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

(0) Obligation:

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(X1)).
less(s(X), s(Y)) :- less(X, Y).

Queries:

insert(g,a,a).

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. 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(X1)) → less_out_ga(0, s(X1))
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(X1)) → less_out_ag(0, s(X1))
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

(2) Obligation:

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(X1)) → less_out_ga(0, s(X1))
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(X1)) → less_out_ag(0, s(X1))
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)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] 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(X1)) → less_out_ga(0, s(X1))
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(X1)) → less_out_ag(0, s(X1))
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)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x6)
LESS_IN_GA(x1, x2)  =  LESS_IN_GA(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)
U5_AG(x1, x2, x3)  =  U5_AG(x3)
U4_GAA(x1, x2, x3, x4, x5, x6)  =  U4_GAA(x6)

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

(4) Obligation:

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(X1)) → less_out_ga(0, s(X1))
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(X1)) → less_out_ag(0, s(X1))
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)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x6)
LESS_IN_GA(x1, x2)  =  LESS_IN_GA(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)
U5_AG(x1, x2, x3)  =  U5_AG(x3)
U4_GAA(x1, x2, x3, x4, x5, x6)  =  U4_GAA(x6)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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(X1)) → less_out_ga(0, s(X1))
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(X1)) → less_out_ag(0, s(X1))
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

(8) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(9) Obligation:

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

(10) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(11) Obligation:

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.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

• LESS_IN_AG(s(Y)) → LESS_IN_AG(Y)
  The graph contains the following edges 1 > 1

(13) TRUE

(14) Obligation:

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(X1)) → less_out_ga(0, s(X1))
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(X1)) → less_out_ag(0, s(X1))
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

(15) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(16) Obligation:

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

(17) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(18) Obligation:

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.

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

• LESS_IN_GA(s(X)) → LESS_IN_GA(X)
  The graph contains the following edges 1 > 1

(20) TRUE

(21) Obligation:

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

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, Left1, Right)) → U1_GAA(X, Y, Left, Right, Left1, less_in_ga(X, Y))
INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_GAA(X, Y, Left, Right, Right1, less_in_ag(Y, X))
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(X1)) → less_out_ga(0, s(X1))
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(X1)) → less_out_ag(0, s(X1))
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)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x6)
U3_GAA(x1, x2, x3, x4, x5, x6)  =  U3_GAA(x1, x6)

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

(22) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(23) Obligation:

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

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

less_in_ga(0, s(X1)) → less_out_ga(0, s(X1))
less_in_ga(s(X), s(Y)) → U5_ga(X, Y, less_in_ga(X, Y))
less_in_ag(0, s(X1)) → less_out_ag(0, s(X1))
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)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x6)
U3_GAA(x1, x2, x3, x4, x5, x6)  =  U3_GAA(x1, x6)

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

(24) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(25) Obligation:

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

U1_GAA(X, less_out_ga) → INSERT_IN_GAA(X)
INSERT_IN_GAA(X) → U1_GAA(X, less_in_ga(X))
INSERT_IN_GAA(X) → U3_GAA(X, less_in_ag(X))
U3_GAA(X, less_out_ag(Y)) → 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(X1)) → 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.

(26) Narrowing (SOUND transformation)

By narrowing [LPAR04] the rule INSERT_IN_GAA(X) → U1_GAA(X, less_in_ga(X)) at position [1] we obtained the following new rules [LPAR04]:

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

(27) Obligation:

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

U1_GAA(X, less_out_ga) → INSERT_IN_GAA(X)
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(0) → U1_GAA(0, less_out_ga)
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(X1)) → 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.

(28) Narrowing (SOUND transformation)

By narrowing [LPAR04] the rule INSERT_IN_GAA(X) → U3_GAA(X, less_in_ag(X)) at position [1] we obtained the following new rules [LPAR04]:

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)))

(29) Obligation:

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

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

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(X1)) → 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.

(30) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule U1_GAA(X, less_out_ga) → INSERT_IN_GAA(X) we obtained the following new rules [LPAR04]:

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

(31) Obligation:

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

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

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(X1)) → 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.

(32) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.

(33) Complex Obligation (AND)

(34) Obligation:

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(X1)) → 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.

(35) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(36) Obligation:

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.

(37) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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

(38) Obligation:

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.

(39) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

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:

• Semiunifier: [ ]
• Matcher: [ ]

---

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.

(40) FALSE

(41) Obligation:

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

INSERT_IN_GAA(s(x0)) → U1_GAA(s(x0), U5_ga(less_in_ga(x0)))
U1_GAA(s(z0), less_out_ga) → INSERT_IN_GAA(s(z0))
INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), less_out_ag(0))
U3_GAA(X, less_out_ag(Y)) → INSERT_IN_GAA(X)
INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), U5_ag(less_in_ag(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(X1)) → 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.

(42) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule U3_GAA(X, less_out_ag(Y)) → INSERT_IN_GAA(X) we obtained the following new rules [LPAR04]:

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))

(43) Obligation:

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

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

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(X1)) → 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.

(44) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

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:

• Semiunifier: [ ]
• Matcher: [ ]

---

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.

