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

(0) Obligation:

Clauses:

q(X, Y) :- m(X, Y, Z).
m(X, 0, X).
m(0, Y, 0) :- !.
m(X, Y, Z) :- ','(p(X, A), ','(p(Y, B), m(A, B, Z))).
p(0, 0).
p(s(X), X).

Query: q(g,g)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: q(T1, T2)\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: q(T1, T2), SCOPE: 1, CLAUSE: 0\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: m(T5, T6, X5)\n\nT5 is ground\nT6 is ground\nX5 is free", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: m(T5, T6, X5), SCOPE: 2, CLAUSE: 1 | m(T5, T6, X5), SCOPE: 2, CLAUSE: 2 | m(T5, T6, X5), SCOPE: 2, CLAUSE: 3\n\nT5 is ground\nT6 is ground\nX5 is free", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: true | m(T9, 0, X5), SCOPE: 2, CLAUSE: 2 | m(T9, 0, X5), SCOPE: 2, CLAUSE: 3\n\nT9 is ground\nX5 is free", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: m(T5, T6, X5), SCOPE: 2, CLAUSE: 2 | m(T5, T6, X5), SCOPE: 2, CLAUSE: 3\n\nT5 is ground\nT6 is ground\nX5 is free\nX8 is free\nm(T5, T6, X5) !=? m(X8, 0, X8)", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: m(T9, 0, X5), SCOPE: 2, CLAUSE: 2 | m(T9, 0, X5), SCOPE: 2, CLAUSE: 3\n\nT9 is ground\nX5 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: !_2 | m(0, 0, X5), SCOPE: 2, CLAUSE: 3\n\nX5 is free", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: m(T9, 0, X5), SCOPE: 2, CLAUSE: 3\n\nT9 is ground\nX5 is free\nX11 is free\nm(T9, 0, X5) !=? m(0, X11, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(p(T12, X27), ','(p(0, X28), m(X27, X28, X29)))\n\nT12 is ground\nX5 is free\nX11 is free\nX29 is free\nX27 is free\nX28 is free\nm(T12, 0, X5) !=? m(0, X11, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(p(T12, X27), ','(p(0, X28), m(X27, X28, X29))), SCOPE: 3, CLAUSE: 4 | ','(p(T12, X27), ','(p(0, X28), m(X27, X28, X29))), SCOPE: 3, CLAUSE: 5\n\nT12 is ground\nX5 is free\nX11 is free\nX29 is free\nX27 is free\nX28 is free\nm(T12, 0, X5) !=? m(0, X11, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(p(T12, X27), ','(p(0, X28), m(X27, X28, X29))), SCOPE: 3, CLAUSE: 5\n\nT12 is ground\nX5 is free\nX11 is free\nX29 is free\nX27 is free\nX28 is free\nm(T12, 0, X5) !=? m(0, X11, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(p(0, X28), m(T15, X28, X29))\n\nT15 is ground\nX29 is free\nX28 is free", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(p(0, X28), m(T15, X28, X29)), SCOPE: 4, CLAUSE: 4 | ','(p(0, X28), m(T15, X28, X29)), SCOPE: 4, CLAUSE: 5\n\nT15 is ground\nX29 is free\nX28 is free", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(p(0, X28), m(T15, X28, X29)), SCOPE: 4, CLAUSE: 4\n\nT15 is ground\nX29 is free\nX28 is free", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: ','(p(0, X28), m(T15, X28, X29)), SCOPE: 4, CLAUSE: 5\n\nT15 is ground\nX29 is free\nX28 is free", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: m(T15, 0, X29)\n\nT15 is ground\nX29 is free", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: m(T15, 0, X29), SCOPE: 5, CLAUSE: 1 | m(T15, 0, X29), SCOPE: 5, CLAUSE: 2 | m(T15, 0, X29), SCOPE: 5, CLAUSE: 3\n\nT15 is ground\nX29 is free", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: true | m(T22, 0, X29), SCOPE: 5, CLAUSE: 2 | m(T22, 0, X29), SCOPE: 5, CLAUSE: 3\n\nT22 is ground\nX29 is free", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: m(T22, 0, X29), SCOPE: 5, CLAUSE: 2 | m(T22, 0, X29), SCOPE: 5, CLAUSE: 3\n\nT22 is ground\nX29 is free", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: !_5 | m(0, 0, X29), SCOPE: 5, CLAUSE: 3\n\nX29 is free", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: m(T22, 0, X29), SCOPE: 5, CLAUSE: 3\n\nT22 is ground\nX29 is free\nX44 is free\nm(T22, 0, X29) !=? m(0, X44, 0)", 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: ','(p(T25, X60), ','(p(0, X61), m(X60, X61, X62)))\n\nT25 is ground\nX29 is free\nX44 is free\nX62 is free\nX60 is free\nX61 is free\nm(T25, 0, X29) !