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

(0) Obligation:

Clauses:

overlap(Xs, Ys) :- ','(member(X, Xs), member(X, Ys)).
member(X1, []) :- ','(!, failure(a)).
member(X, Y) :- head(Y, X).
member(X, Y) :- ','(tail(Y, T), member(X, T)).
head([], X2).
head(.(H, X3), H).
tail([], []).
tail(.(X4, T), T).
failure(b).

Query: overlap(g,g)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: overlap(T1, T2)\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: overlap(T1, T2), SCOPE: 1, CLAUSE: 0\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(member(X9, T5), member(X9, T6))\n\nT5 is ground\nT6 is ground\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: ','(member(X9, T5), member(X9, T6)), SCOPE: 2, CLAUSE: 1 | ','(member(X9, T5), member(X9, T6)), SCOPE: 2, CLAUSE: 2 | ','(member(X9, T5), member(X9, T6)), SCOPE: 2, CLAUSE: 3\n\nT5 is ground\nT6 is ground\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ','(','(!_2, failure(a)), member(X15, T6)) | ','(member(X9, []), member(X9, T6)), SCOPE: 2, CLAUSE: 2 | ','(member(X9, []), member(X9, T6)), SCOPE: 2, CLAUSE: 3\n\nT6 is ground\nX9 is free\nX15 is free", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ','(member(X9, T5), member(X9, T6)), SCOPE: 2, CLAUSE: 2 | ','(member(X9, T5), member(X9, T6)), SCOPE: 2, CLAUSE: 3\n\nT5 is ground\nT6 is ground\nX9 is free\nX14 is free\nmember(X9, T5) !=? member(X14, [])", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(failure(a), member(X15, T6))\n\nT6 is ground\nX15 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(failure(a), member(X15, T6)), SCOPE: 3, CLAUSE: 8\n\nT6 is ground\nX15 is free", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(member(X9, T5), member(X9, T6)), SCOPE: 2, CLAUSE: 2\n\nT5 is ground\nT6 is ground\nX9 is free\nX14 is free\nmember(X9, T5) !=? member(X14, [])", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(member(X9, T5), member(X9, T6)), SCOPE: 2, CLAUSE: 3\n\nT5 is ground\nT6 is ground\nX9 is free\nX14 is free\nmember(X9, T5) !=? member(X14, [])", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(head(T11, X36), member(X36, T6))\n\nT6 is ground\nT11 is ground\nX9 is free\nX14 is free\nX36 is free\nmember(X9, T11) !=? member(X14, [])", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(head(T11, X36), member(X36, T6)), SCOPE: 4, CLAUSE: 4 | ','(head(T11, X36), member(X36, T6)), SCOPE: 4, CLAUSE: 5\n\nT6 is ground\nT11 is ground\nX9 is free\nX14 is free\nX36 is free\nmember(X9, T11) !=? member(X14, [])", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(head(T11, X36), member(X36, T6)), SCOPE: 4, CLAUSE: 5\n\nT6 is ground\nT11 is ground\nX9 is free\nX14 is free\nX36 is free\nmember(X9, T11) !=? member(X14, [])", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: member(T16, T6)\n\nT6 is ground\nT16 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: member(T16, T6), SCOPE: 5, CLAUSE: 1 | member(T16, T6), SCOPE: 5, CLAUSE: 2 | member(T16, T6), SCOPE: 5, CLAUSE: 3\n\nT6 is ground\nT16 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(!