Left Termination of the query pattern balance_in_2(a, g) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

balance(T, TB) :- balance(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])).
balance(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)).
balance(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) :- ','(balance(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)), balance(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))).

Queries:

balance(a,g).

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

balance_in_ag(T, TB) → U1_ag(T, TB, balance_in_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT))
U1_ag(T, TB, balance_out_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out_ag(T, TB)

The argument filtering Pi contains the following mapping:
balance_in_ag(x1, x2)  =  balance_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
balance_in_agggg(x1, x2, x3, x4, x5)  =  balance_in_agggg(x2, x3, x4, x5)
-(x1, x2)  =  -
.(x1, x2)  =  .(x1)
balance_out_agggg(x1, x2, x3, x4, x5)  =  balance_out_agggg(x1)
U2_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_agggg(x18)
U3_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_agggg(x1, x18)
tree(x1, x2, x3)  =  tree(x1, x3)
balance_out_ag(x1, x2)  =  balance_out_ag(x1)

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:

balance_in_ag(T, TB) → U1_ag(T, TB, balance_in_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT))
U1_ag(T, TB, balance_out_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out_ag(T, TB)

The argument filtering Pi contains the following mapping:
balance_in_ag(x1, x2)  =  balance_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
balance_in_agggg(x1, x2, x3, x4, x5)  =  balance_in_agggg(x2, x3, x4, x5)
-(x1, x2)  =  -
.(x1, x2)  =  .(x1)
balance_out_agggg(x1, x2, x3, x4, x5)  =  balance_out_agggg(x1)
U2_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_agggg(x18)
U3_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_agggg(x1, x18)
tree(x1, x2, x3)  =  tree(x1, x3)
balance_out_ag(x1, x2)  =  balance_out_ag(x1)


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

BALANCE_IN_AG(T, TB) → U1_AG(T, TB, balance_in_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
BALANCE_IN_AG(T, TB) → BALANCE_IN_AGGGG(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))
BALANCE_IN_AGGGG(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
BALANCE_IN_AGGGG(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → BALANCE_IN_AGGGG(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN_AGGGG(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))

The TRS R consists of the following rules:

balance_in_ag(T, TB) → U1_ag(T, TB, balance_in_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT))
U1_ag(T, TB, balance_out_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out_ag(T, TB)

The argument filtering Pi contains the following mapping:
balance_in_ag(x1, x2)  =  balance_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
balance_in_agggg(x1, x2, x3, x4, x5)  =  balance_in_agggg(x2, x3, x4, x5)
-(x1, x2)  =  -
.(x1, x2)  =  .(x1)
balance_out_agggg(x1, x2, x3, x4, x5)  =  balance_out_agggg(x1)
U2_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_agggg(x18)
U3_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_agggg(x1, x18)
tree(x1, x2, x3)  =  tree(x1, x3)
balance_out_ag(x1, x2)  =  balance_out_ag(x1)
U3_AGGGG(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_AGGGG(x1, x18)
U2_AGGGG(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_AGGGG(x18)
BALANCE_IN_AGGGG(x1, x2, x3, x4, x5)  =  BALANCE_IN_AGGGG(x2, x3, x4, x5)
U1_AG(x1, x2, x3)  =  U1_AG(x3)
BALANCE_IN_AG(x1, x2)  =  BALANCE_IN_AG(x2)

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:

BALANCE_IN_AG(T, TB) → U1_AG(T, TB, balance_in_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
BALANCE_IN_AG(T, TB) → BALANCE_IN_AGGGG(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))
BALANCE_IN_AGGGG(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
BALANCE_IN_AGGGG(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → BALANCE_IN_AGGGG(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN_AGGGG(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))

The TRS R consists of the following rules:

balance_in_ag(T, TB) → U1_ag(T, TB, balance_in_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT))
U1_ag(T, TB, balance_out_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out_ag(T, TB)

