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) :- ','(member2(X, Xs), member1(X, Ys)).
has_a_or_b(Xs) :- overlap(Xs, .(a, .(b, []))).
member1(X, .(Y, Xs)) :- member1(X, Xs).
member1(X, .(X, Xs)).
member2(X, .(Y, Xs)) :- member2(X, Xs).
member2(X, .(X, Xs)).

Query: overlap(g,g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="C: 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="A: ','(member2(X5, T5), member1(X5, T6))\n\nT5 is ground\nT6 is ground\nX5 is free", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: ','(member2(X5, T5), member1(X5, T6)), SCOPE: 2, CLAUSE: 4 | ','(member2(X5, T5), member1(X5, T6)), SCOPE: 2, CLAUSE: 5\n\nT5 is ground\nT6 is ground\nX5 is free", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ','(member2(X5, T5), member1(X5, T6)), SCOPE: 2, CLAUSE: 4\n\nT5 is ground\nT6 is ground\nX5 is free", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ','(member2(X5, T5), member1(X5, T6)), SCOPE: 2, CLAUSE: 5\n\nT5 is ground\nT6 is ground\nX5 is free", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(member2(X25, T16), member1(X25, T6))\n\nT6 is ground\nT16 is ground\nX25 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="B: member1(T23, T6)\n\nT6 is ground\nT23 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: member1(T23, T6), SCOPE: 3, CLAUSE: 2 | member1(T23, T6), SCOPE: 3, CLAUSE: 3\n\nT6 is ground\nT23 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: member1(T23, T6), SCOPE: 3, CLAUSE: 2\n\nT6 is ground\nT23 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: member1(T23, T6), SCOPE: 3, CLAUSE: 3\n\nT6 is ground\nT23 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: member1(T40, T42)\n\nT40 is ground\nT42 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 19 [arrowhead = none ];
19 -> 2;

20 [label="ONLY EVAL with clause\noverlap(X3, X4) :- ','(member2(X5, X3), member1(X5, X4)).\nand substitutionT1 -> T5,\nX3 -> T5,\nT2 -> T6,\nX4 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 20 [arrowhead = none ];
20 -> 3;

21 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 21 [arrowhead = none ];
21 -> 4;

22 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 22 [arrowhead = none ];
22 -> 5;

23 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 23 [arrowhead = none ];
23 -> 6;

24 [label="EVAL with clause\nmember2(X22, .(X23, X24)) :- member2(X22, X24).\nand substitutionX5 -> X25,\nX22 -> X25,\nX23 -> T15,\nX24 -> T16,\nT5 -> .(T15, T16)", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 24 [arrowhead = none ];
24 -> 7;

25 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 25 [arrowhead = none ];
25 -> 8;

26 [label="EVAL with clause\nmember2(X36, .(X36, X37)).\nand substitutionX5 -> T23,\nX36 -> T23,\nX38 -> T23,\nX37 -> T24,\nT5 -> .(T23, T24)", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 26 [arrowhead = none ];
26 -> 9;

27 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 27 [arrowhead = none ];
27 -> 10;

28 [label="INSTANCE with matching:\nX5 -> X25\nT5 -> T16", fontsize=14, style = filled, fillcolor = lightgrey];
7 -> 28 [arrowhead = none , style=dashed];
28 -> 3[style=dashed];

29 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
9 -> 29 [arrowhead = none ];
29 -> 11;

30 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
11 -> 30 [arrowhead = none ];
30 -> 12;

31 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
11 -> 31 [arrowhead = none ];
31 -> 13;

32 [label="EVAL with clause\nmember1(X54, .(X55, X56)) :- member1(X54, X56).\nand substitutionT23 -> T40,\nX54 -> T40,\nX55 -> T41,\nX56 -> T42,\nT6 -> .(T41, T42)", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 32 [arrowhead = none ];
32 -> 14;

33 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 33 [arrowhead = none ];
33 -> 15;

34 [label="EVAL with clause\nmember1(X64, .(X64, X65)).\nand substitutionT23 -> T50,\nX64 -> T50,\nX65 -> T51,\nT6 -> .(T50, T51)", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 34 [arrowhead = none ];
34 -> 16;

35 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 35 [arrowhead = none ];
35 -> 17;

36 [label="INSTANCE with matching:\nT23 -> T40\nT6 -> T42", fontsize=14, style = filled, fillcolor = lightgrey];
14 -> 36 [arrowhead = none , style=dashed];
36 -> 9[style=dashed];

37 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 37 [arrowhead = none ];
37 -> 18;

}

(2) Obligation:

Triples:

pA(X1, .(X2, X3), X4) :- pA(X1, X3, X4).
pA(X1, .(X1, X2), X3) :- member1B(X1, X3).
member1B(X1, .(X2, X3)) :- member1B(X1, X3).
overlapC(X1, X2) :- pA(X3, X1, X2).

Clauses:

qcA(X1, .(X2, X3), X4) :- qcA(X1, X3, X4).
qcA(X1, .(X1, X2), X3) :- member1cB(X1, X3).
member1cB(X1, .(X2, X3)) :- member1cB(X1, X3).
member1cB(X1, .(X1, X2)).

