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

(0) Obligation:

Clauses:

div(X, 0, Z) :- ','(!, fail).
div(0, Y, Z) :- ','(!, =(Z, 0)).
div(X, Y, s(Z)) :- ','(minus(X, Y, U), div(U, Y, Z)).
minus(0, Y, 0).
minus(X, 0, X).
minus(s(X), s(Y), Z) :- minus(X, Y, Z).
=(X, X).

Query: div(g,g,a)

(1) BuiltinConflictTransformerProof (EQUIVALENT transformation)

Renamed defined predicates conflicting with built-in predicates [PROLOG].

(2) Obligation:

Clauses:

div(X, 0, Z) :- ','(!, fail).
div(0, Y, Z) :- ','(!, user_defined_=(Z, 0)).
div(X, Y, s(Z)) :- ','(minus(X, Y, U), div(U, Y, Z)).
minus(0, Y, 0).
minus(X, 0, X).
minus(s(X), s(Y), Z) :- minus(X, Y, Z).
user_defined_=(X, X).

Query: div(g,g,a)

(3) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: div(T1, T2, T3)\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: div(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | div(T1, T2, T3), SCOPE: 1, CLAUSE: 1 | div(T1, T2, T3), SCOPE: 1, CLAUSE: 2\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(!_1, fail) | div(T6, 0, T3), SCOPE: 1, CLAUSE: 1 | div(T6, 0, T3), SCOPE: 1, CLAUSE: 2\n\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: div(T1, T2, T3), SCOPE: 1, CLAUSE: 1 | div(T1, T2, T3), SCOPE: 1, CLAUSE: 2\n\nT1 is ground\nT2 is ground\nX3 is free\nX4 is free\ndiv(T1, T2, T3) !=? div(X3, 0, X4)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: fail\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(!_1, user_defined_=(T12, 0)) | div(0, T10, T3), SCOPE: 1, CLAUSE: 2\n\nT10 is ground\nX3 is free\nX4 is free\ndiv(0, T10, T3) !=? div(X3, 0, X4)", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: div(T1, T2, T3), SCOPE: 1, CLAUSE: 2\n\nT1 is ground\nT2 is ground\nX3 is free\nX4 is free\nX7 is free\nX8 is free\ndiv(T1, T2, T3) !=? div(X3, 0, X4)\ndiv(T1, T2, T3) !=? div(0, X7, X8)", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: user_defined_=(T12, 0)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: user_defined_=(T12, 0), SCOPE: 2, CLAUSE: 6\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(minus(T19, T20, X18), div(X18, T20, T22))\n\nT19 is ground\nT20 is ground\nX3 is free\nX4 is free\nX7 is free\nX8 is free\nX18 is free\ndiv(T19, T20, T3) !=? div(X3, 0, X4)\ndiv(T19, T20, T3) !=? div(0, X7, X8)", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(minus(T19, T20, X18), div(X18, T20, T22)), SCOPE: 3, CLAUSE: 3 | ','(minus(T19, T20, X18), div(X18, T20, T22)), SCOPE: 3, CLAUSE: 4 | ','(minus(T19, T20, X18), div(X18, T20, T22)), SCOPE: 3, CLAUSE: 5\n\nT19 is ground\nT20 is ground\nX3 is free\nX4 is free\nX7 is free\nX8 is free\nX18 is free\ndiv(T19, T20, T3) !=? div(X3, 0, X4)\ndiv(T19, T20, T3) !=? div(0, X7, X8)", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(minus(T19, T20, X18), div(X18, T20, T22)), SCOPE: 3, CLAUSE: 4 | ','(minus(T19, T20, X18), div(X18, T20, T22)), SCOPE: 3, CLAUSE: 5\n\nT19 is ground\nT20 is ground\nX3 is free\nX4 is free\nX7 is free\nX8 is free\nX18 is free\ndiv(T19, T20, T3) !=? div(X3, 0, X4)\ndiv(T19, T20, T3) !=? div(0, X7, X8)", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(minus(T19, T20, X18), div(X18, T20, T22)), SCOPE: 3, CLAUSE: 5\n\nT19 is ground\nT20 is ground\nX3 is free\nX4 is free\nX7 is free\nX8 is free\nX18 is free\ndiv(T19, T20, T3) !=? div(X3, 0, X4)\ndiv(T19, T20, T3) !=? div(0, X7, X8)", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: ','(minus(T29, T30, X32), div(X32, s(T30), T22))\n\nT29 is ground\nT30 is ground\nX32 is free", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: minus(T29, T30, X32)\n\nT29 is ground\nT30 is ground\nX32 is free", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: div(T33, s(T30), T22)\n\nT30 is ground\nT33 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: minus(T29, T30, X32), SCOPE: 4, CLAUSE: 3 | minus(T29, T30, X32), SCOPE: 4, CLAUSE: 4 | minus(T29, T30, X32), SCOPE: 4, CLAUSE: 5\n\nT29 is ground\nT30 is ground\nX32 is free", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: minus(T29, T30, X32), SCOPE: 4, CLAUSE: 3\n\nT29 is ground\nT30 is ground\nX32 is free", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: minus(T29, T30, X32), SCOPE: 4, CLAUSE: 4 | minus(T29, T30, X32), SCOPE: 4, CLAUSE: 5\n\nT29 is ground\nT30 is ground\nX32 is free", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: minus(T29, T30, X32), SCOPE: 4, CLAUSE: 4\n\nT29 is ground\nT30 is ground\nX32 is free", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: minus(T29, T30, X32), SCOPE: 4, CLAUSE: 5\n\nT29 is ground\nT30 is ground\nX32 is free", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: minus(T50, T51, X58)\n\nT50 is ground\nT51 is ground\nX58 is free", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 36 [arrowhead = none ];
36 -> 2;

