT_IN_GAA(0)
U1_GAA(s(z0), less_out_ga(s(z0))) → INSERT_IN_GAA(s(z0))



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof

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

INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), U5_ag(x0, less_in_ag(x0)))
U1_GAA(0, less_out_ga(0)) → INSERT_IN_GAA(0)
U3_GAA(X, less_out_ag(Y, X)) → INSERT_IN_GAA(X)
INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), less_out_ag(0, s(x0)))
INSERT_IN_GAA(s(x0)) → U1_GAA(s(x0), U5_ga(x0, less_in_ga(x0)))
INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga(0))
U1_GAA(s(z0), less_out_ga(s(z0))) → INSERT_IN_GAA(s(z0))

The TRS R consists of the following rules:

less_in_ga(0) → less_out_ga(0)
less_in_ga(s(X)) → U5_ga(X, less_in_ga(X))
less_in_ag(s(X)) → less_out_ag(0, s(X))
less_in_ag(s(Y)) → U5_ag(Y, less_in_ag(Y))
U5_ga(X, less_out_ga(X)) → less_out_ga(s(X))
U5_ag(Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))

The set Q consists of the following terms:

less_in_ga(x0)
less_in_ag(x0)
U5_ga(x0, x1)
U5_ag(x0, x1)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ AND
                                        ↳ QDP
                                          ↳ UsableRulesProof
                                        ↳ QDP

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

U1_GAA(0, less_out_ga(0)) → INSERT_IN_GAA(0)
INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga(0))

The TRS R consists of the following rules:

less_in_ga(0) → less_out_ga(0)
less_in_ga(s(X)) → U5_ga(X, less_in_ga(X))
less_in_ag(s(X)) → less_out_ag(0, s(X))
less_in_ag(s(Y)) → U5_ag(Y, less_in_ag(Y))
U5_ga(X, less_out_ga(X)) → less_out_ga(s(X))
U5_ag(Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))

The set Q consists of the following terms:

less_in_ga(x0)
less_in_ag(x0)
U5_ga(x0, x1)
U5_ag(x0, x1)

We have to consider all (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ AND
                                        ↳ QDP
                                          ↳ UsableRulesProof
                                            ↳ QDP
                                              ↳ QReductionProof
                                        ↳ QDP

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

U1_GAA(0, less_out_ga(0)) → INSERT_IN_GAA(0)
INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga(0))

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

less_in_ga(x0)
less_in_ag(x0)
U5_ga(x0, x1)
U5_ag(x0, x1)

We have to consider all (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

less_in_ga(x0)
less_in_ag(x0)
U5_ga(x0, x1)
U5_ag(x0, x1)



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ AND
                                        ↳ QDP
                                          ↳ UsableRulesProof
                                            ↳ QDP
                                              ↳ QReductionProof
                                                ↳ QDP
                                                  ↳ NonTerminationProof
                                        ↳ QDP

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

U1_GAA(0, less_out_ga(0)) → INSERT_IN_GAA(0)
INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga(0))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

U1_GAA(0, less_out_ga(0)) → INSERT_IN_GAA(0)
INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga(0))

The TRS R consists of the following rules:none


s = INSERT_IN_GAA(0) evaluates to t =INSERT_IN_GAA(0)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga(0))
with rule INSERT_IN_GAA(0) → U1_GAA(0, less_out_ga(0)) at position [] and matcher [ ]

U1_GAA(0, less_out_ga(0)) → INSERT_IN_GAA(0)
with rule U1_GAA(0, less_out_ga(0)) → INSERT_IN_GAA(0)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.





↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ AND
                                        ↳ QDP
                                        ↳ QDP
                                          ↳ Instantiation

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

INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), U5_ag(x0, less_in_ag(x0)))
INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), less_out_ag(0, s(x0)))
U3_GAA(X, less_out_ag(Y, X)) → INSERT_IN_GAA(X)
INSERT_IN_GAA(s(x0)) → U1_GAA(s(x0), U5_ga(x0, less_in_ga(x0)))
U1_GAA(s(z0), less_out_ga(s(z0))) → INSERT_IN_GAA(s(z0))

The TRS R consists of the following rules:

less_in_ga(0) → less_out_ga(0)
less_in_ga(s(X)) → U5_ga(X, less_in_ga(X))
less_in_ag(s(X)) → less_out_ag(0, s(X))
less_in_ag(s(Y)) → U5_ag(Y, less_in_ag(Y))
U5_ga(X, less_out_ga(X)) → less_out_ga(s(X))
U5_ag(Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))

The set Q consists of the following terms:

less_in_ga(x0)
less_in_ag(x0)
U5_ga(x0, x1)
U5_ag(x0, x1)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U3_GAA(X, less_out_ag(Y, X)) → INSERT_IN_GAA(X) we obtained the following new rules:

U3_GAA(s(z0), less_out_ag(0, s(z0))) → INSERT_IN_GAA(s(z0))
U3_GAA(s(z0), less_out_ag(x1, s(z0))) → INSERT_IN_GAA(s(z0))



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ AND
                                        ↳ QDP
                                        ↳ QDP
                                          ↳ Instantiation
                                            ↳ QDP
                                              ↳ NonTerminationProof

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

INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), U5_ag(x0, less_in_ag(x0)))
INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), less_out_ag(0, s(x0)))
U3_GAA(s(z0), less_out_ag(0, s(z0))) → INSERT_IN_GAA(s(z0))
INSERT_IN_GAA(s(x0)) → U1_GAA(s(x0), U5_ga(x0, less_in_ga(x0)))
U3_GAA(s(z0), less_out_ag(x1, s(z0))) → INSERT_IN_GAA(s(z0))
U1_GAA(s(z0), less_out_ga(s(z0))) → INSERT_IN_GAA(s(z0))

The TRS R consists of the following rules:

less_in_ga(0) → less_out_ga(0)
less_in_ga(s(X)) → U5_ga(X, less_in_ga(X))
less_in_ag(s(X)) → less_out_ag(0, s(X))
less_in_ag(s(Y)) → U5_ag(Y, less_in_ag(Y))
U5_ga(X, less_out_ga(X)) → less_out_ga(s(X))
U5_ag(Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))

The set Q consists of the following terms:

less_in_ga(x0)
less_in_ag(x0)
U5_ga(x0, x1)
U5_ag(x0, x1)

We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), U5_ag(x0, less_in_ag(x0)))
INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), less_out_ag(0, s(x0)))
U3_GAA(s(z0), less_out_ag(0, s(z0))) → INSERT_IN_GAA(s(z0))
INSERT_IN_GAA(s(x0)) → U1_GAA(s(x0), U5_ga(x0, less_in_ga(x0)))
U3_GAA(s(z0), less_out_ag(x1, s(z0))) → INSERT_IN_GAA(s(z0))
U1_GAA(s(z0), less_out_ga(s(z0))) → INSERT_IN_GAA(s(z0))

The TRS R consists of the following rules:

less_in_ga(0) → less_out_ga(0)
less_in_ga(s(X)) → U5_ga(X, less_in_ga(X))
less_in_ag(s(X)) → less_out_ag(0, s(X))
less_in_ag(s(Y)) → U5_ag(Y, less_in_ag(Y))
U5_ga(X, less_out_ga(X)) → less_out_ga(s(X))
U5_ag(Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))


s = U3_GAA(s(z0), less_out_ag(0, s(z0))) evaluates to t =U3_GAA(s(z0), less_out_ag(0, s(z0)))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

U3_GAA(s(z0), less_out_ag(0, s(z0))) → INSERT_IN_GAA(s(z0))
with rule U3_GAA(s(z0'), less_out_ag(0, s(z0'))) → INSERT_IN_GAA(s(z0')) at position [] and matcher [z0' / z0]

INSERT_IN_GAA(s(z0)) → U3_GAA(s(z0), less_out_ag(0, s(z0)))
with rule INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), less_out_ag(0, s(x0)))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.