_5, failure(a)) | member(T21, []), SCOPE: 5, CLAUSE: 2 | member(T21, []), SCOPE: 5, CLAUSE: 3\n\nT21 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: member(T16, T6), SCOPE: 5, CLAUSE: 2 | member(T16, T6), SCOPE: 5, CLAUSE: 3\n\nT6 is ground\nT16 is ground\nX50 is free\nmember(T16, T6) !=? member(X50, [])", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: failure(a)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: failure(a), SCOPE: 6, CLAUSE: 8\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: member(T16, T6), SCOPE: 5, CLAUSE: 2\n\nT6 is ground\nT16 is ground\nX50 is free\nmember(T16, T6) !=? member(X50, [])", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: member(T16, T6), SCOPE: 5, CLAUSE: 3\n\nT6 is ground\nT16 is ground\nX50 is free\nmember(T16, T6) !=? member(X50, [])", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: head(T31, T30)\n\nT30 is ground\nT31 is ground\nX50 is free\nmember(T30, T31) !=? member(X50, [])", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: head(T31, T30), SCOPE: 7, CLAUSE: 4 | head(T31, T30), SCOPE: 7, CLAUSE: 5\n\nT30 is ground\nT31 is ground\nX50 is free\nmember(T30, T31) !=? member(X50, [])", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: head(T31, T30), SCOPE: 7, CLAUSE: 5\n\nT30 is ground\nT31 is ground\nX50 is free\nmember(T30, T31) !=? member(X50, [])", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: ','(tail(T45, X76), member(T44, X76))\n\nT44 is ground\nT45 is ground\nX50 is free\nX76 is free\nmember(T44, T45) !=? member(X50, [])", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: ','(tail(T45, X76), member(T44, X76)), SCOPE: 8, CLAUSE: 6 | ','(tail(T45, X76), member(T44, X76)), SCOPE: 8, CLAUSE: 7\n\nT44 is ground\nT45 is ground\nX50 is free\nX76 is free\nmember(T44, T45) !=? member(X50, [])", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: ','(tail(T45, X76), member(T44, X76)), SCOPE: 8, CLAUSE: 7\n\nT44 is ground\nT45 is ground\nX50 is free\nX76 is free\nmember(T44, T45) !=? member(X50, [])", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: member(T44, T52)\n\nT44 is ground\nT52 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="36: ','(','(tail(T56, X96), member(X97, X96)), member(X97, T6))\n\nT6 is ground\nT56 is ground\nX9 is free\nX14 is free\nX97 is free\nX96 is free\nmember(X9, T56) !=? member(X14, [])", fontsize=16, style=filled, fillcolor=lightyellow];
37 [label="37: ','(','(tail(T56, X96), member(X97, X96)), member(X97, T6)), SCOPE: 9, CLAUSE: 6 | ','(','(tail(T56, X96), member(X97, X96)), member(X97, T6)), SCOPE: 9, CLAUSE: 7\n\nT6 is ground\nT56 is ground\nX9 is free\nX14 is free\nX97 is free\nX96 is free\nmember(X9, T56) !=? member(X14, [])", fontsize=16, style=filled, fillcolor=lightyellow];
38 [label="38: ','(','(tail(T56, X96), member(X97, X96)), member(X97, T6)), SCOPE: 9, CLAUSE: 7\n\nT6 is ground\nT56 is ground\nX9 is free\nX14 is free\nX97 is free\nX96 is free\nmember(X9, T56) !=? member(X14, [])", fontsize=16, style=filled, fillcolor=lightyellow];
39 [label="39: ','(member(X97, T62), member(X97, T6))\n\nT6 is ground\nT62 is ground\nX97 is free", fontsize=16, style=filled, fillcolor=lightyellow];
40 [label="40: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
41 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 41 [arrowhead = none ];
41 -> 2;