37 [label="EVAL with clause\ndiv(X3, 0, X4) :- ','(!_1, fail).\nand substitutionT1 -> T6,\nX3 -> T6,\nT2 -> 0,\nT3 -> T7,\nX4 -> T7", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 37 [arrowhead = none ];
37 -> 3;

38 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 38 [arrowhead = none ];
38 -> 4;

39 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 39 [arrowhead = none ];
39 -> 5;

40 [label="EVAL with clause\ndiv(0, X7, X8) :- ','(!_1, user_defined_=(X8, 0)).\nand substitutionT1 -> 0,\nT2 -> T10,\nX7 -> T10,\nT3 -> T12,\nX8 -> T12,\nT11 -> T12", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 40 [arrowhead = none ];
40 -> 7;

41 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 41 [arrowhead = none ];
41 -> 8;

42 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 42 [arrowhead = none ];
42 -> 6;

43 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 43 [arrowhead = none ];
43 -> 9;

44 [label="EVAL with clause\ndiv(X15, X16, s(X17)) :- ','(minus(X15, X16, X18), div(X18, X16, X17)).\nand substitutionT1 -> T19,\nX15 -> T19,\nT2 -> T20,\nX16 -> T20,\nX17 -> T22,\nT3 -> s(T22),\nT21 -> T22", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 44 [arrowhead = none ];
44 -> 14;

45 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 45 [arrowhead = none ];
45 -> 15;

46 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
9 -> 46 [arrowhead = none ];
46 -> 10;

47 [label="EVAL with clause\nuser_defined_=(X11, X11).\nand substitutionT12 -> 0,\nX11 -> 0,\nT15 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 47 [arrowhead = none ];
47 -> 11;

48 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 48 [arrowhead = none ];
48 -> 12;

49 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 49 [arrowhead = none ];
49 -> 13;

50 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
14 -> 50 [arrowhead = none ];
50 -> 16;

51 [label="BACKTRACK\nfor clause: minus(0, Y, 0)\nwith clash: (div(T19, T20, T3), div(0, X7, X8))", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 51 [arrowhead = none ];
51 -> 17;

52 [label="BACKTRACK\nfor clause: minus(X, 0, X)\nwith clash: (div(T19, T20, T3), div(X3, 0, X4))", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 52 [arrowhead = none ];
52 -> 18;

53 [label="EVAL with clause\nminus(s(X29), s(X30), X31) :- minus(X29, X30, X31).\nand substitutionX29 -> T29,\nT19 -> s(T29),\nX30 -> T30,\nT20 -> s(T30),\nX18 -> X32,\nX31 -> X32", fontsize=14, style = filled, fillcolor = lightblue];
18 -> 53 [arrowhead = none ];
53 -> 19;

