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) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: overlap(T1, T2)\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
6 [label="6: overlap(T1, T2), SCOPE: 1, CLAUSE: 0\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="B: ','(member2(X5, T5), member1(X5, T6))\n\nT5 is ground\nT6 is ground\nX5 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(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];
13 [label="13: ','(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];
15 [label="15: ','(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];
17 [label="17: ','(member2(X25, T16), member1(X25, T6))\n\nT6 is ground\nT16 is ground\nX25 is free", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="C: member1(T23, T6)\n\nT6 is ground\nT23 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: 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];
37 [label="37: member1(T23, T6), SCOPE: 3, CLAUSE: 2\n\nT6 is ground\nT23 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
38 [label="38: member1(T23, T6), SCOPE: 3, CLAUSE: 3\n\nT6 is ground\nT23 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
39 [label="39: member1(T40, T42)\n\nT40 is ground\nT42 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
40 [label="40: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
44 [label="44: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
45 [label="45: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
46 [label="46: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
47 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 47 [arrowhead = none ];
47 -> 6;

48 [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];
6 -> 48 [arrowhead = none ];
48 -> 7;

49 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 49 [arrowhead = none ];
49 -> 11;

50 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
11 -> 50 [arrowhead = none ];
50 -> 13;

51 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
11 -> 51 [arrowhead = none ];
51 -> 15;

52 [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];
13 -> 52 [arrowhead = none ];
52 -> 17;

53 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 53 [arrowhead = none ];
53 -> 19;

54 [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];
15 -> 54 [arrowhead = none ];
54 -> 25;

55 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 55 [arrowhead = none ];
55 -> 33;

56 [label="INSTANCE with matching:\nX5 -> X25\nT5 -> T16", fontsize=14, style = filled, fillcolor = lightgrey];
17 -> 56 [arrowhead = none , style=dashed];
56 -> 7[style=dashed];

57 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
25 -> 57 [arrowhead = none ];
57 -> 34;

58 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
34 -> 58 [arrowhead = none ];
58 -> 37;

59 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
34 -> 59 [arrowhead = none ];
59 -> 38;

60 [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];
37 -> 60 [arrowhead = none ];
60 -> 39;

61 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
37 -> 61 [arrowhead = none ];
61 -> 40;

62 [label="EVAL with clause\nmember1(X64, .(X64, X65)).\nand substitutionT23 -> T50,\nX64 -> T50,\nX65 -> T51,\nT6 -> .(T50, T51)", fontsize=14, style = filled, fillcolor = lightblue];
38 -> 62 [arrowhead = none ];
62 -> 44;

63 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
38 -> 63 [arrowhead = none ];
63 -> 45;

64 [label="INSTANCE with matching:\nT23 -> T40\nT6 -> T42", fontsize=14, style = filled, fillcolor = lightgrey];
39 -> 64 [arrowhead = none , style=dashed];
64 -> 25[style=dashed];

65 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
44 -> 65 [arrowhead = none ];
65 -> 46;

}

(2) Obligation:

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

overlapA_in_gg(T5, T6) → U1_gg(T5, T6, pB_in_agg(X5, T5, T6))
pB_in_agg(X25, .(T15, T16), T6) → U2_agg(X25, T15, T16, T6, pB_in_agg(X25, T16, T6))
pB_in_agg(T23, .(T23, T24), T6) → U3_agg(T23, T24, T6, member1C_in_gg(T23, T6))
member1C_in_gg(T40, .(T41, T42)) → U4_gg(T40, T41, T42, member1C_in_gg(T40, T42))
member1C_in_gg(T50, .(T50, T51)) → member1C_out_gg(T50, .(T50, T51))
U4_gg(T40, T41, T42, member1C_out_gg(T40, T42)) → member1C_out_gg(T40, .(T41, T42))
U3_agg(T23, T24, T6, member1C_out_gg(T23, T6)) → pB_out_agg(T23, .(T23, T24), T6)
U2_agg(X25, T15, T16, T6, pB_out_agg(X25, T16, T6)) → pB_out_agg(X25, .(T15, T16), T6)
U1_gg(T5, T6, pB_out_agg(X5, T5, T6)) → overlapA_out_gg(T5, T6)