=? m(0, X44, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: ','(p(T25, X60), ','(p(0, X61), m(X60, X61, X62))), SCOPE: 6, CLAUSE: 4 | ','(p(T25, X60), ','(p(0, X61), m(X60, X61, X62))), SCOPE: 6, CLAUSE: 5\n\nT25 is ground\nX29 is free\nX44 is free\nX62 is free\nX60 is free\nX61 is free\nm(T25, 0, X29) !=? m(0, X44, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: ','(p(T25, X60), ','(p(0, X61), m(X60, X61, X62))), SCOPE: 6, CLAUSE: 5\n\nT25 is ground\nX29 is free\nX44 is free\nX62 is free\nX60 is free\nX61 is free\nm(T25, 0, X29) !=? m(0, X44, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: ','(p(0, X61), m(T28, X61, X62))\n\nT28 is ground\nX62 is free\nX61 is free", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: ','(p(0, X61), m(T28, X61, X62)), SCOPE: 7, CLAUSE: 4 | ','(p(0, X61), m(T28, X61, X62)), SCOPE: 7, CLAUSE: 5\n\nT28 is ground\nX62 is free\nX61 is free", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: ','(p(0, X61), m(T28, X61, X62)), SCOPE: 7, CLAUSE: 4\n\nT28 is ground\nX62 is free\nX61 is free", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: ','(p(0, X61), m(T28, X61, X62)), SCOPE: 7, CLAUSE: 5\n\nT28 is ground\nX62 is free\nX61 is free", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="36: m(T28, 0, X62)\n\nT28 is ground\nX62 is free", fontsize=16, style=filled, fillcolor=lightyellow];
37 [label="37: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
38 [label="38: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
39 [label="39: !_2 | m(0, T35, X5), SCOPE: 2, CLAUSE: 3\n\nT35 is ground\nX5 is free\nX8 is free\nm(0, T35, X5) !=? m(X8, 0, X8)", fontsize=16, style=filled, fillcolor=lightyellow];
40 [label="40: m(T5, T6, X5), SCOPE: 2, CLAUSE: 3\n\nT5 is ground\nT6 is ground\nX5 is free\nX8 is free\nX76 is free\nm(T5, T6, X5) !=? m(X8, 0, X8)\nm(T5, T6, X5) !=? m(0, X76, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
41 [label="41: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
42 [label="42: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
43 [label="43: ','(p(T40, X92), ','(p(T41, X93), m(X92, X93, X94)))\n\nT40 is ground\nT41 is ground\nX5 is free\nX8 is free\nX76 is free\nX94 is free\nX92 is free\nX93 is free\nm(T40, T41, X5) !=? m(X8, 0, X8)\nm(T40, T41, X5) !=? m(0, X76, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
44 [label="44: ','(p(T40, X92), ','(p(T41, X93), m(X92, X93, X94))), SCOPE: 8, CLAUSE: 4 | ','(p(T40, X92), ','(p(T41, X93), m(X92, X93, X94))), SCOPE: 8, CLAUSE: 5\n\nT40 is ground\nT41 is ground\nX5 is free\nX8 is free\nX76 is free\nX94 is free\nX92 is free\nX93 is free\nm(T40, T41, X5) !=? m(X8, 0, X8)\nm(T40, T41, X5) !=? m(0, X76, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
45 [label="45: ','(p(T40, X92), ','(p(T41, X93), m(X92, X93, X94))), SCOPE: 8, CLAUSE: 5\n\nT40 is ground\nT41 is ground\nX5 is free\nX8 is free\nX76 is free\nX94 is free\nX92 is free\nX93 is free\nm(T40, T41, X5) !=? m(X8, 0, X8)\nm(T40, T41, X5) !=? m(0, X76, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
46 [label="46: ','(p(T41, X93), m(T44, X93, X94))\n\nT41 is ground\nT44 is ground\nX5 is free\nX8 is free\nX94 is free\nX93 is free\nm(s(T44), T41, X5) !=? m(X8, 0, X8)", fontsize=16, style=filled, fillcolor=lightyellow];
47 [label="47: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
48 [label="48: ','(p(T41, X93), m(T44, X93, X94)), SCOPE: 9, CLAUSE: 4 | ','(p(T41, X93), m(T44, X93, X94)), SCOPE: 9, CLAUSE: 5\n\nT41 is ground\nT44 is ground\nX5 is free\nX8 is free\nX94 is free\nX93 is free\nm(s(T44), T41, X5) !=? m(X8, 0, X8)", fontsize=16, style=filled, fillcolor=lightyellow];
49 [label="49: ','(p(T41, X93), m(T44, X93, X94)), SCOPE: 9, CLAUSE: 5\n\nT41 is ground\nT44 is ground\nX5 is free\nX8 is free\nX94 is free\nX93 is free\nm(s(T44), T41, X5) !=? m(X8, 0, X8)", fontsize=16, style=filled, fillcolor=lightyellow];
50 [label="50: m(T44, T48, X94)\n\nT44 is ground\nT48 is ground\nX94 is free", fontsize=16, style=filled, fillcolor=lightyellow];
51 [label="51: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
52 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 52 [arrowhead = none ];
52 -> 2;