54 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
18 -> 54 [arrowhead = none ];
54 -> 20;

55 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
19 -> 55 [arrowhead = none ];
55 -> 21;

56 [label="SPLIT 2\nnew knowledge:\nT29 is ground\nT30 is ground\nT33 is ground\nreplacements:X32 -> T33", fontsize=14, style = filled, fillcolor = violet];
19 -> 56 [arrowhead = none ];
56 -> 22;

57 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
21 -> 57 [arrowhead = none ];
57 -> 23;

58 [label="INSTANCE with matching:\nT1 -> T33\nT2 -> s(T30)\nT3 -> T22", fontsize=14, style = filled, fillcolor = lightgrey];
22 -> 58 [arrowhead = none , style=dashed];
58 -> 1[style=dashed];

59 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
23 -> 59 [arrowhead = none ];
59 -> 24;

60 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
23 -> 60 [arrowhead = none ];
60 -> 25;

61 [label="EVAL with clause\nminus(0, X41, 0).\nand substitutionT29 -> 0,\nT30 -> T40,\nX41 -> T40,\nX32 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
24 -> 61 [arrowhead = none ];
61 -> 26;

62 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
24 -> 62 [arrowhead = none ];
62 -> 27;

63 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
25 -> 63 [arrowhead = none ];
63 -> 29;

64 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
25 -> 64 [arrowhead = none ];
64 -> 30;

65 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
26 -> 65 [arrowhead = none ];
65 -> 28;

66 [label="EVAL with clause\nminus(X46, 0, X46).\nand substitutionT29 -> T45,\nX46 -> T45,\nT30 -> 0,\nX32 -> T45", fontsize=14, style = filled, fillcolor = lightblue];
29 -> 66 [arrowhead = none ];
66 -> 31;

67 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
29 -> 67 [arrowhead = none ];
67 -> 32;

68 [label="EVAL with clause\nminus(s(X55), s(X56), X57) :- minus(X55, X56, X57).\nand substitutionX55 -> T50,\nT29 -> s(T50),\nX56 -> T51,\nT30 -> s(T51),\nX32 -> X58,\nX57 -> X58", fontsize=14, style = filled, fillcolor = lightblue];
30 -> 68 [arrowhead = none ];
68 -> 34;

69 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
30 -> 69 [arrowhead = none ];
69 -> 35;

70 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
31 -> 70 [arrowhead = none ];
70 -> 33;

71 [label="INSTANCE with matching:\nT29 -> T50\nT30 -> T51\nX32 -> X58", fontsize=14, style = filled, fillcolor = lightgrey];
34 -> 71 [arrowhead = none , style=dashed];
71 -> 21[style=dashed];

}

(4) Obligation:

Clauses:

minusA(0, T40, 0).
minusA(T45, 0, T45).
minusA(s(T50), s(T51), X58) :- minusA(T50, T51, X58).
divB(0, T10, 0).
divB(s(T29), s(T30), s(T22)) :- minusA(T29, T30, X32).
divB(s(T29), s(T30), s(T22)) :- ','(minusA(T29, T30, T33), divB(T33, s(T30), T22)).

Query: divB(g,g,a)

(5) PrologToPiTRSProof (SOUND transformation)

We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
divB_in: (b,b,f)
minusA_in: (b,b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

divB_in_gga(0, T10, 0) → divB_out_gga(0, T10, 0)
divB_in_gga(s(T29), s(T30), s(T22)) → U2_gga(T29, T30, T22, minusA_in_gga(T29, T30, X32))
minusA_in_gga(0, T40, 0) → minusA_out_gga(0, T40, 0)
minusA_in_gga(T45, 0, T45) → minusA_out_gga(T45, 0, T45)
minusA_in_gga(s(T50), s(T51), X58) → U1_gga(T50, T51, X58, minusA_in_gga(T50, T51, X58))
U1_gga(T50, T51, X58, minusA_out_gga(T50, T51, X58)) → minusA_out_gga(s(T50), s(T51), X58)
U2_gga(T29, T30, T22, minusA_out_gga(T29, T30, X32)) → divB_out_gga(s(T29), s(T30), s(T22))
divB_in_gga(s(T29), s(T30), s(T22)) → U3_gga(T29, T30, T22, minusA_in_gga(T29, T30, T33))
U3_gga(T29, T30, T22, minusA_out_gga(T29, T30, T33)) → U4_gga(T29, T30, T22, divB_in_gga(T33, s(T30), T22))
U4_gga(T29, T30, T22, divB_out_gga(T33, s(T30), T22)) → divB_out_gga(s(T29), s(T30), s(T22))