42 [label="ONLY EVAL with clause\noverlap(X7, X8) :- ','(member(X9, X7), member(X9, X8)).\nand substitutionT1 -> T5,\nX7 -> T5,\nT2 -> T6,\nX8 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 42 [arrowhead = none ];
42 -> 3;

43 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 43 [arrowhead = none ];
43 -> 4;

44 [label="EVAL with clause\nmember(X14, []) :- ','(!_2, failure(a)).\nand substitutionX9 -> X15,\nX14 -> X15,\nT5 -> []", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 44 [arrowhead = none ];
44 -> 5;

45 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 45 [arrowhead = none ];
45 -> 6;

46 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 46 [arrowhead = none ];
46 -> 7;

47 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
6 -> 47 [arrowhead = none ];
47 -> 10;

48 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
6 -> 48 [arrowhead = none ];
48 -> 11;

49 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 49 [arrowhead = none ];
49 -> 8;

50 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 50 [arrowhead = none ];
50 -> 9;

51 [label="ONLY EVAL with clause\nmember(X34, X35) :- head(X35, X34).\nand substitutionX9 -> X36,\nX34 -> X36,\nT5 -> T11,\nX35 -> T11", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 51 [arrowhead = none ];
51 -> 12;

52 [label="ONLY EVAL with clause\nmember(X94, X95) :- ','(tail(X95, X96), member(X94, X96)).\nand substitutionX9 -> X97,\nX94 -> X97,\nT5 -> T56,\nX95 -> T56", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 52 [arrowhead = none ];
52 -> 36;

53 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 53 [arrowhead = none ];
53 -> 13;

54 [label="BACKTRACK\nfor clause: head([], X2)\nwith clash: (member(X9, T11), member(X14, []))", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 54 [arrowhead = none ];
54 -> 14;

55 [label="EVAL with clause\nhead(.(X45, X46), X45).\nand substitutionX45 -> T16,\nX46 -> T17,\nT11 -> .(T16, T17),\nX36 -> T16", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 55 [arrowhead = none ];
55 -> 15;

56 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 56 [arrowhead = none ];
56 -> 16;

57 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
15 -> 57 [arrowhead = none ];
57 -> 17;

58 [label="EVAL with clause\nmember(X50, []) :- ','(!_5, failure(a)).\nand substitutionT16 -> T21,\nX50 -> T21,\nT6 -> []", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 58 [arrowhead = none ];
58 -> 18;

59 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 59 [arrowhead = none ];
59 -> 19;

60 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
18 -> 60 [arrowhead = none ];
60 -> 20;

61 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
19 -> 61 [arrowhead = none ];
61 -> 23;

62 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
19 -> 62 [arrowhead = none ];
62 -> 24;

63 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
20 -> 63 [arrowhead = none ];
63 -> 21;

64 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
21 -> 64 [arrowhead = none ];
64 -> 22;

65 [label="ONLY EVAL with clause\nmember(X59, X60) :- head(X60, X59).\nand substitutionT16 -> T30,\nX59 -> T30,\nT6 -> T31,\nX60 -> T31", fontsize=14, style = filled, fillcolor = lightblue];
23 -> 65 [arrowhead = none ];
65 -> 25;

66 [label="ONLY EVAL with clause\nmember(X74, X75) :- ','(tail(X75, X76), member(X74, X76)).\nand substitutionT16 -> T44,\nX74 -> T44,\nT6 -> T45,\nX75 -> T45", fontsize=14, style = filled, fillcolor = lightblue];
24 -> 66 [arrowhead = none ];
66 -> 31;

67 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
25 -> 67 [arrowhead = none ];
67 -> 26;

68 [label="BACKTRACK\nfor clause: head([], X2)\nwith clash: (member(T30, T31), member(X50, []))", fontsize=14, style = filled, fillcolor = lightgreen];
26 -> 68 [arrowhead = none ];
68 -> 27;

69 [label="EVAL with clause\nhead(.(X66, X67), X66).\nand substitutionX66 -> T37,\nX67 -> T38,\nT31 -> .(T37, T38),\nT30 -> T37", fontsize=14, style = filled, fillcolor = lightblue];
27 -> 69 [arrowhead = none ];
69 -> 28;

70 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
27 -> 70 [arrowhead = none ];
70 -> 29;

71 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
28 -> 71 [arrowhead = none ];
71 -> 30;

72 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
31 -> 72 [arrowhead = none ];
72 -> 32;