The argument filtering Pi contains the following mapping:
balance_in_ag(x1, x2)  =  balance_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
balance_in_agggg(x1, x2, x3, x4, x5)  =  balance_in_agggg(x2, x3, x4, x5)
-(x1, x2)  =  -
.(x1, x2)  =  .(x1)
balance_out_agggg(x1, x2, x3, x4, x5)  =  balance_out_agggg(x1)
U2_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_agggg(x18)
U3_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_agggg(x1, x18)
tree(x1, x2, x3)  =  tree(x1, x3)
balance_out_ag(x1, x2)  =  balance_out_ag(x1)
U3_AGGGG(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_AGGGG(x1, x18)
U2_AGGGG(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_AGGGG(x18)
BALANCE_IN_AGGGG(x1, x2, x3, x4, x5)  =  BALANCE_IN_AGGGG(x2, x3, x4, x5)
U1_AG(x1, x2, x3)  =  U1_AG(x3)
BALANCE_IN_AG(x1, x2)  =  BALANCE_IN_AG(x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 3 less nodes.

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

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

U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN_AGGGG(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))
BALANCE_IN_AGGGG(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → BALANCE_IN_AGGGG(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
BALANCE_IN_AGGGG(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))

The TRS R consists of the following rules:

balance_in_ag(T, TB) → U1_ag(T, TB, balance_in_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT))
U1_ag(T, TB, balance_out_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out_ag(T, TB)

The argument filtering Pi contains the following mapping:
balance_in_ag(x1, x2)  =  balance_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
balance_in_agggg(x1, x2, x3, x4, x5)  =  balance_in_agggg(x2, x3, x4, x5)
-(x1, x2)  =  -
.(x1, x2)  =  .(x1)
balance_out_agggg(x1, x2, x3, x4, x5)  =  balance_out_agggg(x1)
U2_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_agggg(x18)
U3_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_agggg(x1, x18)
tree(x1, x2, x3)  =  tree(x1, x3)
balance_out_ag(x1, x2)  =  balance_out_ag(x1)
U2_AGGGG(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_AGGGG(x18)
BALANCE_IN_AGGGG(x1, x2, x3, x4, x5)  =  BALANCE_IN_AGGGG(x2, x3, x4, x5)

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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

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

U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN_AGGGG(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))
BALANCE_IN_AGGGG(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → BALANCE_IN_AGGGG(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
BALANCE_IN_AGGGG(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))

The TRS R consists of the following rules:

balance_in_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT))

The argument filtering Pi contains the following mapping:
balance_in_agggg(x1, x2, x3, x4, x5)  =  balance_in_agggg(x2, x3, x4, x5)
-(x1, x2)  =  -
.(x1, x2)  =  .(x1)
balance_out_agggg(x1, x2, x3, x4, x5)  =  balance_out_agggg(x1)
U2_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_agggg(x18)
U3_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_agggg(x1, x18)
tree(x1, x2, x3)  =  tree(x1, x3)
U2_AGGGG(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_AGGGG(x18)
BALANCE_IN_AGGGG(x1, x2, x3, x4, x5)  =  BALANCE_IN_AGGGG(x2, x3, x4, x5)

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

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

U2_AGGGG(balance_out_agggg(L)) → BALANCE_IN_AGGGG(-, -, -, -)
BALANCE_IN_AGGGG(-, -, -, -) → BALANCE_IN_AGGGG(-, -, -, -)
BALANCE_IN_AGGGG(-, -, -, -) → U2_AGGGG(balance_in_agggg(-, -, -, -))

The TRS R consists of the following rules:

balance_in_agggg(-, -, -, -) → balance_out_agggg(nil)
balance_in_agggg(-, -, -, -) → U2_agggg(balance_in_agggg(-, -, -, -))
U2_agggg(balance_out_agggg(L)) → U3_agggg(L, balance_in_agggg(-, -, -, -))
U3_agggg(L, balance_out_agggg(R)) → balance_out_agggg(tree(L, R))

The set Q consists of the following terms:

balance_in_agggg(x0, x1, x2, x3)
U2_agggg(x0)
U3_agggg(x0, x1)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule BALANCE_IN_AGGGG(-, -, -, -) → U2_AGGGG(balance_in_agggg(-, -, -, -)) at position [0] we obtained the following new rules:

BALANCE_IN_AGGGG(-, -, -, -) → U2_AGGGG(balance_out_agggg(nil))
BALANCE_IN_AGGGG(-, -, -, -) → U2_AGGGG(U2_agggg(balance_in_agggg(-, -, -, -)))



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

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

BALANCE_IN_AGGGG(-, -, -, -) → U2_AGGGG(balance_out_agggg(nil))
BALANCE_IN_AGGGG(-, -, -, -) → U2_AGGGG(U2_agggg(balance_in_agggg(-, -, -, -)))
U2_AGGGG(balance_out_agggg(L)) → BALANCE_IN_AGGGG(-, -, -, -)
BALANCE_IN_AGGGG(-, -, -, -) → BALANCE_IN_AGGGG(-, -, -, -)

The TRS R consists of the following rules:

balance_in_agggg(-, -, -, -) → balance_out_agggg(nil)
balance_in_agggg(-, -, -, -) → U2_agggg(balance_in_agggg(-, -, -, -))
U2_agggg(balance_out_agggg(L)) → U3_agggg(L, balance_in_agggg(-, -, -, -))
U3_agggg(L, balance_out_agggg(R)) → balance_out_agggg(tree(L, R))

The set Q consists of the following terms:

balance_in_agggg(x0, x1, x2, x3)
U2_agggg(x0)
U3_agggg(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 semiunifying a rule from P directly.

The TRS P consists of the following rules:

BALANCE_IN_AGGGG(-, -, -, -) → U2_AGGGG(balance_out_agggg(nil))
BALANCE_IN_AGGGG(-, -, -, -) → U2_AGGGG(U2_agggg(balance_in_agggg(-, -, -, -)))
U2_AGGGG(balance_out_agggg(L)) → BALANCE_IN_AGGGG(-, -, -, -)
BALANCE_IN_AGGGG(-, -, -, -) → BALANCE_IN_AGGGG(-, -, -, -)

The TRS R consists of the following rules:

balance_in_agggg(-, -, -, -) → balance_out_agggg(nil)
balance_in_agggg(-, -, -, -) → U2_agggg(balance_in_agggg(-, -, -, -))
U2_agggg(balance_out_agggg(L)) → U3_agggg(L, balance_in_agggg(-, -, -, -))
U3_agggg(L, balance_out_agggg(R)) → balance_out_agggg(tree(L, R))


s = BALANCE_IN_AGGGG(-, -, -, -) evaluates to t =BALANCE_IN_AGGGG(-, -, -, -)

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



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from BALANCE_IN_AGGGG(-, -, -, -) to BALANCE_IN_AGGGG(-, -, -, -).




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

balance_in_ag(T, TB) → U1_ag(T, TB, balance_in_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT))
U1_ag(T, TB, balance_out_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out_ag(T, TB)

The argument filtering Pi contains the following mapping:
balance_in_ag(x1, x2)  =  balance_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
balance_in_agggg(x1, x2, x3, x4, x5)  =  balance_in_agggg(x2, x3, x4, x5)
-(x1, x2)  =  -
.(x1, x2)  =  .(x1)
balance_out_agggg(x1, x2, x3, x4, x5)  =  balance_out_agggg(x1, x2, x3, x4, x5)
U2_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_agggg(x18)
U3_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_agggg(x1, x18)
tree(x1, x2, x3)  =  tree(x1, x3)
balance_out_ag(x1, x2)  =  balance_out_ag(x1, x2)

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:

balance_in_ag(T, TB) → U1_ag(T, TB, balance_in_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT))
U1_ag(T, TB, balance_out_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out_ag(T, TB)

The argument filtering Pi contains the following mapping:
balance_in_ag(x1, x2)  =  balance_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
balance_in_agggg(x1, x2, x3, x4, x5)  =  balance_in_agggg(x2, x3, x4, x5)
-(x1, x2)  =  -
.(x1, x2)  =  .(x1)
balance_out_agggg(x1, x2, x3, x4, x5)  =  balance_out_agggg(x1, x2, x3, x4, x5)
U2_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_agggg(x18)
U3_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_agggg(x1, x18)
tree(x1, x2, x3)  =  tree(x1, x3)
balance_out_ag(x1, x2)  =  balance_out_ag(x1, x2)