The argument filtering Pi contains the following mapping:
divB_in_gga(x1, x2, x3)  =  divB_in_gga(x1, x2)
0  =  0
divB_out_gga(x1, x2, x3)  =  divB_out_gga(x1, x2)
s(x1)  =  s(x1)
U2_gga(x1, x2, x3, x4)  =  U2_gga(x1, x2, x4)
minusA_in_gga(x1, x2, x3)  =  minusA_in_gga(x1, x2)
minusA_out_gga(x1, x2, x3)  =  minusA_out_gga(x1, x2, x3)
U1_gga(x1, x2, x3, x4)  =  U1_gga(x1, x2, x4)
U3_gga(x1, x2, x3, x4)  =  U3_gga(x1, x2, x4)
U4_gga(x1, x2, x3, x4)  =  U4_gga(x1, x2, x4)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(6) Obligation:

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

divB_in_gga(0, T10, 0) → divB_out_gga(0, T10, 0)
divB_in_gga(s(T29), s(T30), s(T22)) → U2_gga(T29, T30, T22, minusA_in_gga(T29, T30, X32))
minusA_in_gga(0, T40, 0) → minusA_out_gga(0, T40, 0)
minusA_in_gga(T45, 0, T45) → minusA_out_gga(T45, 0, T45)
minusA_in_gga(s(T50), s(T51), X58) → U1_gga(T50, T51, X58, minusA_in_gga(T50, T51, X58))
U1_gga(T50, T51, X58, minusA_out_gga(T50, T51, X58)) → minusA_out_gga(s(T50), s(T51), X58)
U2_gga(T29, T30, T22, minusA_out_gga(T29, T30, X32)) → divB_out_gga(s(T29), s(T30), s(T22))
divB_in_gga(s(T29), s(T30), s(T22)) → U3_gga(T29, T30, T22, minusA_in_gga(T29, T30, T33))
U3_gga(T29, T30, T22, minusA_out_gga(T29, T30, T33)) → U4_gga(T29, T30, T22, divB_in_gga(T33, s(T30), T22))
U4_gga(T29, T30, T22, divB_out_gga(T33, s(T30), T22)) → divB_out_gga(s(T29), s(T30), s(T22))

The argument filtering Pi contains the following mapping:
divB_in_gga(x1, x2, x3)  =  divB_in_gga(x1, x2)
0  =  0
divB_out_gga(x1, x2, x3)  =  divB_out_gga(x1, x2)
s(x1)  =  s(x1)
U2_gga(x1, x2, x3, x4)  =  U2_gga(x1, x2, x4)
minusA_in_gga(x1, x2, x3)  =  minusA_in_gga(x1, x2)
minusA_out_gga(x1, x2, x3)  =  minusA_out_gga(x1, x2, x3)
U1_gga(x1, x2, x3, x4)  =  U1_gga(x1, x2, x4)
U3_gga(x1, x2, x3, x4)  =  U3_gga(x1, x2, x4)
U4_gga(x1, x2, x3, x4)  =  U4_gga(x1, x2, x4)

(7) DependencyPairsProof (EQUIVALENT transformation)

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

DIVB_IN_GGA(s(T29), s(T30), s(T22)) → U2_GGA(T29, T30, T22, minusA_in_gga(T29, T30, X32))
DIVB_IN_GGA(s(T29), s(T30), s(T22)) → MINUSA_IN_GGA(T29, T30, X32)
MINUSA_IN_GGA(s(T50), s(T51), X58) → U1_GGA(T50, T51, X58, minusA_in_gga(T50, T51, X58))
MINUSA_IN_GGA(s(T50), s(T51), X58) → MINUSA_IN_GGA(T50, T51, X58)
DIVB_IN_GGA(s(T29), s(T30), s(T22)) → U3_GGA(T29, T30, T22, minusA_in_gga(T29, T30, T33))
U3_GGA(T29, T30, T22, minusA_out_gga(T29, T30, T33)) → U4_GGA(T29, T30, T22, divB_in_gga(T33, s(T30), T22))
U3_GGA(T29, T30, T22, minusA_out_gga(T29, T30, T33)) → DIVB_IN_GGA(T33, s(T30), T22)