73 [label="BACKTRACK\nfor clause: tail([], [])\nwith clash: (member(T44, T45), member(X50, []))", fontsize=14, style = filled, fillcolor = lightgreen];
32 -> 73 [arrowhead = none ];
73 -> 33;

74 [label="EVAL with clause\ntail(.(X82, X83), X83).\nand substitutionX82 -> T51,\nX83 -> T52,\nT45 -> .(T51, T52),\nX76 -> T52", fontsize=14, style = filled, fillcolor = lightblue];
33 -> 74 [arrowhead = none ];
74 -> 34;

75 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
33 -> 75 [arrowhead = none ];
75 -> 35;

76 [label="INSTANCE with matching:\nT16 -> T44\nT6 -> T52", fontsize=14, style = filled, fillcolor = lightgrey];
34 -> 76 [arrowhead = none , style=dashed];
76 -> 15[style=dashed];

77 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
36 -> 77 [arrowhead = none ];
77 -> 37;

78 [label="BACKTRACK\nfor clause: tail([], [])\nwith clash: (member(X9, T56), member(X14, []))", fontsize=14, style = filled, fillcolor = lightgreen];
37 -> 78 [arrowhead = none ];
78 -> 38;

79 [label="EVAL with clause\ntail(.(X104, X105), X105).\nand substitutionX104 -> T61,\nX105 -> T62,\nT56 -> .(T61, T62),\nX96 -> T62", fontsize=14, style = filled, fillcolor = lightblue];
38 -> 79 [arrowhead = none ];
79 -> 39;

80 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
38 -> 80 [arrowhead = none ];
80 -> 40;

81 [label="INSTANCE with matching:\nX9 -> X97\nT5 -> T62", fontsize=14, style = filled, fillcolor = lightgrey];
39 -> 81 [arrowhead = none , style=dashed];
81 -> 3[style=dashed];

}

(2) Obligation:

Clauses:

memberA(T37, .(T37, T38)).
memberA(T44, .(T51, T52)) :- memberA(T44, T52).
pB(T16, .(T16, T17), T6) :- memberA(T16, T6).
pB(X97, .(T61, T62), T6) :- pB(X97, T62, T6).
overlapC(T5, T6) :- pB(X9, T5, T6).

Query: overlapC(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:
overlapC_in: (b,b)
pB_in: (f,b,b)
memberA_in: (b,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

overlapC_in_gg(T5, T6) → U4_gg(T5, T6, pB_in_agg(X9, T5, T6))
pB_in_agg(T16, .(T16, T17), T6) → U2_agg(T16, T17, T6, memberA_in_gg(T16, T6))
memberA_in_gg(T37, .(T37, T38)) → memberA_out_gg(T37, .(T37, T38))
memberA_in_gg(T44, .(T51, T52)) → U1_gg(T44, T51, T52, memberA_in_gg(T44, T52))
U1_gg(T44, T51, T52, memberA_out_gg(T44, T52)) → memberA_out_gg(T44, .(T51, T52))
U2_agg(T16, T17, T6, memberA_out_gg(T16, T6)) → pB_out_agg(T16, .(T16, T17), T6)
pB_in_agg(X97, .(T61, T62), T6) → U3_agg(X97, T61, T62, T6, pB_in_agg(X97, T62, T6))
U3_agg(X97, T61, T62, T6, pB_out_agg(X97, T62, T6)) → pB_out_agg(X97, .(T61, T62), T6)
U4_gg(T5, T6, pB_out_agg(X9, T5, T6)) → overlapC_out_gg(T5, T6)