53 [label="ONLY EVAL with clause\nq(X3, X4) :- m(X3, X4, X5).\nand substitutionT1 -> T5,\nX3 -> T5,\nT2 -> T6,\nX4 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 53 [arrowhead = none ];
53 -> 3;

54 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 54 [arrowhead = none ];
54 -> 4;

55 [label="EVAL with clause\nm(X8, 0, X8).\nand substitutionT5 -> T9,\nX8 -> T9,\nT6 -> 0,\nX5 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 55 [arrowhead = none ];
55 -> 5;

56 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 56 [arrowhead = none ];
56 -> 6;

57 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 57 [arrowhead = none ];
57 -> 7;

58 [label="EVAL with clause\nm(0, X76, 0) :- !_2.\nand substitutionT5 -> 0,\nT6 -> T35,\nX76 -> T35,\nX5 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 58 [arrowhead = none ];
58 -> 39;

59 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 59 [arrowhead = none ];
59 -> 40;

60 [label="EVAL with clause\nm(0, X11, 0) :- !_2.\nand substitutionT9 -> 0,\nX11 -> 0,\nX5 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
7 -> 60 [arrowhead = none ];
60 -> 8;

61 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 61 [arrowhead = none ];
61 -> 9;

62 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 62 [arrowhead = none ];
62 -> 10;

63 [label="ONLY EVAL with clause\nm(X24, X25, X26) :- ','(p(X24, X27), ','(p(X25, X28), m(X27, X28, X26))).\nand substitutionT9 -> T12,\nX24 -> T12,\nX25 -> 0,\nX5 -> X29,\nX26 -> X29", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 63 [arrowhead = none ];
63 -> 12;

64 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 64 [arrowhead = none ];
64 -> 11;

65 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 65 [arrowhead = none ];
65 -> 13;

66 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (m(T12, 0, X5), m(0, X11, 0))", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 66 [arrowhead = none ];
66 -> 14;

67 [label="EVAL with clause\np(s(X34), X34).\nand substitutionX34 -> T15,\nT12 -> s(T15),\nX27 -> T15", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 67 [arrowhead = none ];
67 -> 15;

68 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 68 [arrowhead = none ];
68 -> 16;

69 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
15 -> 69 [arrowhead = none ];
69 -> 17;

70 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
17 -> 70 [arrowhead = none ];
70 -> 18;

71 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
17 -> 71 [arrowhead = none ];
71 -> 19;

72 [label="ONLY EVAL with clause\np(0, 0).\nand substitutionX28 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
18 -> 72 [arrowhead = none ];
72 -> 20;

73 [label="BACKTRACK\nfor clause: p(s(X), X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
19 -> 73 [arrowhead = none ];
73 -> 38;

74 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
20 -> 74 [arrowhead = none ];
74 -> 21;

75 [label="ONLY EVAL with clause\nm(X41, 0, X41).\nand substitutionT15 -> T22,\nX41 -> T22,\nX29 -> T22", fontsize=14, style = filled, fillcolor = lightblue];
21 -> 75 [arrowhead = none ];
75 -> 22;

76 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
22 -> 76 [arrowhead = none ];
76 -> 23;