(45) FALSE

(46) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. 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(X1)) → less_out_ga(0, s(X1))
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(X1)) → less_out_ag(0, s(X1))
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

(47) Obligation:

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(X1)) → less_out_ga(0, s(X1))
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(X1)) → less_out_ag(0, s(X1))
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)

(48) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] 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(X1)) → less_out_ga(0, s(X1))
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(X1)) → less_out_ag(0, s(X1))
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)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x6)
LESS_IN_GA(x1, x2)  =  LESS_IN_GA(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)
U5_AG(x1, x2, x3)  =  U5_AG(x2, x3)
U4_GAA(x1, x2, x3, x4, x5, x6)  =  U4_GAA(x1, x6)

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

(49) Obligation:

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(X1)) → less_out_ga(0, s(X1))
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(X1)) → less_out_ag(0, s(X1))
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)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x6)
LESS_IN_GA(x1, x2)  =  LESS_IN_GA(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)
U5_AG(x1, x2, x3)  =  U5_AG(x2, x3)
U4_GAA(x1, x2, x3, x4, x5, x6)  =  U4_GAA(x1, x6)

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

(50) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 6 less nodes.

(51) Complex Obligation (AND)

(52) Obligation:

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(X1)) → less_out_ga(0, s(X1))
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(X1)) → less_out_ag(0, s(X1))
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

(53) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(54) Obligation:

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

(55) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(56) Obligation:

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.

(57) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

• LESS_IN_AG(s(Y)) → LESS_IN_AG(Y)
  The graph contains the following edges 1 > 1

(58) TRUE

(59) Obligation:

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(X1)) → less_out_ga(0, s(X1))
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(X1)) → less_out_ag(0, s(X1))
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

(60) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(61) Obligation:

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

(62) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(63) Obligation:

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.

(64) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

• LESS_IN_GA(s(X)) → LESS_IN_GA(X)
  The graph contains the following edges 1 > 1

(65) TRUE

(66) Obligation:

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

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, Left1, Right)) → U1_GAA(X, Y, Left, Right, Left1, less_in_ga(X, Y))
INSERT_IN_GAA(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U3_GAA(X, Y, Left, Right, Right1, less_in_ag(Y, X))
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(X1)) → less_out_ga(0, s(X1))
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(X1)) → less_out_ag(0, s(X1))
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)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x6)
U3_GAA(x1, x2, x3, x4, x5, x6)  =  U3_GAA(x1, x6)

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

(67) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(68) Obligation:

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

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

less_in_ga(0, s(X1)) → less_out_ga(0, s(X1))
less_in_ga(s(X), s(Y)) → U5_ga(X, Y, less_in_ga(X, Y))
less_in_ag(0, s(X1)) → less_out_ag(0, s(X1))
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)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x6)
U3_GAA(x1, x2, x3, x4, x5, x6)  =  U3_GAA(x1, x6)

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

(69) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(70) Obligation:

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

U1_GAA(X, less_out_ga(X)) → INSERT_IN_GAA(X)
INSERT_IN_GAA(X) → U1_GAA(X, less_in_ga(X))
INSERT_IN_GAA(X) → U3_GAA(X, less_in_ag(X))
U3_GAA(X, less_out_ag(Y, X)) → INSERT_IN_GAA(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(X1)) → less_out_ag(0, s(X1))
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.

(71) Narrowing (SOUND transformation)

By narrowing [LPAR04] the rule INSERT_IN_GAA(X) → U1_GAA(X, less_in_ga(X)) at position [1] we obtained the following new rules [LPAR04]:

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)))

(72) Obligation:

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

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

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(X1)) → less_out_ag(0, s(X1))
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.

(73) Narrowing (SOUND transformation)

By narrowing [LPAR04] the rule INSERT_IN_GAA(X) → U3_GAA(X, less_in_ag(X)) at position [1] we obtained the following new rules [LPAR04]:

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

(74) Obligation:

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

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

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(X1)) → less_out_ag(0, s(X1))
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.

(75) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule U1_GAA(X, less_out_ga(X)) → INSERT_IN_GAA(X) we obtained the following new rules [LPAR04]:

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))

(76) Obligation:

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(0) → U1_GAA(0, less_out_ga(0))
INSERT_IN_GAA(s(x0)) → U1_GAA(s(x0), U5_ga(x0, less_in_ga(x0)))
INSERT_IN_GAA(s(x0)) → U3_GAA(s(x0), less_out_ag(0, s(x0)))
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)
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(X1)) → less_out_ag(0, s(X1))
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.

(77) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.

(78) Complex Obligation (AND)

(79) Obligation:

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(X1)) → less_out_ag(0, s(X1))
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.

(80) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(81) Obligation:

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.

(82) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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

(83) Obligation:

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.

(84) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

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:

• Semiunifier: [ ]
• Matcher: [ ]

---

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.

(85) FALSE

(86) Obligation:

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

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))
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)) → U3_GAA(s(x0), U5_ag(x0, less_in_ag(x0)))

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(X1)) → less_out_ag(0, s(X1))
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.

(87) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule U3_GAA(X, less_out_ag(Y, X)) → INSERT_IN_GAA(X) we obtained the following new rules [LPAR04]:

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))

(88) Obligation:

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

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

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(X1)) → less_out_ag(0, s(X1))
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.

(89) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

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:

• Matcher: [ ]
• Semiunifier: [ ]

---

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.

(90) FALSE