The argument filtering Pi contains the following mapping:
overlapC_in_gg(x1, x2)  =  overlapC_in_gg(x1, x2)
U4_gg(x1, x2, x3)  =  U4_gg(x3)
pB_in_agg(x1, x2, x3)  =  pB_in_agg(x2, x3)
.(x1, x2)  =  .(x1, x2)
U2_agg(x1, x2, x3, x4)  =  U2_agg(x1, x4)
memberA_in_gg(x1, x2)  =  memberA_in_gg(x1, x2)
memberA_out_gg(x1, x2)  =  memberA_out_gg
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
pB_out_agg(x1, x2, x3)  =  pB_out_agg(x1)
U3_agg(x1, x2, x3, x4, x5)  =  U3_agg(x5)
overlapC_out_gg(x1, x2)  =  overlapC_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:

overlapC_in_gg(T5, T6) → U4_gg(T5, T6, pB_in_agg(X9, T5, T6))
pB_in_agg(T16, .(T16, T17), T6) → U2_agg(T16, T17, T6, memberA_in_gg(T16, T6))
memberA_in_gg(T37, .(T37, T38)) → memberA_out_gg(T37, .(T37, T38))
memberA_in_gg(T44, .(T51, T52)) → U1_gg(T44, T51, T52, memberA_in_gg(T44, T52))
U1_gg(T44, T51, T52, memberA_out_gg(T44, T52)) → memberA_out_gg(T44, .(T51, T52))
U2_agg(T16, T17, T6, memberA_out_gg(T16, T6)) → pB_out_agg(T16, .(T16, T17), T6)
pB_in_agg(X97, .(T61, T62), T6) → U3_agg(X97, T61, T62, T6, pB_in_agg(X97, T62, T6))
U3_agg(X97, T61, T62, T6, pB_out_agg(X97, T62, T6)) → pB_out_agg(X97, .(T61, T62), T6)
U4_gg(T5, T6, pB_out_agg(X9, T5, T6)) → overlapC_out_gg(T5, T6)

The argument filtering Pi contains the following mapping:
overlapC_in_gg(x1, x2)  =  overlapC_in_gg(x1, x2)
U4_gg(x1, x2, x3)  =  U4_gg(x3)
pB_in_agg(x1, x2, x3)  =  pB_in_agg(x2, x3)
.(x1, x2)  =  .(x1, x2)
U2_agg(x1, x2, x3, x4)  =  U2_agg(x1, x4)
memberA_in_gg(x1, x2)  =  memberA_in_gg(x1, x2)
memberA_out_gg(x1, x2)  =  memberA_out_gg
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
pB_out_agg(x1, x2, x3)  =  pB_out_agg(x1)
U3_agg(x1, x2, x3, x4, x5)  =  U3_agg(x5)
overlapC_out_gg(x1, x2)  =  overlapC_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:

OVERLAPC_IN_GG(T5, T6) → U4_GG(T5, T6, pB_in_agg(X9, T5, T6))
OVERLAPC_IN_GG(T5, T6) → PB_IN_AGG(X9, T5, T6)
PB_IN_AGG(T16, .(T16, T17), T6) → U2_AGG(T16, T17, T6, memberA_in_gg(T16, T6))
PB_IN_AGG(T16, .(T16, T17), T6) → MEMBERA_IN_GG(T16, T6)
MEMBERA_IN_GG(T44, .(T51, T52)) → U1_GG(T44, T51, T52, memberA_in_gg(T44, T52))
MEMBERA_IN_GG(T44, .(T51, T52)) → MEMBERA_IN_GG(T44, T52)
PB_IN_AGG(X97, .(T61, T62), T6) → U3_AGG(X97, T61, T62, T6, pB_in_agg(X97, T62, T6))
PB_IN_AGG(X97, .(T61, T62), T6) → PB_IN_AGG(X97, T62, T6)