77 [label="EVAL with clause\nm(0, X44, 0) :- !_5.\nand substitutionT22 -> 0,\nX44 -> 0,\nX29 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
23 -> 77 [arrowhead = none ];
77 -> 24;

78 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
23 -> 78 [arrowhead = none ];
78 -> 25;

79 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
24 -> 79 [arrowhead = none ];
79 -> 26;

80 [label="ONLY EVAL with clause\nm(X57, X58, X59) :- ','(p(X57, X60), ','(p(X58, X61), m(X60, X61, X59))).\nand substitutionT22 -> T25,\nX57 -> T25,\nX58 -> 0,\nX29 -> X62,\nX59 -> X62", fontsize=14, style = filled, fillcolor = lightblue];
25 -> 80 [arrowhead = none ];
80 -> 28;

81 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
26 -> 81 [arrowhead = none ];
81 -> 27;

82 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
28 -> 82 [arrowhead = none ];
82 -> 29;

83 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (m(T25, 0, X29), m(0, X44, 0))", fontsize=14, style = filled, fillcolor = lightgreen];
29 -> 83 [arrowhead = none ];
83 -> 30;

84 [label="EVAL with clause\np(s(X67), X67).\nand substitutionX67 -> T28,\nT25 -> s(T28),\nX60 -> T28", fontsize=14, style = filled, fillcolor = lightblue];
30 -> 84 [arrowhead = none ];
84 -> 31;

85 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
30 -> 85 [arrowhead = none ];
85 -> 32;

86 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
31 -> 86 [arrowhead = none ];
86 -> 33;

87 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
33 -> 87 [arrowhead = none ];
87 -> 34;

88 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
33 -> 88 [arrowhead = none ];
88 -> 35;

89 [label="ONLY EVAL with clause\np(0, 0).\nand substitutionX61 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
34 -> 89 [arrowhead = none ];
89 -> 36;

90 [label="BACKTRACK\nfor clause: p(s(X), X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
35 -> 90 [arrowhead = none ];
90 -> 37;

91 [label="INSTANCE with matching:\nT15 -> T28\nX29 -> X62", fontsize=14, style = filled, fillcolor = lightgrey];
36 -> 91 [arrowhead = none , style=dashed];
91 -> 20[style=dashed];

92 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
39 -> 92 [arrowhead = none ];
92 -> 41;

93 [label="ONLY EVAL with clause\nm(X89, X90, X91) :- ','(p(X89, X92), ','(p(X90, X93), m(X92, X93, X91))).\nand substitutionT5 -> T40,\nX89 -> T40,\nT6 -> T41,\nX90 -> T41,\nX5 -> X94,\nX91 -> X94", fontsize=14, style = filled, fillcolor = lightblue];
40 -> 93 [arrowhead = none ];
93 -> 43;

94 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
41 -> 94 [arrowhead = none ];
94 -> 42;

95 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
43 -> 95 [arrowhead = none ];
95 -> 44;

96 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (m(T40, T41, X5), m(0, X76, 0))", fontsize=14, style = filled, fillcolor = lightgreen];
44 -> 96 [arrowhead = none ];
96 -> 45;

97 [label="EVAL with clause\np(s(X99), X99).\nand substitutionX99 -> T44,\nT40 -> s(T44),\nX92 -> T44", fontsize=14, style = filled, fillcolor = lightblue];
45 -> 97 [arrowhead = none ];
97 -> 46;

98 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
45 -> 98 [arrowhead = none ];
98 -> 47;

99 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
46 -> 99 [arrowhead = none ];
99 -> 48;

100 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (m(s(T44), T41, X5), m(X8, 0, X8))", fontsize=14, style = filled, fillcolor = lightgreen];
48 -> 100 [arrowhead = none ];
100 -> 49;

101 [label="EVAL with clause\np(s(X103), X103).\nand substitutionX103 -> T48,\nT41 -> s(T48),\nX93 -> T48", fontsize=14, style = filled, fillcolor = lightblue];
49 -> 101 [arrowhead = none ];
101 -> 50;

102 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
49 -> 102 [arrowhead = none ];
102 -> 51;

103 [label="INSTANCE with matching:\nT5 -> T44\nT6 -> T48\nX5 -> X94", fontsize=14, style = filled, fillcolor = lightgrey];
50 -> 103 [arrowhead = none , style=dashed];
103 -> 3[style=dashed];

}

(2) Obligation:

Clauses:

mA(T22, T22).
mA(0, 0).
mA(s(T28), X62) :- mA(T28, X62).
mB(T9, 0, T9).
mB(0, 0, 0).
mB(s(T15), 0, X29) :- mA(T15, X29).
mB(0, T35, 0).
mB(s(T44), s(T48), X94) :- mB(T44, T48, X94).
qC(T5, T6) :- mB(T5, T6, X5).