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

BALANCE_IN_AG(T, TB) → U1_AG(T, TB, balance_in_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
BALANCE_IN_AG(T, TB) → BALANCE_IN_AGGGG(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))
BALANCE_IN_AGGGG(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
BALANCE_IN_AGGGG(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → BALANCE_IN_AGGGG(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN_AGGGG(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))

The TRS R consists of the following rules:

balance_in_ag(T, TB) → U1_ag(T, TB, balance_in_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT))
U1_ag(T, TB, balance_out_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out_ag(T, TB)

The argument filtering Pi contains the following mapping:
balance_in_ag(x1, x2)  =  balance_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
balance_in_agggg(x1, x2, x3, x4, x5)  =  balance_in_agggg(x2, x3, x4, x5)
-(x1, x2)  =  -
.(x1, x2)  =  .(x1)
balance_out_agggg(x1, x2, x3, x4, x5)  =  balance_out_agggg(x1, x2, x3, x4, x5)
U2_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_agggg(x18)
U3_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_agggg(x1, x18)
tree(x1, x2, x3)  =  tree(x1, x3)
balance_out_ag(x1, x2)  =  balance_out_ag(x1, x2)
U3_AGGGG(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_AGGGG(x1, x18)
U2_AGGGG(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_AGGGG(x18)
BALANCE_IN_AGGGG(x1, x2, x3, x4, x5)  =  BALANCE_IN_AGGGG(x2, x3, x4, x5)
U1_AG(x1, x2, x3)  =  U1_AG(x2, x3)
BALANCE_IN_AG(x1, x2)  =  BALANCE_IN_AG(x2)

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:

BALANCE_IN_AG(T, TB) → U1_AG(T, TB, balance_in_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
BALANCE_IN_AG(T, TB) → BALANCE_IN_AGGGG(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))
BALANCE_IN_AGGGG(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
BALANCE_IN_AGGGG(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → BALANCE_IN_AGGGG(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN_AGGGG(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))

The TRS R consists of the following rules:

balance_in_ag(T, TB) → U1_ag(T, TB, balance_in_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT))
U1_ag(T, TB, balance_out_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out_ag(T, TB)

The argument filtering Pi contains the following mapping:
balance_in_ag(x1, x2)  =  balance_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
balance_in_agggg(x1, x2, x3, x4, x5)  =  balance_in_agggg(x2, x3, x4, x5)
-(x1, x2)  =  -
.(x1, x2)  =  .(x1)
balance_out_agggg(x1, x2, x3, x4, x5)  =  balance_out_agggg(x1, x2, x3, x4, x5)
U2_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_agggg(x18)
U3_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_agggg(x1, x18)
tree(x1, x2, x3)  =  tree(x1, x3)
balance_out_ag(x1, x2)  =  balance_out_ag(x1, x2)
U3_AGGGG(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_AGGGG(x1, x18)
U2_AGGGG(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_AGGGG(x18)
BALANCE_IN_AGGGG(x1, x2, x3, x4, x5)  =  BALANCE_IN_AGGGG(x2, x3, x4, x5)
U1_AG(x1, x2, x3)  =  U1_AG(x2, x3)
BALANCE_IN_AG(x1, x2)  =  BALANCE_IN_AG(x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 3 less nodes.

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

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

U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN_AGGGG(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))
BALANCE_IN_AGGGG(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → BALANCE_IN_AGGGG(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
BALANCE_IN_AGGGG(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))

The TRS R consists of the following rules:

balance_in_ag(T, TB) → U1_ag(T, TB, balance_in_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT))
U1_ag(T, TB, balance_out_agggg(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out_ag(T, TB)

The argument filtering Pi contains the following mapping:
balance_in_ag(x1, x2)  =  balance_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
balance_in_agggg(x1, x2, x3, x4, x5)  =  balance_in_agggg(x2, x3, x4, x5)
-(x1, x2)  =  -
.(x1, x2)  =  .(x1)
balance_out_agggg(x1, x2, x3, x4, x5)  =  balance_out_agggg(x1, x2, x3, x4, x5)
U2_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_agggg(x18)
U3_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_agggg(x1, x18)
tree(x1, x2, x3)  =  tree(x1, x3)
balance_out_ag(x1, x2)  =  balance_out_ag(x1, x2)
U2_AGGGG(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_AGGGG(x18)
BALANCE_IN_AGGGG(x1, x2, x3, x4, x5)  =  BALANCE_IN_AGGGG(x2, x3, x4, x5)