The TRS R consists of the following rules:

overlapC_in_gg(T5, T6) → U4_gg(T5, T6, pB_in_agg(X9, T5, T6))
pB_in_agg(T16, .(T16, T17), T6) → U2_agg(T16, T17, T6, memberA_in_gg(T16, T6))
memberA_in_gg(T37, .(T37, T38)) → memberA_out_gg(T37, .(T37, T38))
memberA_in_gg(T44, .(T51, T52)) → U1_gg(T44, T51, T52, memberA_in_gg(T44, T52))
U1_gg(T44, T51, T52, memberA_out_gg(T44, T52)) → memberA_out_gg(T44, .(T51, T52))
U2_agg(T16, T17, T6, memberA_out_gg(T16, T6)) → pB_out_agg(T16, .(T16, T17), T6)
pB_in_agg(X97, .(T61, T62), T6) → U3_agg(X97, T61, T62, T6, pB_in_agg(X97, T62, T6))
U3_agg(X97, T61, T62, T6, pB_out_agg(X97, T62, T6)) → pB_out_agg(X97, .(T61, T62), T6)
U4_gg(T5, T6, pB_out_agg(X9, T5, T6)) → overlapC_out_gg(T5, T6)

The argument filtering Pi contains the following mapping:
overlapC_in_gg(x1, x2)  =  overlapC_in_gg(x1, x2)
U4_gg(x1, x2, x3)  =  U4_gg(x3)
pB_in_agg(x1, x2, x3)  =  pB_in_agg(x2, x3)
.(x1, x2)  =  .(x1, x2)
U2_agg(x1, x2, x3, x4)  =  U2_agg(x1, x4)
memberA_in_gg(x1, x2)  =  memberA_in_gg(x1, x2)
memberA_out_gg(x1, x2)  =  memberA_out_gg
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
pB_out_agg(x1, x2, x3)  =  pB_out_agg(x1)
U3_agg(x1, x2, x3, x4, x5)  =  U3_agg(x5)
overlapC_out_gg(x1, x2)  =  overlapC_out_gg
OVERLAPC_IN_GG(x1, x2)  =  OVERLAPC_IN_GG(x1, x2)
U4_GG(x1, x2, x3)  =  U4_GG(x3)
PB_IN_AGG(x1, x2, x3)  =  PB_IN_AGG(x2, x3)
U2_AGG(x1, x2, x3, x4)  =  U2_AGG(x1, x4)
MEMBERA_IN_GG(x1, x2)  =  MEMBERA_IN_GG(x1, x2)
U1_GG(x1, x2, x3, x4)  =  U1_GG(x4)
U3_AGG(x1, x2, x3, x4, x5)  =  U3_AGG(x5)

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

(6) Obligation:

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

OVERLAPC_IN_GG(T5, T6) → U4_GG(T5, T6, pB_in_agg(X9, T5, T6))
OVERLAPC_IN_GG(T5, T6) → PB_IN_AGG(X9, T5, T6)
PB_IN_AGG(T16, .(T16, T17), T6) → U2_AGG(T16, T17, T6, memberA_in_gg(T16, T6))
PB_IN_AGG(T16, .(T16, T17), T6) → MEMBERA_IN_GG(T16, T6)
MEMBERA_IN_GG(T44, .(T51, T52)) → U1_GG(T44, T51, T52, memberA_in_gg(T44, T52))
MEMBERA_IN_GG(T44, .(T51, T52)) → MEMBERA_IN_GG(T44, T52)
PB_IN_AGG(X97, .(T61, T62), T6) → U3_AGG(X97, T61, T62, T6, pB_in_agg(X97, T62, T6))
PB_IN_AGG(X97, .(T61, T62), T6) → PB_IN_AGG(X97, T62, T6)