Query: qC(g,g)

(3) PrologToPiTRSProof (SOUND transformation)

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

qC_in_gg(T5, T6) → U4_gg(T5, T6, mB_in_gga(T5, T6, X5))
mB_in_gga(T9, 0, T9) → mB_out_gga(T9, 0, T9)
mB_in_gga(0, 0, 0) → mB_out_gga(0, 0, 0)
mB_in_gga(s(T15), 0, X29) → U2_gga(T15, X29, mA_in_ga(T15, X29))
mA_in_ga(T22, T22) → mA_out_ga(T22, T22)
mA_in_ga(0, 0) → mA_out_ga(0, 0)
mA_in_ga(s(T28), X62) → U1_ga(T28, X62, mA_in_ga(T28, X62))
U1_ga(T28, X62, mA_out_ga(T28, X62)) → mA_out_ga(s(T28), X62)
U2_gga(T15, X29, mA_out_ga(T15, X29)) → mB_out_gga(s(T15), 0, X29)
mB_in_gga(0, T35, 0) → mB_out_gga(0, T35, 0)
mB_in_gga(s(T44), s(T48), X94) → U3_gga(T44, T48, X94, mB_in_gga(T44, T48, X94))
U3_gga(T44, T48, X94, mB_out_gga(T44, T48, X94)) → mB_out_gga(s(T44), s(T48), X94)
U4_gg(T5, T6, mB_out_gga(T5, T6, X5)) → qC_out_gg(T5, T6)

The argument filtering Pi contains the following mapping:
qC_in_gg(x1, x2)  =  qC_in_gg(x1, x2)
U4_gg(x1, x2, x3)  =  U4_gg(x3)
mB_in_gga(x1, x2, x3)  =  mB_in_gga(x1, x2)
0  =  0
mB_out_gga(x1, x2, x3)  =  mB_out_gga(x3)
s(x1)  =  s(x1)
U2_gga(x1, x2, x3)  =  U2_gga(x3)
mA_in_ga(x1, x2)  =  mA_in_ga(x1)
mA_out_ga(x1, x2)  =  mA_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U3_gga(x1, x2, x3, x4)  =  U3_gga(x4)
qC_out_gg(x1, x2)  =  qC_out_gg

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

qC_in_gg(T5, T6) → U4_gg(T5, T6, mB_in_gga(T5, T6, X5))
mB_in_gga(T9, 0, T9) → mB_out_gga(T9, 0, T9)
mB_in_gga(0, 0, 0) → mB_out_gga(0, 0, 0)
mB_in_gga(s(T15), 0, X29) → U2_gga(T15, X29, mA_in_ga(T15, X29))
mA_in_ga(T22, T22) → mA_out_ga(T22, T22)
mA_in_ga(0, 0) → mA_out_ga(0, 0)
mA_in_ga(s(T28), X62) → U1_ga(T28, X62, mA_in_ga(T28, X62))
U1_ga(T28, X62, mA_out_ga(T28, X62)) → mA_out_ga(s(T28), X62)
U2_gga(T15, X29, mA_out_ga(T15, X29)) → mB_out_gga(s(T15), 0, X29)
mB_in_gga(0, T35, 0) → mB_out_gga(0, T35, 0)
mB_in_gga(s(T44), s(T48), X94) → U3_gga(T44, T48, X94, mB_in_gga(T44, T48, X94))
U3_gga(T44, T48, X94, mB_out_gga(T44, T48, X94)) → mB_out_gga(s(T44), s(T48), X94)
U4_gg(T5, T6, mB_out_gga(T5, T6, X5)) → qC_out_gg(T5, T6)

The argument filtering Pi contains the following mapping:
qC_in_gg(x1, x2)  =  qC_in_gg(x1, x2)
U4_gg(x1, x2, x3)  =  U4_gg(x3)
mB_in_gga(x1, x2, x3)  =  mB_in_gga(x1, x2)
0  =  0
mB_out_gga(x1, x2, x3)  =  mB_out_gga(x3)
s(x1)  =  s(x1)
U2_gga(x1, x2, x3)  =  U2_gga(x3)
mA_in_ga(x1, x2)  =  mA_in_ga(x1)
mA_out_ga(x1, x2)  =  mA_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U3_gga(x1, x2, x3, x4)  =  U3_gga(x4)
qC_out_gg(x1, x2)  =  qC_out_gg