Afs:

overlapC(x1, x2)  =  overlapC(x1, x2)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes:
overlapC_in: (b,b)
pA_in: (f,b,b)
member1B_in: (b,b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

OVERLAPC_IN_GG(X1, X2) → U4_GG(X1, X2, pA_in_agg(X3, X1, X2))
OVERLAPC_IN_GG(X1, X2) → PA_IN_AGG(X3, X1, X2)
PA_IN_AGG(X1, .(X2, X3), X4) → U1_AGG(X1, X2, X3, X4, pA_in_agg(X1, X3, X4))
PA_IN_AGG(X1, .(X2, X3), X4) → PA_IN_AGG(X1, X3, X4)
PA_IN_AGG(X1, .(X1, X2), X3) → U2_AGG(X1, X2, X3, member1B_in_gg(X1, X3))
PA_IN_AGG(X1, .(X1, X2), X3) → MEMBER1B_IN_GG(X1, X3)
MEMBER1B_IN_GG(X1, .(X2, X3)) → U3_GG(X1, X2, X3, member1B_in_gg(X1, X3))
MEMBER1B_IN_GG(X1, .(X2, X3)) → MEMBER1B_IN_GG(X1, X3)

R is empty.
The argument filtering Pi contains the following mapping:
pA_in_agg(x1, x2, x3)  =  pA_in_agg(x2, x3)
.(x1, x2)  =  .(x1, x2)
member1B_in_gg(x1, x2)  =  member1B_in_gg(x1, x2)
OVERLAPC_IN_GG(x1, x2)  =  OVERLAPC_IN_GG(x1, x2)
U4_GG(x1, x2, x3)  =  U4_GG(x1, x2, x3)
PA_IN_AGG(x1, x2, x3)  =  PA_IN_AGG(x2, x3)
U1_AGG(x1, x2, x3, x4, x5)  =  U1_AGG(x2, x3, x4, x5)
U2_AGG(x1, x2, x3, x4)  =  U2_AGG(x1, x2, x3, x4)
MEMBER1B_IN_GG(x1, x2)  =  MEMBER1B_IN_GG(x1, x2)
U3_GG(x1, x2, x3, x4)  =  U3_GG(x1, x2, x3, x4)

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

Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES

(4) Obligation:

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

OVERLAPC_IN_GG(X1, X2) → U4_GG(X1, X2, pA_in_agg(X3, X1, X2))
OVERLAPC_IN_GG(X1, X2) → PA_IN_AGG(X3, X1, X2)
PA_IN_AGG(X1, .(X2, X3), X4) → U1_AGG(X1, X2, X3, X4, pA_in_agg(X1, X3, X4))
PA_IN_AGG(X1, .(X2, X3), X4) → PA_IN_AGG(X1, X3, X4)
PA_IN_AGG(X1, .(X1, X2), X3) → U2_AGG(X1, X2, X3, member1B_in_gg(X1, X3))
PA_IN_AGG(X1, .(X1, X2), X3) → MEMBER1B_IN_GG(X1, X3)
MEMBER1B_IN_GG(X1, .(X2, X3)) → U3_GG(X1, X2, X3, member1B_in_gg(X1, X3))
MEMBER1B_IN_GG(X1, .(X2, X3)) → MEMBER1B_IN_GG(X1, X3)

R is empty.
The argument filtering Pi contains the following mapping:
pA_in_agg(x1, x2, x3)  =  pA_in_agg(x2, x3)
.(x1, x2)  =  .(x1, x2)
member1B_in_gg(x1, x2)  =  member1B_in_gg(x1, x2)
OVERLAPC_IN_GG(x1, x2)  =  OVERLAPC_IN_GG(x1, x2)
U4_GG(x1, x2, x3)  =  U4_GG(x1, x2, x3)
PA_IN_AGG(x1, x2, x3)  =  PA_IN_AGG(x2, x3)
U1_AGG(x1, x2, x3, x4, x5)  =  U1_AGG(x2, x3, x4, x5)
U2_AGG(x1, x2, x3, x4)  =  U2_AGG(x1, x2, x3, x4)
MEMBER1B_IN_GG(x1, x2)  =  MEMBER1B_IN_GG(x1, x2)
U3_GG(x1, x2, x3, x4)  =  U3_GG(x1, x2, x3, x4)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

MEMBER1B_IN_GG(X1, .(X2, X3)) → MEMBER1B_IN_GG(X1, X3)

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

(8) PiDPToQDPProof (EQUIVALENT transformation)

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

(9) Obligation:

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

MEMBER1B_IN_GG(X1, .(X2, X3)) → MEMBER1B_IN_GG(X1, X3)

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

(10) 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:

• MEMBER1B_IN_GG(X1, .(X2, X3)) → MEMBER1B_IN_GG(X1, X3)
  The graph contains the following edges 1 >= 1, 2 > 2

(11) YES

(12) Obligation:

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

PA_IN_AGG(X1, .(X2, X3), X4) → PA_IN_AGG(X1, X3, X4)

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

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

(13) PiDPToQDPProof (SOUND transformation)

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

(14) Obligation:

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

PA_IN_AGG(.(X2, X3), X4) → PA_IN_AGG(X3, X4)

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

(15) 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:

• PA_IN_AGG(.(X2, X3), X4) → PA_IN_AGG(X3, X4)
  The graph contains the following edges 1 > 1, 2 >= 2

(16) YES