YES
proof of /hpcwork/ff862203/termcomp26/benchmarks/bNmw0.pl
# AProVE Commit ID: 23a904c96b029b0a549cde0d0d17dbccf967db59 jckassing 20260626 unpublished dirty


Left Termination of the query pattern

balance(g,a)

w.r.t. the given Prolog program could successfully be proven:

(0) Prolog
(1) PrologToPiTRSProof [SOUND, 0 ms]
(2) PiTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) PiDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) PiDP
(7) UsableRulesProof [EQUIVALENT, 0 ms]
(8) PiDP
(9) PiDPToQDPProof [SOUND, 0 ms]
(10) QDP
(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(12) YES


----------------------------------------

(0)
Obligation:
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))).


Query: balance(g,a)
----------------------------------------

(1) PrologToPiTRSProof (SOUND)
We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:

balance_in_2: (b,f)

balance_in_5: (b,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_ga(T, TB) -> U1_ga(T, TB, balance_in_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
   balance_in_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) -> balance_out_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
   balance_in_gaaaa(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_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
   U2_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
   U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) -> balance_out_gaaaa(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_ga(T, TB, balance_out_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) -> balance_out_ga(T, TB)

The argument filtering Pi contains the following mapping:
balance_in_ga(x1, x2)  =  balance_in_ga(x1)

U1_ga(x1, x2, x3)  =  U1_ga(x3)

balance_in_gaaaa(x1, x2, x3, x4, x5)  =  balance_in_gaaaa(x1)

nil  =  nil

balance_out_gaaaa(x1, x2, x3, x4, x5)  =  balance_out_gaaaa

tree(x1, x2, x3)  =  tree(x1, x2, x3)

U2_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_gaaaa(x3, x18)

U3_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_gaaaa(x18)

balance_out_ga(x1, x2)  =  balance_out_ga





Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



----------------------------------------

(2)
Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:

   balance_in_ga(T, TB) -> U1_ga(T, TB, balance_in_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
   balance_in_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) -> balance_out_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
   balance_in_gaaaa(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_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
   U2_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
   U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) -> balance_out_gaaaa(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_ga(T, TB, balance_out_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) -> balance_out_ga(T, TB)

The argument filtering Pi contains the following mapping:
balance_in_ga(x1, x2)  =  balance_in_ga(x1)

U1_ga(x1, x2, x3)  =  U1_ga(x3)

balance_in_gaaaa(x1, x2, x3, x4, x5)  =  balance_in_gaaaa(x1)

nil  =  nil

balance_out_gaaaa(x1, x2, x3, x4, x5)  =  balance_out_gaaaa

tree(x1, x2, x3)  =  tree(x1, x2, x3)

U2_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_gaaaa(x3, x18)

U3_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_gaaaa(x18)

balance_out_ga(x1, x2)  =  balance_out_ga



----------------------------------------

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

   BALANCE_IN_GA(T, TB) -> U1_GA(T, TB, balance_in_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
   BALANCE_IN_GA(T, TB) -> BALANCE_IN_GAAAA(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))
   BALANCE_IN_GAAAA(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_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
   BALANCE_IN_GAAAA(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_GAAAA(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
   U2_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> U3_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
   U2_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> BALANCE_IN_GAAAA(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))

The TRS R consists of the following rules:

   balance_in_ga(T, TB) -> U1_ga(T, TB, balance_in_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
   balance_in_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) -> balance_out_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
   balance_in_gaaaa(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_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
   U2_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
   U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) -> balance_out_gaaaa(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_ga(T, TB, balance_out_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) -> balance_out_ga(T, TB)

The argument filtering Pi contains the following mapping:
balance_in_ga(x1, x2)  =  balance_in_ga(x1)

U1_ga(x1, x2, x3)  =  U1_ga(x3)

balance_in_gaaaa(x1, x2, x3, x4, x5)  =  balance_in_gaaaa(x1)

nil  =  nil

balance_out_gaaaa(x1, x2, x3, x4, x5)  =  balance_out_gaaaa

tree(x1, x2, x3)  =  tree(x1, x2, x3)

U2_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_gaaaa(x3, x18)

U3_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_gaaaa(x18)

balance_out_ga(x1, x2)  =  balance_out_ga

BALANCE_IN_GA(x1, x2)  =  BALANCE_IN_GA(x1)

U1_GA(x1, x2, x3)  =  U1_GA(x3)

BALANCE_IN_GAAAA(x1, x2, x3, x4, x5)  =  BALANCE_IN_GAAAA(x1)

U2_GAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_GAAAA(x3, x18)

U3_GAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_GAAAA(x18)


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

(4)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   BALANCE_IN_GA(T, TB) -> U1_GA(T, TB, balance_in_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
   BALANCE_IN_GA(T, TB) -> BALANCE_IN_GAAAA(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))
   BALANCE_IN_GAAAA(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_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
   BALANCE_IN_GAAAA(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_GAAAA(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))
   U2_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> U3_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
   U2_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> BALANCE_IN_GAAAA(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))

The TRS R consists of the following rules:

   balance_in_ga(T, TB) -> U1_ga(T, TB, balance_in_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
   balance_in_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) -> balance_out_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
   balance_in_gaaaa(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_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
   U2_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
   U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) -> balance_out_gaaaa(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_ga(T, TB, balance_out_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) -> balance_out_ga(T, TB)