(5) 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:

QC_IN_GG(T5, T6) → U4_GG(T5, T6, mB_in_gga(T5, T6, X5))
QC_IN_GG(T5, T6) → MB_IN_GGA(T5, T6, X5)
MB_IN_GGA(s(T15), 0, X29) → U2_GGA(T15, X29, mA_in_ga(T15, X29))
MB_IN_GGA(s(T15), 0, X29) → MA_IN_GA(T15, X29)
MA_IN_GA(s(T28), X62) → U1_GA(T28, X62, mA_in_ga(T28, X62))
MA_IN_GA(s(T28), X62) → MA_IN_GA(T28, X62)
MB_IN_GGA(s(T44), s(T48), X94) → U3_GGA(T44, T48, X94, mB_in_gga(T44, T48, X94))
MB_IN_GGA(s(T44), s(T48), X94) → MB_IN_GGA(T44, T48, X94)

The TRS R consists of the following rules:

qC_in_gg(T5, T6) → U4_gg(T5, T6, mB_in_gga(T5, T6, X5))
mB_in_gga(T9, 0, T9) → mB_out_gga(T9, 0, T9)
mB_in_gga(0, 0, 0) → mB_out_gga(0, 0, 0)
mB_in_gga(s(T15), 0, X29) → U2_gga(T15, X29, mA_in_ga(T15, X29))
mA_in_ga(T22, T22) → mA_out_ga(T22, T22)
mA_in_ga(0, 0) → mA_out_ga(0, 0)
mA_in_ga(s(T28), X62) → U1_ga(T28, X62, mA_in_ga(T28, X62))
U1_ga(T28, X62, mA_out_ga(T28, X62)) → mA_out_ga(s(T28), X62)
U2_gga(T15, X29, mA_out_ga(T15, X29)) → mB_out_gga(s(T15), 0, X29)
mB_in_gga(0, T35, 0) → mB_out_gga(0, T35, 0)
mB_in_gga(s(T44), s(T48), X94) → U3_gga(T44, T48, X94, mB_in_gga(T44, T48, X94))
U3_gga(T44, T48, X94, mB_out_gga(T44, T48, X94)) → mB_out_gga(s(T44), s(T48), X94)
U4_gg(T5, T6, mB_out_gga(T5, T6, X5)) → qC_out_gg(T5, T6)

The argument filtering Pi contains the following mapping:
qC_in_gg(x1, x2)  =  qC_in_gg(x1, x2)
U4_gg(x1, x2, x3)  =  U4_gg(x3)
mB_in_gga(x1, x2, x3)  =  mB_in_gga(x1, x2)
0  =  0
mB_out_gga(x1, x2, x3)  =  mB_out_gga(x3)
s(x1)  =  s(x1)
U2_gga(x1, x2, x3)  =  U2_gga(x3)
mA_in_ga(x1, x2)  =  mA_in_ga(x1)
mA_out_ga(x1, x2)  =  mA_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U3_gga(x1, x2, x3, x4)  =  U3_gga(x4)
qC_out_gg(x1, x2)  =  qC_out_gg
QC_IN_GG(x1, x2)  =  QC_IN_GG(x1, x2)
U4_GG(x1, x2, x3)  =  U4_GG(x3)
MB_IN_GGA(x1, x2, x3)  =  MB_IN_GGA(x1, x2)
U2_GGA(x1, x2, x3)  =  U2_GGA(x3)
MA_IN_GA(x1, x2)  =  MA_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
U3_GGA(x1, x2, x3, x4)  =  U3_GGA(x4)

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

(6) Obligation:

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

QC_IN_GG(T5, T6) → U4_GG(T5, T6, mB_in_gga(T5, T6, X5))
QC_IN_GG(T5, T6) → MB_IN_GGA(T5, T6, X5)
MB_IN_GGA(s(T15), 0, X29) → U2_GGA(T15, X29, mA_in_ga(T15, X29))
MB_IN_GGA(s(T15), 0, X29) → MA_IN_GA(T15, X29)
MA_IN_GA(s(T28), X62) → U1_GA(T28, X62, mA_in_ga(T28, X62))
MA_IN_GA(s(T28), X62) → MA_IN_GA(T28, X62)
MB_IN_GGA(s(T44), s(T48), X94) → U3_GGA(T44, T48, X94, mB_in_gga(T44, T48, X94))
MB_IN_GGA(s(T44), s(T48), X94) → MB_IN_GGA(T44, T48, X94)

