Left Termination proof of /tmp/tmpshKkRc/balance

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,f,f,f,f)
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_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_aaaaa(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_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_aaaaa(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_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out_ag(T, TB)

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_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_aaaaa(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_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_aaaaa(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_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out_ag(T, TB)

BALANCE_IN_AG(T, TB) → U1_AG(T, TB, balance_in_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
BALANCE_IN_AG(T, TB) → BALANCE_IN_AAAAA(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))
BALANCE_IN_AAAAA(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_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
BALANCE_IN_AAAAA(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_AAAAA(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
U2_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U2_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN_AAAAA(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_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_aaaaa(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_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_aaaaa(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_aaaaa(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_aaaaa(x1, x2, x3, x4, x5) = balance_in_aaaaa
balance_out_aaaaa(x1, x2, x3, x4, x5) = balance_out_aaaaa
U2_aaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_aaaaa(x18)
U3_aaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3_aaaaa(x18)
balance_out_ag(x1, x2) = balance_out_ag
U1_AG(x1, x2, x3) = U1_AG(x3)
BALANCE_IN_AAAAA(x1, x2, x3, x4, x5) = BALANCE_IN_AAAAA
BALANCE_IN_AG(x1, x2) = BALANCE_IN_AG(x2)
U3_AAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3_AAAAA(x18)
U2_AAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_AAAAA(x18)

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_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
BALANCE_IN_AG(T, TB) → BALANCE_IN_AAAAA(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))
BALANCE_IN_AAAAA(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_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
BALANCE_IN_AAAAA(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_AAAAA(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
U2_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U2_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN_AAAAA(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_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_aaaaa(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_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_aaaaa(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_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out_ag(T, TB)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

BALANCE_IN_AAAAA(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_AAAAA(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
U2_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN_AAAAA(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))
BALANCE_IN_AAAAA(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_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(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_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_aaaaa(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_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_aaaaa(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_aaaaa(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_aaaaa(x1, x2, x3, x4, x5) = balance_in_aaaaa
balance_out_aaaaa(x1, x2, x3, x4, x5) = balance_out_aaaaa
U2_aaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_aaaaa(x18)
U3_aaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3_aaaaa(x18)
balance_out_ag(x1, x2) = balance_out_ag
BALANCE_IN_AAAAA(x1, x2, x3, x4, x5) = BALANCE_IN_AAAAA
U2_AAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_AAAAA(x18)

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:

BALANCE_IN_AAAAA(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_AAAAA(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
U2_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN_AAAAA(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))
BALANCE_IN_AAAAA(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_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))

The TRS R consists of the following rules:

balance_in_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_aaaaa(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_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_aaaaa(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_aaaaa(x1, x2, x3, x4, x5) = balance_in_aaaaa
balance_out_aaaaa(x1, x2, x3, x4, x5) = balance_out_aaaaa
U2_aaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_aaaaa(x18)
U3_aaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3_aaaaa(x18)
BALANCE_IN_AAAAA(x1, x2, x3, x4, x5) = BALANCE_IN_AAAAA
U2_AAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_AAAAA(x18)

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:

BALANCE_IN_AAAAA → U2_AAAAA(balance_in_aaaaa)
U2_AAAAA(balance_out_aaaaa) → BALANCE_IN_AAAAA
BALANCE_IN_AAAAA → BALANCE_IN_AAAAA

The TRS R consists of the following rules:

balance_in_aaaaa → balance_out_aaaaa
balance_in_aaaaa → U2_aaaaa(balance_in_aaaaa)
U2_aaaaa(balance_out_aaaaa) → U3_aaaaa(balance_in_aaaaa)
U3_aaaaa(balance_out_aaaaa) → balance_out_aaaaa

The set Q consists of the following terms:

balance_in_aaaaa
U2_aaaaa(x0)
U3_aaaaa(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule BALANCE_IN_AAAAA → U2_AAAAA(balance_in_aaaaa) at position [0] we obtained the following new rules:

BALANCE_IN_AAAAA → U2_AAAAA(balance_out_aaaaa)
BALANCE_IN_AAAAA → U2_AAAAA(U2_aaaaa(balance_in_aaaaa))

↳ 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_AAAAA → U2_AAAAA(balance_out_aaaaa)
BALANCE_IN_AAAAA → U2_AAAAA(U2_aaaaa(balance_in_aaaaa))
U2_AAAAA(balance_out_aaaaa) → BALANCE_IN_AAAAA
BALANCE_IN_AAAAA → BALANCE_IN_AAAAA

The TRS R consists of the following rules:

balance_in_aaaaa → balance_out_aaaaa
balance_in_aaaaa → U2_aaaaa(balance_in_aaaaa)
U2_aaaaa(balance_out_aaaaa) → U3_aaaaa(balance_in_aaaaa)
U3_aaaaa(balance_out_aaaaa) → balance_out_aaaaa

The set Q consists of the following terms:

balance_in_aaaaa
U2_aaaaa(x0)
U3_aaaaa(x0)

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

The TRS P consists of the following rules:

BALANCE_IN_AAAAA → U2_AAAAA(balance_out_aaaaa)
BALANCE_IN_AAAAA → U2_AAAAA(U2_aaaaa(balance_in_aaaaa))
U2_AAAAA(balance_out_aaaaa) → BALANCE_IN_AAAAA
BALANCE_IN_AAAAA → BALANCE_IN_AAAAA

The TRS R consists of the following rules:

balance_in_aaaaa → balance_out_aaaaa
balance_in_aaaaa → U2_aaaaa(balance_in_aaaaa)
U2_aaaaa(balance_out_aaaaa) → U3_aaaaa(balance_in_aaaaa)
U3_aaaaa(balance_out_aaaaa) → balance_out_aaaaa

s = BALANCE_IN_AAAAA evaluates to t =BALANCE_IN_AAAAA

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from BALANCE_IN_AAAAA to BALANCE_IN_AAAAA.