The TRS R consists of the following rules:

divB_in_gga(0, T10, 0) → divB_out_gga(0, T10, 0)
divB_in_gga(s(T29), s(T30), s(T22)) → U2_gga(T29, T30, T22, minusA_in_gga(T29, T30, X32))
minusA_in_gga(0, T40, 0) → minusA_out_gga(0, T40, 0)
minusA_in_gga(T45, 0, T45) → minusA_out_gga(T45, 0, T45)
minusA_in_gga(s(T50), s(T51), X58) → U1_gga(T50, T51, X58, minusA_in_gga(T50, T51, X58))
U1_gga(T50, T51, X58, minusA_out_gga(T50, T51, X58)) → minusA_out_gga(s(T50), s(T51), X58)
U2_gga(T29, T30, T22, minusA_out_gga(T29, T30, X32)) → divB_out_gga(s(T29), s(T30), s(T22))
divB_in_gga(s(T29), s(T30), s(T22)) → U3_gga(T29, T30, T22, minusA_in_gga(T29, T30, T33))
U3_gga(T29, T30, T22, minusA_out_gga(T29, T30, T33)) → U4_gga(T29, T30, T22, divB_in_gga(T33, s(T30), T22))
U4_gga(T29, T30, T22, divB_out_gga(T33, s(T30), T22)) → divB_out_gga(s(T29), s(T30), s(T22))

The argument filtering Pi contains the following mapping:
divB_in_gga(x1, x2, x3)  =  divB_in_gga(x1, x2)
0  =  0
divB_out_gga(x1, x2, x3)  =  divB_out_gga(x1, x2)
s(x1)  =  s(x1)
U2_gga(x1, x2, x3, x4)  =  U2_gga(x1, x2, x4)
minusA_in_gga(x1, x2, x3)  =  minusA_in_gga(x1, x2)
minusA_out_gga(x1, x2, x3)  =  minusA_out_gga(x1, x2, x3)
U1_gga(x1, x2, x3, x4)  =  U1_gga(x1, x2, x4)
U3_gga(x1, x2, x3, x4)  =  U3_gga(x1, x2, x4)
U4_gga(x1, x2, x3, x4)  =  U4_gga(x1, x2, x4)
DIVB_IN_GGA(x1, x2, x3)  =  DIVB_IN_GGA(x1, x2)
U2_GGA(x1, x2, x3, x4)  =  U2_GGA(x1, x2, x4)
MINUSA_IN_GGA(x1, x2, x3)  =  MINUSA_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4)  =  U1_GGA(x1, x2, x4)
U3_GGA(x1, x2, x3, x4)  =  U3_GGA(x1, x2, x4)
U4_GGA(x1, x2, x3, x4)  =  U4_GGA(x1, x2, x4)

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

(8) Obligation:

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

DIVB_IN_GGA(s(T29), s(T30), s(T22)) → U2_GGA(T29, T30, T22, minusA_in_gga(T29, T30, X32))
DIVB_IN_GGA(s(T29), s(T30), s(T22)) → MINUSA_IN_GGA(T29, T30, X32)
MINUSA_IN_GGA(s(T50), s(T51), X58) → U1_GGA(T50, T51, X58, minusA_in_gga(T50, T51, X58))
MINUSA_IN_GGA(s(T50), s(T51), X58) → MINUSA_IN_GGA(T50, T51, X58)
DIVB_IN_GGA(s(T29), s(T30), s(T22)) → U3_GGA(T29, T30, T22, minusA_in_gga(T29, T30, T33))
U3_GGA(T29, T30, T22, minusA_out_gga(T29, T30, T33)) → U4_GGA(T29, T30, T22, divB_in_gga(T33, s(T30), T22))
U3_GGA(T29, T30, T22, minusA_out_gga(T29, T30, T33)) → DIVB_IN_GGA(T33, s(T30), T22)