The TRS R consists of the following rules:

overlapC_in_gg(T5, T6) → U4_gg(T5, T6, pB_in_agg(X9, T5, T6))
pB_in_agg(T16, .(T16, T17), T6) → U2_agg(T16, T17, T6, memberA_in_gg(T16, T6))
memberA_in_gg(T37, .(T37, T38)) → memberA_out_gg(T37, .(T37, T38))
memberA_in_gg(T44, .(T51, T52)) → U1_gg(T44, T51, T52, memberA_in_gg(T44, T52))
U1_gg(T44, T51, T52, memberA_out_gg(T44, T52)) → memberA_out_gg(T44, .(T51, T52))
U2_agg(T16, T17, T6, memberA_out_gg(T16, T6)) → pB_out_agg(T16, .(T16, T17), T6)
pB_in_agg(X97, .(T61, T62), T6) → U3_agg(X97, T61, T62, T6, pB_in_agg(X97, T62, T6))
U3_agg(X97, T61, T62, T6, pB_out_agg(X97, T62, T6)) → pB_out_agg(X97, .(T61, T62), T6)
U4_gg(T5, T6, pB_out_agg(X9, T5, T6)) → overlapC_out_gg(T5, T6)

The argument filtering Pi contains the following mapping:
overlapC_in_gg(x1, x2)  =  overlapC_in_gg(x1, x2)
U4_gg(x1, x2, x3)  =  U4_gg(x3)
pB_in_agg(x1, x2, x3)  =  pB_in_agg(x2, x3)
.(x1, x2)  =  .(x1, x2)
U2_agg(x1, x2, x3, x4)  =  U2_agg(x1, x4)
memberA_in_gg(x1, x2)  =  memberA_in_gg(x1, x2)
memberA_out_gg(x1, x2)  =  memberA_out_gg
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
pB_out_agg(x1, x2, x3)  =  pB_out_agg(x1)
U3_agg(x1, x2, x3, x4, x5)  =  U3_agg(x5)
overlapC_out_gg(x1, x2)  =  overlapC_out_gg
OVERLAPC_IN_GG(x1, x2)  =  OVERLAPC_IN_GG(x1, x2)
U4_GG(x1, x2, x3)  =  U4_GG(x3)
PB_IN_AGG(x1, x2, x3)  =  PB_IN_AGG(x2, x3)
U2_AGG(x1, x2, x3, x4)  =  U2_AGG(x1, x4)
MEMBERA_IN_GG(x1, x2)  =  MEMBERA_IN_GG(x1, x2)
U1_GG(x1, x2, x3, x4)  =  U1_GG(x4)
U3_AGG(x1, x2, x3, x4, x5)  =  U3_AGG(x5)

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:

MEMBERA_IN_GG(T44, .(T51, T52)) → MEMBERA_IN_GG(T44, T52)

The TRS R consists of the following rules:

overlapC_in_gg(T5, T6) → U4_gg(T5, T6, pB_in_agg(X9, T5, T6))
pB_in_agg(T16, .(T16, T17), T6) → U2_agg(T16, T17, T6, memberA_in_gg(T16, T6))
memberA_in_gg(T37, .(T37, T38)) → memberA_out_gg(T37, .(T37, T38))
memberA_in_gg(T44, .(T51, T52)) → U1_gg(T44, T51, T52, memberA_in_gg(T44, T52))
U1_gg(T44, T51, T52, memberA_out_gg(T44, T52)) → memberA_out_gg(T44, .(T51, T52))
U2_agg(T16, T17, T6, memberA_out_gg(T16, T6)) → pB_out_agg(T16, .(T16, T17), T6)
pB_in_agg(X97, .(T61, T62), T6) → U3_agg(X97, T61, T62, T6, pB_in_agg(X97, T62, T6))
U3_agg(X97, T61, T62, T6, pB_out_agg(X97, T62, T6)) → pB_out_agg(X97, .(T61, T62), T6)
U4_gg(T5, T6, pB_out_agg(X9, T5, T6)) → overlapC_out_gg(T5, T6)