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,f,f,f,f)
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_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_aaaaa(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_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_aaaaa(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_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out_ag(T, TB)

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_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_aaaaa(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_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_aaaaa(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_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out_ag(T, TB)

BALANCE_IN_AG(T, TB) → U1_AG(T, TB, balance_in_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
BALANCE_IN_AG(T, TB) → BALANCE_IN_AAAAA(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))
BALANCE_IN_AAAAA(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_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
BALANCE_IN_AAAAA(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_AAAAA(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
U2_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U2_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN_AAAAA(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_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_aaaaa(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_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_aaaaa(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_aaaaa(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_aaaaa(x1, x2, x3, x4, x5) = balance_in_aaaaa
balance_out_aaaaa(x1, x2, x3, x4, x5) = balance_out_aaaaa
U2_aaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_aaaaa(x18)
U3_aaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3_aaaaa(x18)
balance_out_ag(x1, x2) = balance_out_ag(x2)
U1_AG(x1, x2, x3) = U1_AG(x2, x3)
BALANCE_IN_AAAAA(x1, x2, x3, x4, x5) = BALANCE_IN_AAAAA
BALANCE_IN_AG(x1, x2) = BALANCE_IN_AG(x2)
U3_AAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3_AAAAA(x18)
U2_AAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_AAAAA(x18)

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_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
BALANCE_IN_AG(T, TB) → BALANCE_IN_AAAAA(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))
BALANCE_IN_AAAAA(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_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
BALANCE_IN_AAAAA(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_AAAAA(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
U2_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U2_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN_AAAAA(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_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_aaaaa(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_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_aaaaa(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_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out_ag(T, TB)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

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

BALANCE_IN_AAAAA(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_AAAAA(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
U2_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN_AAAAA(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))
BALANCE_IN_AAAAA(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_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(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_aaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_aaaaa(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_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_aaaaa(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_aaaaa(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_aaaaa(x1, x2, x3, x4, x5) = balance_in_aaaaa
balance_out_aaaaa(x1, x2, x3, x4, x5) = balance_out_aaaaa
U2_aaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_aaaaa(x18)
U3_aaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3_aaaaa(x18)
balance_out_ag(x1, x2) = balance_out_ag(x2)
BALANCE_IN_AAAAA(x1, x2, x3, x4, x5) = BALANCE_IN_AAAAA
U2_AAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_AAAAA(x18)

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:

BALANCE_IN_AAAAA(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_AAAAA(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
U2_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN_AAAAA(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))
BALANCE_IN_AAAAA(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_AAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))

The TRS R consists of the following rules:

balance_in_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out_aaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
balance_in_aaaaa(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_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3_aaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_aaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out_aaaaa(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))

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ Narrowing

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

BALANCE_IN_AAAAA → U2_AAAAA(balance_in_aaaaa)
U2_AAAAA(balance_out_aaaaa) → BALANCE_IN_AAAAA
BALANCE_IN_AAAAA → BALANCE_IN_AAAAA

The TRS R consists of the following rules:

balance_in_aaaaa → balance_out_aaaaa
balance_in_aaaaa → U2_aaaaa(balance_in_aaaaa)
U2_aaaaa(balance_out_aaaaa) → U3_aaaaa(balance_in_aaaaa)
U3_aaaaa(balance_out_aaaaa) → balance_out_aaaaa

The set Q consists of the following terms:

balance_in_aaaaa
U2_aaaaa(x0)
U3_aaaaa(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule BALANCE_IN_AAAAA → U2_AAAAA(balance_in_aaaaa) at position [0] we obtained the following new rules:

BALANCE_IN_AAAAA → U2_AAAAA(balance_out_aaaaa)
BALANCE_IN_AAAAA → U2_AAAAA(U2_aaaaa(balance_in_aaaaa))

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ NonTerminationProof

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

BALANCE_IN_AAAAA → U2_AAAAA(balance_out_aaaaa)
BALANCE_IN_AAAAA → U2_AAAAA(U2_aaaaa(balance_in_aaaaa))
U2_AAAAA(balance_out_aaaaa) → BALANCE_IN_AAAAA
BALANCE_IN_AAAAA → BALANCE_IN_AAAAA

The TRS R consists of the following rules:

balance_in_aaaaa → balance_out_aaaaa
balance_in_aaaaa → U2_aaaaa(balance_in_aaaaa)
U2_aaaaa(balance_out_aaaaa) → U3_aaaaa(balance_in_aaaaa)
U3_aaaaa(balance_out_aaaaa) → balance_out_aaaaa

The set Q consists of the following terms:

balance_in_aaaaa
U2_aaaaa(x0)
U3_aaaaa(x0)

BALANCE_IN_AAAAA → U2_AAAAA(balance_out_aaaaa)
BALANCE_IN_AAAAA → U2_AAAAA(U2_aaaaa(balance_in_aaaaa))
U2_AAAAA(balance_out_aaaaa) → BALANCE_IN_AAAAA
BALANCE_IN_AAAAA → BALANCE_IN_AAAAA

The TRS R consists of the following rules:

balance_in_aaaaa → balance_out_aaaaa
balance_in_aaaaa → U2_aaaaa(balance_in_aaaaa)
U2_aaaaa(balance_out_aaaaa) → U3_aaaaa(balance_in_aaaaa)
U3_aaaaa(balance_out_aaaaa) → balance_out_aaaaa

s = BALANCE_IN_AAAAA evaluates to t =BALANCE_IN_AAAAA

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from BALANCE_IN_AAAAA to BALANCE_IN_AAAAA.