The TRS R consists of the following rules:

divB_in_gga(0, T10, 0) → divB_out_gga(0, T10, 0)
divB_in_gga(s(T29), s(T30), s(T22)) → U2_gga(T29, T30, T22, minusA_in_gga(T29, T30, X32))
minusA_in_gga(0, T40, 0) → minusA_out_gga(0, T40, 0)
minusA_in_gga(T45, 0, T45) → minusA_out_gga(T45, 0, T45)
minusA_in_gga(s(T50), s(T51), X58) → U1_gga(T50, T51, X58, minusA_in_gga(T50, T51, X58))
U1_gga(T50, T51, X58, minusA_out_gga(T50, T51, X58)) → minusA_out_gga(s(T50), s(T51), X58)
U2_gga(T29, T30, T22, minusA_out_gga(T29, T30, X32)) → divB_out_gga(s(T29), s(T30), s(T22))
divB_in_gga(s(T29), s(T30), s(T22)) → U3_gga(T29, T30, T22, minusA_in_gga(T29, T30, T33))
U3_gga(T29, T30, T22, minusA_out_gga(T29, T30, T33)) → U4_gga(T29, T30, T22, divB_in_gga(T33, s(T30), T22))
U4_gga(T29, T30, T22, divB_out_gga(T33, s(T30), T22)) → divB_out_gga(s(T29), s(T30), s(T22))

The argument filtering Pi contains the following mapping:
divB_in_gga(x1, x2, x3)  =  divB_in_gga(x1, x2)
0  =  0
divB_out_gga(x1, x2, x3)  =  divB_out_gga(x1, x2)
s(x1)  =  s(x1)
U2_gga(x1, x2, x3, x4)  =  U2_gga(x1, x2, x4)
minusA_in_gga(x1, x2, x3)  =  minusA_in_gga(x1, x2)
minusA_out_gga(x1, x2, x3)  =  minusA_out_gga(x1, x2, x3)
U1_gga(x1, x2, x3, x4)  =  U1_gga(x1, x2, x4)
U3_gga(x1, x2, x3, x4)  =  U3_gga(x1, x2, x4)
U4_gga(x1, x2, x3, x4)  =  U4_gga(x1, x2, x4)
DIVB_IN_GGA(x1, x2, x3)  =  DIVB_IN_GGA(x1, x2)
U2_GGA(x1, x2, x3, x4)  =  U2_GGA(x1, x2, x4)
MINUSA_IN_GGA(x1, x2, x3)  =  MINUSA_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4)  =  U1_GGA(x1, x2, x4)
U3_GGA(x1, x2, x3, x4)  =  U3_GGA(x1, x2, x4)
U4_GGA(x1, x2, x3, x4)  =  U4_GGA(x1, x2, x4)

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 4 less nodes.

(10) Complex Obligation (AND)

(11) Obligation:

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

MINUSA_IN_GGA(s(T50), s(T51), X58) → MINUSA_IN_GGA(T50, T51, X58)

The TRS R consists of the following rules:

divB_in_gga(0, T10, 0) → divB_out_gga(0, T10, 0)
divB_in_gga(s(T29), s(T30), s(T22)) → U2_gga(T29, T30, T22, minusA_in_gga(T29, T30, X32))
minusA_in_gga(0, T40, 0) → minusA_out_gga(0, T40, 0)
minusA_in_gga(T45, 0, T45) → minusA_out_gga(T45, 0, T45)
minusA_in_gga(s(T50), s(T51), X58) → U1_gga(T50, T51, X58, minusA_in_gga(T50, T51, X58))
U1_gga(T50, T51, X58, minusA_out_gga(T50, T51, X58)) → minusA_out_gga(s(T50), s(T51), X58)
U2_gga(T29, T30, T22, minusA_out_gga(T29, T30, X32)) → divB_out_gga(s(T29), s(T30), s(T22))
divB_in_gga(s(T29), s(T30), s(T22)) → U3_gga(T29, T30, T22, minusA_in_gga(T29, T30, T33))
U3_gga(T29, T30, T22, minusA_out_gga(T29, T30, T33)) → U4_gga(T29, T30, T22, divB_in_gga(T33, s(T30), T22))
U4_gga(T29, T30, T22, divB_out_gga(T33, s(T30), T22)) → divB_out_gga(s(T29), s(T30), s(T22))

