Left Termination proof of /tmp/tmpu32t9-/balance

Left Termination of the query pattern balance_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ 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(g,a).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

balance_in(T, TB) → U1(T, TB, balance_in(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in(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(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
balance_in(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
U2(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out(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(T, TB, balance_out(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out(T, TB)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

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

balance_in(T, TB) → U1(T, TB, balance_in(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in(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(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
balance_in(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
U2(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out(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(T, TB, balance_out(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out(T, TB)

The argument filtering Pi contains the following mapping:
balance_in(x1, x2) = balance_in(x1)
U1(x1, x2, x3) = U1(x3)
balance_in(x1, x2, x3, x4, x5) = balance_in(x1)
-(x1, x2) = -(x1, x2)
[] = []
.(x1, x2) = .(x1, x2)
','(x1, x2) = ','(x1, x2)
tree(x1, x2, x3) = tree(x1, x2, x3)
U2(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2(x3, x18)
nil = nil
balance_out(x1, x2, x3, x4, x5) = balance_out
U3(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3(x18)
balance_out(x1, x2) = balance_out

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(T, TB) → U1¹(T, TB, balance_in(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
BALANCE_IN(T, TB) → BALANCE_IN(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))
BALANCE_IN(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)) → U2¹(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
BALANCE_IN(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(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
U2¹(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3¹(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U2¹(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))

The TRS R consists of the following rules:

balance_in(T, TB) → U1(T, TB, balance_in(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in(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(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
balance_in(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
U2(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out(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(T, TB, balance_out(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out(T, TB)

The argument filtering Pi contains the following mapping:
balance_in(x1, x2) = balance_in(x1)
U1(x1, x2, x3) = U1(x3)
balance_in(x1, x2, x3, x4, x5) = balance_in(x1)
-(x1, x2) = -(x1, x2)
[] = []
.(x1, x2) = .(x1, x2)
','(x1, x2) = ','(x1, x2)
tree(x1, x2, x3) = tree(x1, x2, x3)
U2(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2(x3, x18)
nil = nil
balance_out(x1, x2, x3, x4, x5) = balance_out
U3(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3(x18)
balance_out(x1, x2) = balance_out
BALANCE_IN(x1, x2) = BALANCE_IN(x1)
U3¹(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3¹(x18)
BALANCE_IN(x1, x2, x3, x4, x5) = BALANCE_IN(x1)
U2¹(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2¹(x3, x18)
U1¹(x1, x2, x3) = U1¹(x3)

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

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

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

BALANCE_IN(T, TB) → U1¹(T, TB, balance_in(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
BALANCE_IN(T, TB) → BALANCE_IN(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))
BALANCE_IN(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)) → U2¹(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
BALANCE_IN(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(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
U2¹(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3¹(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U2¹(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))

The TRS R consists of the following rules:

balance_in(T, TB) → U1(T, TB, balance_in(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in(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(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
balance_in(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
U2(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out(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(T, TB, balance_out(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out(T, TB)

The argument filtering Pi contains the following mapping:
balance_in(x1, x2) = balance_in(x1)
U1(x1, x2, x3) = U1(x3)
balance_in(x1, x2, x3, x4, x5) = balance_in(x1)
-(x1, x2) = -(x1, x2)
[] = []
.(x1, x2) = .(x1, x2)
','(x1, x2) = ','(x1, x2)
tree(x1, x2, x3) = tree(x1, x2, x3)
U2(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2(x3, x18)
nil = nil
balance_out(x1, x2, x3, x4, x5) = balance_out
U3(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3(x18)
balance_out(x1, x2) = balance_out
BALANCE_IN(x1, x2) = BALANCE_IN(x1)
U3¹(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3¹(x18)
BALANCE_IN(x1, x2, x3, x4, x5) = BALANCE_IN(x1)
U2¹(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2¹(x3, x18)
U1¹(x1, x2, x3) = U1¹(x3)

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

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

BALANCE_IN(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(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
BALANCE_IN(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)) → U2¹(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2¹(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))

The TRS R consists of the following rules:

balance_in(T, TB) → U1(T, TB, balance_in(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
balance_in(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(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
balance_in(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
U2(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out(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(T, TB, balance_out(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) → balance_out(T, TB)

The argument filtering Pi contains the following mapping:
balance_in(x1, x2) = balance_in(x1)
U1(x1, x2, x3) = U1(x3)
balance_in(x1, x2, x3, x4, x5) = balance_in(x1)
-(x1, x2) = -(x1, x2)
[] = []
.(x1, x2) = .(x1, x2)
','(x1, x2) = ','(x1, x2)
tree(x1, x2, x3) = tree(x1, x2, x3)
U2(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2(x3, x18)
nil = nil
balance_out(x1, x2, x3, x4, x5) = balance_out
U3(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3(x18)
balance_out(x1, x2) = balance_out
BALANCE_IN(x1, x2, x3, x4, x5) = BALANCE_IN(x1)
U2¹(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2¹(x3, 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

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

BALANCE_IN(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(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
BALANCE_IN(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)) → U2¹(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
U2¹(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → BALANCE_IN(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))

The TRS R consists of the following rules:

balance_in(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(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
balance_in(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) → balance_out(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
U2(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) → U3(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
U3(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) → balance_out(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(x1, x2, x3, x4, x5) = balance_in(x1)
-(x1, x2) = -(x1, x2)
.(x1, x2) = .(x1, x2)
','(x1, x2) = ','(x1, x2)
tree(x1, x2, x3) = tree(x1, x2, x3)
U2(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2(x3, x18)
nil = nil
balance_out(x1, x2, x3, x4, x5) = balance_out
U3(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3(x18)
BALANCE_IN(x1, x2, x3, x4, x5) = BALANCE_IN(x1)
U2¹(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2¹(x3, 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

                      ↳ QDPSizeChangeProof

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

BALANCE_IN(tree(L, V, R)) → BALANCE_IN(L)
BALANCE_IN(tree(L, V, R)) → U2¹(R, balance_in(L))
U2¹(R, balance_out) → BALANCE_IN(R)

The TRS R consists of the following rules:

balance_in(tree(L, V, R)) → U2(R, balance_in(L))
balance_in(nil) → balance_out
U2(R, balance_out) → U3(balance_in(R))
U3(balance_out) → balance_out

The set Q consists of the following terms:

balance_in(x0)
U2(x0, x1)
U3(x0)

We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

BALANCE_IN(tree(L, V, R)) → BALANCE_IN(L)
The graph contains the following edges 1 > 1
BALANCE_IN(tree(L, V, R)) → U2¹(R, balance_in(L))
The graph contains the following edges 1 > 1
U2¹(R, balance_out) → BALANCE_IN(R)
The graph contains the following edges 1 >= 1