The TRS R consists of the following rules:

qC_in_gg(T5, T6) → U4_gg(T5, T6, mB_in_gga(T5, T6, X5))
mB_in_gga(T9, 0, T9) → mB_out_gga(T9, 0, T9)
mB_in_gga(0, 0, 0) → mB_out_gga(0, 0, 0)
mB_in_gga(s(T15), 0, X29) → U2_gga(T15, X29, mA_in_ga(T15, X29))
mA_in_ga(T22, T22) → mA_out_ga(T22, T22)
mA_in_ga(0, 0) → mA_out_ga(0, 0)
mA_in_ga(s(T28), X62) → U1_ga(T28, X62, mA_in_ga(T28, X62))
U1_ga(T28, X62, mA_out_ga(T28, X62)) → mA_out_ga(s(T28), X62)
U2_gga(T15, X29, mA_out_ga(T15, X29)) → mB_out_gga(s(T15), 0, X29)
mB_in_gga(0, T35, 0) → mB_out_gga(0, T35, 0)
mB_in_gga(s(T44), s(T48), X94) → U3_gga(T44, T48, X94, mB_in_gga(T44, T48, X94))
U3_gga(T44, T48, X94, mB_out_gga(T44, T48, X94)) → mB_out_gga(s(T44), s(T48), X94)
U4_gg(T5, T6, mB_out_gga(T5, T6, X5)) → qC_out_gg(T5, T6)

The argument filtering Pi contains the following mapping:
qC_in_gg(x1, x2)  =  qC_in_gg(x1, x2)
U4_gg(x1, x2, x3)  =  U4_gg(x3)
mB_in_gga(x1, x2, x3)  =  mB_in_gga(x1, x2)
0  =  0
mB_out_gga(x1, x2, x3)  =  mB_out_gga(x3)
s(x1)  =  s(x1)
U2_gga(x1, x2, x3)  =  U2_gga(x3)
mA_in_ga(x1, x2)  =  mA_in_ga(x1)
mA_out_ga(x1, x2)  =  mA_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U3_gga(x1, x2, x3, x4)  =  U3_gga(x4)
qC_out_gg(x1, x2)  =  qC_out_gg
QC_IN_GG(x1, x2)  =  QC_IN_GG(x1, x2)
U4_GG(x1, x2, x3)  =  U4_GG(x3)
MB_IN_GGA(x1, x2, x3)  =  MB_IN_GGA(x1, x2)
U2_GGA(x1, x2, x3)  =  U2_GGA(x3)
MA_IN_GA(x1, x2)  =  MA_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
U3_GGA(x1, x2, x3, x4)  =  U3_GGA(x4)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 6 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

MA_IN_GA(s(T28), X62) → MA_IN_GA(T28, X62)

The TRS R consists of the following rules:

qC_in_gg(T5, T6) → U4_gg(T5, T6, mB_in_gga(T5, T6, X5))
mB_in_gga(T9, 0, T9) → mB_out_gga(T9, 0, T9)
mB_in_gga(0, 0, 0) → mB_out_gga(0, 0, 0)
mB_in_gga(s(T15), 0, X29) → U2_gga(T15, X29, mA_in_ga(T15, X29))
mA_in_ga(T22, T22) → mA_out_ga(T22, T22)
mA_in_ga(0, 0) → mA_out_ga(0, 0)
mA_in_ga(s(T28), X62) → U1_ga(T28, X62, mA_in_ga(T28, X62))
U1_ga(T28, X62, mA_out_ga(T28, X62)) → mA_out_ga(s(T28), X62)
U2_gga(T15, X29, mA_out_ga(T15, X29)) → mB_out_gga(s(T15), 0, X29)
mB_in_gga(0, T35, 0) → mB_out_gga(0, T35, 0)
mB_in_gga(s(T44), s(T48), X94) → U3_gga(T44, T48, X94, mB_in_gga(T44, T48, X94))
U3_gga(T44, T48, X94, mB_out_gga(T44, T48, X94)) → mB_out_gga(s(T44), s(T48), X94)
U4_gg(T5, T6, mB_out_gga(T5, T6, X5)) → qC_out_gg(T5, T6)