The argument filtering Pi contains the following mapping:
divB_in_gga(x1, x2, x3)  =  divB_in_gga(x1, x2)
0  =  0
divB_out_gga(x1, x2, x3)  =  divB_out_gga(x1, x2)
s(x1)  =  s(x1)
U2_gga(x1, x2, x3, x4)  =  U2_gga(x1, x2, x4)
minusA_in_gga(x1, x2, x3)  =  minusA_in_gga(x1, x2)
minusA_out_gga(x1, x2, x3)  =  minusA_out_gga(x1, x2, x3)
U1_gga(x1, x2, x3, x4)  =  U1_gga(x1, x2, x4)
U3_gga(x1, x2, x3, x4)  =  U3_gga(x1, x2, x4)
U4_gga(x1, x2, x3, x4)  =  U4_gga(x1, x2, x4)
MINUSA_IN_GGA(x1, x2, x3)  =  MINUSA_IN_GGA(x1, x2)

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

(12) UsableRulesProof (EQUIVALENT transformation)

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

(13) Obligation:

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

MINUSA_IN_GGA(s(T50), s(T51), X58) → MINUSA_IN_GGA(T50, T51, X58)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
MINUSA_IN_GGA(x1, x2, x3)  =  MINUSA_IN_GGA(x1, x2)

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

(14) PiDPToQDPProof (SOUND transformation)

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

(15) Obligation:

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

MINUSA_IN_GGA(s(T50), s(T51)) → MINUSA_IN_GGA(T50, T51)

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

(16) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

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

• MINUSA_IN_GGA(s(T50), s(T51)) → MINUSA_IN_GGA(T50, T51)
  The graph contains the following edges 1 > 1, 2 > 2

(17) YES

(18) Obligation:

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

DIVB_IN_GGA(s(T29), s(T30), s(T22)) → U3_GGA(T29, T30, T22, minusA_in_gga(T29, T30, T33))
U3_GGA(T29, T30, T22, minusA_out_gga(T29, T30, T33)) → DIVB_IN_GGA(T33, s(T30), T22)

The TRS R consists of the following rules:

divB_in_gga(0, T10, 0) → divB_out_gga(0, T10, 0)
divB_in_gga(s(T29), s(T30), s(T22)) → U2_gga(T29, T30, T22, minusA_in_gga(T29, T30, X32))
minusA_in_gga(0, T40, 0) → minusA_out_gga(0, T40, 0)
minusA_in_gga(T45, 0, T45) → minusA_out_gga(T45, 0, T45)
minusA_in_gga(s(T50), s(T51), X58) → U1_gga(T50, T51, X58, minusA_in_gga(T50, T51, X58))
U1_gga(T50, T51, X58, minusA_out_gga(T50, T51, X58)) → minusA_out_gga(s(T50), s(T51), X58)
U2_gga(T29, T30, T22, minusA_out_gga(T29, T30, X32)) → divB_out_gga(s(T29), s(T30), s(T22))
divB_in_gga(s(T29), s(T30), s(T22)) → U3_gga(T29, T30, T22, minusA_in_gga(T29, T30, T33))
U3_gga(T29, T30, T22, minusA_out_gga(T29, T30, T33)) → U4_gga(T29, T30, T22, divB_in_gga(T33, s(T30), T22))
U4_gga(T29, T30, T22, divB_out_gga(T33, s(T30), T22)) → divB_out_gga(s(T29), s(T30), s(T22))

The argument filtering Pi contains the following mapping:
divB_in_gga(x1, x2, x3)  =  divB_in_gga(x1, x2)
0  =  0
divB_out_gga(x1, x2, x3)  =  divB_out_gga(x1, x2)
s(x1)  =  s(x1)
U2_gga(x1, x2, x3, x4)  =  U2_gga(x1, x2, x4)
minusA_in_gga(x1, x2, x3)  =  minusA_in_gga(x1, x2)
minusA_out_gga(x1, x2, x3)  =  minusA_out_gga(x1, x2, x3)
U1_gga(x1, x2, x3, x4)  =  U1_gga(x1, x2, x4)
U3_gga(x1, x2, x3, x4)  =  U3_gga(x1, x2, x4)
U4_gga(x1, x2, x3, x4)  =  U4_gga(x1, x2, x4)
DIVB_IN_GGA(x1, x2, x3)  =  DIVB_IN_GGA(x1, x2)
U3_GGA(x1, x2, x3, x4)  =  U3_GGA(x1, x2, x4)