The argument filtering Pi contains the following mapping:
balance_in_ga(x1, x2)  =  balance_in_ga(x1)

U1_ga(x1, x2, x3)  =  U1_ga(x3)

balance_in_gaaaa(x1, x2, x3, x4, x5)  =  balance_in_gaaaa(x1)

nil  =  nil

balance_out_gaaaa(x1, x2, x3, x4, x5)  =  balance_out_gaaaa

tree(x1, x2, x3)  =  tree(x1, x2, x3)

U2_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_gaaaa(x3, x18)

U3_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_gaaaa(x18)

balance_out_ga(x1, x2)  =  balance_out_ga

BALANCE_IN_GA(x1, x2)  =  BALANCE_IN_GA(x1)

U1_GA(x1, x2, x3)  =  U1_GA(x3)

BALANCE_IN_GAAAA(x1, x2, x3, x4, x5)  =  BALANCE_IN_GAAAA(x1)

U2_GAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_GAAAA(x3, x18)

U3_GAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_GAAAA(x18)


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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.
----------------------------------------

(6)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   U2_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> BALANCE_IN_GAAAA(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))
   BALANCE_IN_GAAAA(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_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
   BALANCE_IN_GAAAA(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_GAAAA(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))

The TRS R consists of the following rules:

   balance_in_ga(T, TB) -> U1_ga(T, TB, balance_in_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])))
   balance_in_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) -> balance_out_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
   balance_in_gaaaa(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_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
   U2_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
   U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) -> balance_out_gaaaa(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_ga(T, TB, balance_out_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) -> balance_out_ga(T, TB)

The argument filtering Pi contains the following mapping:
balance_in_ga(x1, x2)  =  balance_in_ga(x1)

U1_ga(x1, x2, x3)  =  U1_ga(x3)

balance_in_gaaaa(x1, x2, x3, x4, x5)  =  balance_in_gaaaa(x1)

nil  =  nil

balance_out_gaaaa(x1, x2, x3, x4, x5)  =  balance_out_gaaaa

tree(x1, x2, x3)  =  tree(x1, x2, x3)

U2_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_gaaaa(x3, x18)

U3_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_gaaaa(x18)

balance_out_ga(x1, x2)  =  balance_out_ga

BALANCE_IN_GAAAA(x1, x2, x3, x4, x5)  =  BALANCE_IN_GAAAA(x1)

U2_GAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_GAAAA(x3, x18)


We have to consider all (P,R,Pi)-chains
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(7) UsableRulesProof (EQUIVALENT)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
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(8)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   U2_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> BALANCE_IN_GAAAA(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))
   BALANCE_IN_GAAAA(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_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
   BALANCE_IN_GAAAA(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_GAAAA(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))

The TRS R consists of the following rules:

   balance_in_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) -> balance_out_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T))
   balance_in_gaaaa(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_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)))
   U2_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)))
   U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) -> balance_out_gaaaa(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_gaaaa(x1, x2, x3, x4, x5)  =  balance_in_gaaaa(x1)

nil  =  nil

balance_out_gaaaa(x1, x2, x3, x4, x5)  =  balance_out_gaaaa

tree(x1, x2, x3)  =  tree(x1, x2, x3)

U2_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_gaaaa(x3, x18)

U3_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U3_gaaaa(x18)

BALANCE_IN_GAAAA(x1, x2, x3, x4, x5)  =  BALANCE_IN_GAAAA(x1)

U2_GAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18)  =  U2_GAAAA(x3, x18)


We have to consider all (P,R,Pi)-chains
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(9) PiDPToQDPProof (SOUND)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
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(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U2_GAAAA(R, balance_out_gaaaa) -> BALANCE_IN_GAAAA(R)
   BALANCE_IN_GAAAA(tree(L, V, R)) -> U2_GAAAA(R, balance_in_gaaaa(L))
   BALANCE_IN_GAAAA(tree(L, V, R)) -> BALANCE_IN_GAAAA(L)

The TRS R consists of the following rules:

   balance_in_gaaaa(nil) -> balance_out_gaaaa
   balance_in_gaaaa(tree(L, V, R)) -> U2_gaaaa(R, balance_in_gaaaa(L))
   U2_gaaaa(R, balance_out_gaaaa) -> U3_gaaaa(balance_in_gaaaa(R))
   U3_gaaaa(balance_out_gaaaa) -> balance_out_gaaaa

The set Q consists of the following terms:

   balance_in_gaaaa(x0)
   U2_gaaaa(x0, x1)
   U3_gaaaa(x0)

We have to consider all (P,Q,R)-chains.
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(11) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*BALANCE_IN_GAAAA(tree(L, V, R)) -> U2_GAAAA(R, balance_in_gaaaa(L))
The graph contains the following edges 1 > 1


*BALANCE_IN_GAAAA(tree(L, V, R)) -> BALANCE_IN_GAAAA(L)
The graph contains the following edges 1 > 1


*U2_GAAAA(R, balance_out_gaaaa) -> BALANCE_IN_GAAAA(R)
The graph contains the following edges 1 >= 1


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(12)
YES