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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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

U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN_AGGGG(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))
BALANCE_IN_AGGGG(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → BALANCE_IN_AGGGG(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
BALANCE_IN_AGGGG(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_AGGGG(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))

The TRS R consists of the following rules:

balance_in_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_agggg(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) → U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_agggg(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_agggg(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_agggg(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT))

The argument filtering Pi contains the following mapping:
balance_in_agggg(x1, x2, x3, x4, x5)  =  balance_in_agggg(x2, x3, x4, x5)
-(x1, x2)  =  -
.(x1, x2)  =  .(x1)
balance_out_agggg(x1, x2, x3, x4, x5)  =  balance_out_agggg(x1, x2, x3, x4, x5)
U2_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_agggg(x18)
U3_agggg(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_agggg(x1, x18)
tree(x1, x2, x3)  =  tree(x1, x3)
U2_AGGGG(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_AGGGG(x18)
BALANCE_IN_AGGGG(x1, x2, x3, x4, x5)  =  BALANCE_IN_AGGGG(x2, x3, x4, x5)

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

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

U2_AGGGG(balance_out_agggg(L, -, -, -, -)) → BALANCE_IN_AGGGG(-, -, -, -)
BALANCE_IN_AGGGG(-, -, -, -) → BALANCE_IN_AGGGG(-, -, -, -)
BALANCE_IN_AGGGG(-, -, -, -) → U2_AGGGG(balance_in_agggg(-, -, -, -))

The TRS R consists of the following rules:

balance_in_agggg(-, -, -, -) → balance_out_agggg(nil, -, -, -, -)
balance_in_agggg(-, -, -, -) → U2_agggg(balance_in_agggg(-, -, -, -))
U2_agggg(balance_out_agggg(L, -, -, -, -)) → U3_agggg(L, balance_in_agggg(-, -, -, -))
U3_agggg(L, balance_out_agggg(R, -, -, -, -)) → balance_out_agggg(tree(L, R), -, -, -, -)

The set Q consists of the following terms:

balance_in_agggg(x0, x1, x2, x3)
U2_agggg(x0)
U3_agggg(x0, x1)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule BALANCE_IN_AGGGG(-, -, -, -) → U2_AGGGG(balance_in_agggg(-, -, -, -)) at position [0] we obtained the following new rules:

BALANCE_IN_AGGGG(-, -, -, -) → U2_AGGGG(balance_out_agggg(nil, -, -, -, -))
BALANCE_IN_AGGGG(-, -, -, -) → U2_AGGGG(U2_agggg(balance_in_agggg(-, -, -, -)))



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

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

BALANCE_IN_AGGGG(-, -, -, -) → U2_AGGGG(balance_out_agggg(nil, -, -, -, -))
U2_AGGGG(balance_out_agggg(L, -, -, -, -)) → BALANCE_IN_AGGGG(-, -, -, -)
BALANCE_IN_AGGGG(-, -, -, -) → U2_AGGGG(U2_agggg(balance_in_agggg(-, -, -, -)))
BALANCE_IN_AGGGG(-, -, -, -) → BALANCE_IN_AGGGG(-, -, -, -)

The TRS R consists of the following rules:

balance_in_agggg(-, -, -, -) → balance_out_agggg(nil, -, -, -, -)
balance_in_agggg(-, -, -, -) → U2_agggg(balance_in_agggg(-, -, -, -))
U2_agggg(balance_out_agggg(L, -, -, -, -)) → U3_agggg(L, balance_in_agggg(-, -, -, -))
U3_agggg(L, balance_out_agggg(R, -, -, -, -)) → balance_out_agggg(tree(L, R), -, -, -, -)

The set Q consists of the following terms:

balance_in_agggg(x0, x1, x2, x3)
U2_agggg(x0)
U3_agggg(x0, x1)

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