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

(19) UsableRulesProof (EQUIVALENT transformation)

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

(20) Obligation:

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

DIVB_IN_GGA(s(T29), s(T30), s(T22)) → U3_GGA(T29, T30, T22, minusA_in_gga(T29, T30, T33))
U3_GGA(T29, T30, T22, minusA_out_gga(T29, T30, T33)) → DIVB_IN_GGA(T33, s(T30), T22)

The TRS R consists of the following rules:

minusA_in_gga(0, T40, 0) → minusA_out_gga(0, T40, 0)
minusA_in_gga(T45, 0, T45) → minusA_out_gga(T45, 0, T45)
minusA_in_gga(s(T50), s(T51), X58) → U1_gga(T50, T51, X58, minusA_in_gga(T50, T51, X58))
U1_gga(T50, T51, X58, minusA_out_gga(T50, T51, X58)) → minusA_out_gga(s(T50), s(T51), X58)

The argument filtering Pi contains the following mapping:
0  =  0
s(x1)  =  s(x1)
minusA_in_gga(x1, x2, x3)  =  minusA_in_gga(x1, x2)
minusA_out_gga(x1, x2, x3)  =  minusA_out_gga(x1, x2, x3)
U1_gga(x1, x2, x3, x4)  =  U1_gga(x1, x2, x4)
DIVB_IN_GGA(x1, x2, x3)  =  DIVB_IN_GGA(x1, x2)
U3_GGA(x1, x2, x3, x4)  =  U3_GGA(x1, x2, x4)

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

(21) PiDPToQDPProof (SOUND transformation)

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

(22) Obligation:

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

DIVB_IN_GGA(s(T29), s(T30)) → U3_GGA(T29, T30, minusA_in_gga(T29, T30))
U3_GGA(T29, T30, minusA_out_gga(T29, T30, T33)) → DIVB_IN_GGA(T33, s(T30))

The TRS R consists of the following rules:

minusA_in_gga(0, T40) → minusA_out_gga(0, T40, 0)
minusA_in_gga(T45, 0) → minusA_out_gga(T45, 0, T45)
minusA_in_gga(s(T50), s(T51)) → U1_gga(T50, T51, minusA_in_gga(T50, T51))
U1_gga(T50, T51, minusA_out_gga(T50, T51, X58)) → minusA_out_gga(s(T50), s(T51), X58)

The set Q consists of the following terms:

minusA_in_gga(x0, x1)
U1_gga(x0, x1, x2)

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

DIVB_IN_GGA(s(T29), s(T30)) → U3_GGA(T29, T30, minusA_in_gga(T29, T30))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(DIVB_IN_GGA(x₁, x₂)) = x₁ + x₂   
POL(U1_gga(x₁, x₂, x₃)) = 1 + x₃   
POL(U3_GGA(x₁, x₂, x₃)) = x₂ + x₃   
POL(minusA_in_gga(x₁, x₂)) = 1 + x₁   
POL(minusA_out_gga(x₁, x₂, x₃)) = 1 + x₃   
POL(s(x₁)) = 1 + x₁   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

minusA_in_gga(0, T40) → minusA_out_gga(0, T40, 0)
minusA_in_gga(T45, 0) → minusA_out_gga(T45, 0, T45)
minusA_in_gga(s(T50), s(T51)) → U1_gga(T50, T51, minusA_in_gga(T50, T51))
U1_gga(T50, T51, minusA_out_gga(T50, T51, X58)) → minusA_out_gga(s(T50), s(T51), X58)

(24) Obligation:

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

U3_GGA(T29, T30, minusA_out_gga(T29, T30, T33)) → DIVB_IN_GGA(T33, s(T30))

The TRS R consists of the following rules:

minusA_in_gga(0, T40) → minusA_out_gga(0, T40, 0)
minusA_in_gga(T45, 0) → minusA_out_gga(T45, 0, T45)
minusA_in_gga(s(T50), s(T51)) → U1_gga(T50, T51, minusA_in_gga(T50, T51))
U1_gga(T50, T51, minusA_out_gga(T50, T51, X58)) → minusA_out_gga(s(T50), s(T51), X58)

The set Q consists of the following terms:

minusA_in_gga(x0, x1)
U1_gga(x0, x1, x2)

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

(25) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(26) TRUE