The argument filtering Pi contains the following mapping:
overlapC_in_gg(x1, x2)  =  overlapC_in_gg(x1, x2)
U4_gg(x1, x2, x3)  =  U4_gg(x3)
pB_in_agg(x1, x2, x3)  =  pB_in_agg(x2, x3)
.(x1, x2)  =  .(x1, x2)
U2_agg(x1, x2, x3, x4)  =  U2_agg(x1, x4)
memberA_in_gg(x1, x2)  =  memberA_in_gg(x1, x2)
memberA_out_gg(x1, x2)  =  memberA_out_gg
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
pB_out_agg(x1, x2, x3)  =  pB_out_agg(x1)
U3_agg(x1, x2, x3, x4, x5)  =  U3_agg(x5)
overlapC_out_gg(x1, x2)  =  overlapC_out_gg
MEMBERA_IN_GG(x1, x2)  =  MEMBERA_IN_GG(x1, x2)

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:

MEMBERA_IN_GG(T44, .(T51, T52)) → MEMBERA_IN_GG(T44, T52)

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

(12) PiDPToQDPProof (EQUIVALENT 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:

MEMBERA_IN_GG(T44, .(T51, T52)) → MEMBERA_IN_GG(T44, T52)

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:

• MEMBERA_IN_GG(T44, .(T51, T52)) → MEMBERA_IN_GG(T44, T52)
  The graph contains the following edges 1 >= 1, 2 > 2

(15) YES

(16) Obligation:

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

PB_IN_AGG(X97, .(T61, T62), T6) → PB_IN_AGG(X97, T62, T6)

The TRS R consists of the following rules:

overlapC_in_gg(T5, T6) → U4_gg(T5, T6, pB_in_agg(X9, T5, T6))
pB_in_agg(T16, .(T16, T17), T6) → U2_agg(T16, T17, T6, memberA_in_gg(T16, T6))
memberA_in_gg(T37, .(T37, T38)) → memberA_out_gg(T37, .(T37, T38))
memberA_in_gg(T44, .(T51, T52)) → U1_gg(T44, T51, T52, memberA_in_gg(T44, T52))
U1_gg(T44, T51, T52, memberA_out_gg(T44, T52)) → memberA_out_gg(T44, .(T51, T52))
U2_agg(T16, T17, T6, memberA_out_gg(T16, T6)) → pB_out_agg(T16, .(T16, T17), T6)
pB_in_agg(X97, .(T61, T62), T6) → U3_agg(X97, T61, T62, T6, pB_in_agg(X97, T62, T6))
U3_agg(X97, T61, T62, T6, pB_out_agg(X97, T62, T6)) → pB_out_agg(X97, .(T61, T62), T6)
U4_gg(T5, T6, pB_out_agg(X9, T5, T6)) → overlapC_out_gg(T5, T6)

The argument filtering Pi contains the following mapping:
overlapC_in_gg(x1, x2)  =  overlapC_in_gg(x1, x2)
U4_gg(x1, x2, x3)  =  U4_gg(x3)
pB_in_agg(x1, x2, x3)  =  pB_in_agg(x2, x3)
.(x1, x2)  =  .(x1, x2)
U2_agg(x1, x2, x3, x4)  =  U2_agg(x1, x4)
memberA_in_gg(x1, x2)  =  memberA_in_gg(x1, x2)
memberA_out_gg(x1, x2)  =  memberA_out_gg
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
pB_out_agg(x1, x2, x3)  =  pB_out_agg(x1)
U3_agg(x1, x2, x3, x4, x5)  =  U3_agg(x5)
overlapC_out_gg(x1, x2)  =  overlapC_out_gg
PB_IN_AGG(x1, x2, x3)  =  PB_IN_AGG(x2, x3)

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:

PB_IN_AGG(X97, .(T61, T62), T6) → PB_IN_AGG(X97, T62, T6)

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

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:

PB_IN_AGG(.(T61, T62), T6) → PB_IN_AGG(T62, T6)

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:

• PB_IN_AGG(.(T61, T62), T6) → PB_IN_AGG(T62, T6)
  The graph contains the following edges 1 > 1, 2 >= 2

(22) YES