The argument filtering Pi contains the following mapping:
qC_in_gg(x1, x2)  =  qC_in_gg(x1, x2)
U4_gg(x1, x2, x3)  =  U4_gg(x3)
mB_in_gga(x1, x2, x3)  =  mB_in_gga(x1, x2)
0  =  0
mB_out_gga(x1, x2, x3)  =  mB_out_gga(x3)
s(x1)  =  s(x1)
U2_gga(x1, x2, x3)  =  U2_gga(x3)
mA_in_ga(x1, x2)  =  mA_in_ga(x1)
mA_out_ga(x1, x2)  =  mA_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U3_gga(x1, x2, x3, x4)  =  U3_gga(x4)
qC_out_gg(x1, x2)  =  qC_out_gg
MA_IN_GA(x1, x2)  =  MA_IN_GA(x1)

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

(10) UsableRulesProof (EQUIVALENT transformation)

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

(11) Obligation:

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

MA_IN_GA(s(T28), X62) → MA_IN_GA(T28, X62)

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

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

(12) PiDPToQDPProof (SOUND transformation)

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

(13) Obligation:

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

MA_IN_GA(s(T28)) → MA_IN_GA(T28)

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

(14) 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:

• MA_IN_GA(s(T28)) → MA_IN_GA(T28)
  The graph contains the following edges 1 > 1

(15) YES

(16) Obligation:

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

MB_IN_GGA(s(T44), s(T48), X94) → MB_IN_GGA(T44, T48, X94)

The TRS R consists of the following rules:

qC_in_gg(T5, T6) → U4_gg(T5, T6, mB_in_gga(T5, T6, X5))
mB_in_gga(T9, 0, T9) → mB_out_gga(T9, 0, T9)
mB_in_gga(0, 0, 0) → mB_out_gga(0, 0, 0)
mB_in_gga(s(T15), 0, X29) → U2_gga(T15, X29, mA_in_ga(T15, X29))
mA_in_ga(T22, T22) → mA_out_ga(T22, T22)
mA_in_ga(0, 0) → mA_out_ga(0, 0)
mA_in_ga(s(T28), X62) → U1_ga(T28, X62, mA_in_ga(T28, X62))
U1_ga(T28, X62, mA_out_ga(T28, X62)) → mA_out_ga(s(T28), X62)
U2_gga(T15, X29, mA_out_ga(T15, X29)) → mB_out_gga(s(T15), 0, X29)
mB_in_gga(0, T35, 0) → mB_out_gga(0, T35, 0)
mB_in_gga(s(T44), s(T48), X94) → U3_gga(T44, T48, X94, mB_in_gga(T44, T48, X94))
U3_gga(T44, T48, X94, mB_out_gga(T44, T48, X94)) → mB_out_gga(s(T44), s(T48), X94)
U4_gg(T5, T6, mB_out_gga(T5, T6, X5)) → qC_out_gg(T5, T6)

The argument filtering Pi contains the following mapping:
qC_in_gg(x1, x2)  =  qC_in_gg(x1, x2)
U4_gg(x1, x2, x3)  =  U4_gg(x3)
mB_in_gga(x1, x2, x3)  =  mB_in_gga(x1, x2)
0  =  0
mB_out_gga(x1, x2, x3)  =  mB_out_gga(x3)
s(x1)  =  s(x1)
U2_gga(x1, x2, x3)  =  U2_gga(x3)
mA_in_ga(x1, x2)  =  mA_in_ga(x1)
mA_out_ga(x1, x2)  =  mA_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U3_gga(x1, x2, x3, x4)  =  U3_gga(x4)
qC_out_gg(x1, x2)  =  qC_out_gg
MB_IN_GGA(x1, x2, x3)  =  MB_IN_GGA(x1, x2)

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

(17) UsableRulesProof (EQUIVALENT transformation)

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

(18) Obligation:

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

MB_IN_GGA(s(T44), s(T48), X94) → MB_IN_GGA(T44, T48, X94)

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

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

(19) PiDPToQDPProof (SOUND transformation)

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

(20) Obligation:

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

MB_IN_GGA(s(T44), s(T48)) → MB_IN_GGA(T44, T48)

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

(21) 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:

• MB_IN_GGA(s(T44), s(T48)) → MB_IN_GGA(T44, T48)
  The graph contains the following edges 1 > 1, 2 > 2

(22) YES