The argument filtering Pi contains the following mapping:
overlapA_in_gg(x1, x2)  =  overlapA_in_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x2, x3)
pB_in_agg(x1, x2, x3)  =  pB_in_agg(x2, x3)
.(x1, x2)  =  .(x1, x2)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x2, x3, x4, x5)
U3_agg(x1, x2, x3, x4)  =  U3_agg(x1, x2, x3, x4)
member1C_in_gg(x1, x2)  =  member1C_in_gg(x1, x2)
U4_gg(x1, x2, x3, x4)  =  U4_gg(x1, x2, x3, x4)
member1C_out_gg(x1, x2)  =  member1C_out_gg(x1, x2)
pB_out_agg(x1, x2, x3)  =  pB_out_agg(x1, x2, x3)
overlapA_out_gg(x1, x2)  =  overlapA_out_gg(x1, x2)

(3) 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:

OVERLAPA_IN_GG(T5, T6) → U1_GG(T5, T6, pB_in_agg(X5, T5, T6))
OVERLAPA_IN_GG(T5, T6) → PB_IN_AGG(X5, T5, T6)
PB_IN_AGG(X25, .(T15, T16), T6) → U2_AGG(X25, T15, T16, T6, pB_in_agg(X25, T16, T6))
PB_IN_AGG(X25, .(T15, T16), T6) → PB_IN_AGG(X25, T16, T6)
PB_IN_AGG(T23, .(T23, T24), T6) → U3_AGG(T23, T24, T6, member1C_in_gg(T23, T6))
PB_IN_AGG(T23, .(T23, T24), T6) → MEMBER1C_IN_GG(T23, T6)
MEMBER1C_IN_GG(T40, .(T41, T42)) → U4_GG(T40, T41, T42, member1C_in_gg(T40, T42))
MEMBER1C_IN_GG(T40, .(T41, T42)) → MEMBER1C_IN_GG(T40, T42)

The TRS R consists of the following rules:

overlapA_in_gg(T5, T6) → U1_gg(T5, T6, pB_in_agg(X5, T5, T6))
pB_in_agg(X25, .(T15, T16), T6) → U2_agg(X25, T15, T16, T6, pB_in_agg(X25, T16, T6))
pB_in_agg(T23, .(T23, T24), T6) → U3_agg(T23, T24, T6, member1C_in_gg(T23, T6))
member1C_in_gg(T40, .(T41, T42)) → U4_gg(T40, T41, T42, member1C_in_gg(T40, T42))
member1C_in_gg(T50, .(T50, T51)) → member1C_out_gg(T50, .(T50, T51))
U4_gg(T40, T41, T42, member1C_out_gg(T40, T42)) → member1C_out_gg(T40, .(T41, T42))
U3_agg(T23, T24, T6, member1C_out_gg(T23, T6)) → pB_out_agg(T23, .(T23, T24), T6)
U2_agg(X25, T15, T16, T6, pB_out_agg(X25, T16, T6)) → pB_out_agg(X25, .(T15, T16), T6)
U1_gg(T5, T6, pB_out_agg(X5, T5, T6)) → overlapA_out_gg(T5, T6)

The argument filtering Pi contains the following mapping:
overlapA_in_gg(x1, x2)  =  overlapA_in_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x2, x3)
pB_in_agg(x1, x2, x3)  =  pB_in_agg(x2, x3)
.(x1, x2)  =  .(x1, x2)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x2, x3, x4, x5)
U3_agg(x1, x2, x3, x4)  =  U3_agg(x1, x2, x3, x4)
member1C_in_gg(x1, x2)  =  member1C_in_gg(x1, x2)
U4_gg(x1, x2, x3, x4)  =  U4_gg(x1, x2, x3, x4)
member1C_out_gg(x1, x2)  =  member1C_out_gg(x1, x2)
pB_out_agg(x1, x2, x3)  =  pB_out_agg(x1, x2, x3)
overlapA_out_gg(x1, x2)  =  overlapA_out_gg(x1, x2)
OVERLAPA_IN_GG(x1, x2)  =  OVERLAPA_IN_GG(x1, x2)
U1_GG(x1, x2, x3)  =  U1_GG(x1, x2, x3)
PB_IN_AGG(x1, x2, x3)  =  PB_IN_AGG(x2, x3)
U2_AGG(x1, x2, x3, x4, x5)  =  U2_AGG(x2, x3, x4, x5)
U3_AGG(x1, x2, x3, x4)  =  U3_AGG(x1, x2, x3, x4)
MEMBER1C_IN_GG(x1, x2)  =  MEMBER1C_IN_GG(x1, x2)
U4_GG(x1, x2, x3, x4)  =  U4_GG(x1, x2, x3, x4)

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

(4) Obligation:

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

OVERLAPA_IN_GG(T5, T6) → U1_GG(T5, T6, pB_in_agg(X5, T5, T6))
OVERLAPA_IN_GG(T5, T6) → PB_IN_AGG(X5, T5, T6)
PB_IN_AGG(X25, .(T15, T16), T6) → U2_AGG(X25, T15, T16, T6, pB_in_agg(X25, T16, T6))
PB_IN_AGG(X25, .(T15, T16), T6) → PB_IN_AGG(X25, T16, T6)
PB_IN_AGG(T23, .(T23, T24), T6) → U3_AGG(T23, T24, T6, member1C_in_gg(T23, T6))
PB_IN_AGG(T23, .(T23, T24), T6) → MEMBER1C_IN_GG(T23, T6)
MEMBER1C_IN_GG(T40, .(T41, T42)) → U4_GG(T40, T41, T42, member1C_in_gg(T40, T42))
MEMBER1C_IN_GG(T40, .(T41, T42)) → MEMBER1C_IN_GG(T40, T42)

The TRS R consists of the following rules:

overlapA_in_gg(T5, T6) → U1_gg(T5, T6, pB_in_agg(X5, T5, T6))
pB_in_agg(X25, .(T15, T16), T6) → U2_agg(X25, T15, T16, T6, pB_in_agg(X25, T16, T6))
pB_in_agg(T23, .(T23, T24), T6) → U3_agg(T23, T24, T6, member1C_in_gg(T23, T6))
member1C_in_gg(T40, .(T41, T42)) → U4_gg(T40, T41, T42, member1C_in_gg(T40, T42))
member1C_in_gg(T50, .(T50, T51)) → member1C_out_gg(T50, .(T50, T51))
U4_gg(T40, T41, T42, member1C_out_gg(T40, T42)) → member1C_out_gg(T40, .(T41, T42))
U3_agg(T23, T24, T6, member1C_out_gg(T23, T6)) → pB_out_agg(T23, .(T23, T24), T6)
U2_agg(X25, T15, T16, T6, pB_out_agg(X25, T16, T6)) → pB_out_agg(X25, .(T15, T16), T6)
U1_gg(T5, T6, pB_out_agg(X5, T5, T6)) → overlapA_out_gg(T5, T6)

The argument filtering Pi contains the following mapping:
overlapA_in_gg(x1, x2)  =  overlapA_in_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x2, x3)
pB_in_agg(x1, x2, x3)  =  pB_in_agg(x2, x3)
.(x1, x2)  =  .(x1, x2)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x2, x3, x4, x5)
U3_agg(x1, x2, x3, x4)  =  U3_agg(x1, x2, x3, x4)
member1C_in_gg(x1, x2)  =  member1C_in_gg(x1, x2)
U4_gg(x1, x2, x3, x4)  =  U4_gg(x1, x2, x3, x4)
member1C_out_gg(x1, x2)  =  member1C_out_gg(x1, x2)
pB_out_agg(x1, x2, x3)  =  pB_out_agg(x1, x2, x3)
overlapA_out_gg(x1, x2)  =  overlapA_out_gg(x1, x2)
OVERLAPA_IN_GG(x1, x2)  =  OVERLAPA_IN_GG(x1, x2)
U1_GG(x1, x2, x3)  =  U1_GG(x1, x2, x3)
PB_IN_AGG(x1, x2, x3)  =  PB_IN_AGG(x2, x3)
U2_AGG(x1, x2, x3, x4, x5)  =  U2_AGG(x2, x3, x4, x5)
U3_AGG(x1, x2, x3, x4)  =  U3_AGG(x1, x2, x3, x4)
MEMBER1C_IN_GG(x1, x2)  =  MEMBER1C_IN_GG(x1, x2)
U4_GG(x1, x2, x3, x4)  =  U4_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:

MEMBER1C_IN_GG(T40, .(T41, T42)) → MEMBER1C_IN_GG(T40, T42)

The TRS R consists of the following rules:

overlapA_in_gg(T5, T6) → U1_gg(T5, T6, pB_in_agg(X5, T5, T6))
pB_in_agg(X25, .(T15, T16), T6) → U2_agg(X25, T15, T16, T6, pB_in_agg(X25, T16, T6))
pB_in_agg(T23, .(T23, T24), T6) → U3_agg(T23, T24, T6, member1C_in_gg(T23, T6))
member1C_in_gg(T40, .(T41, T42)) → U4_gg(T40, T41, T42, member1C_in_gg(T40, T42))
member1C_in_gg(T50, .(T50, T51)) → member1C_out_gg(T50, .(T50, T51))
U4_gg(T40, T41, T42, member1C_out_gg(T40, T42)) → member1C_out_gg(T40, .(T41, T42))
U3_agg(T23, T24, T6, member1C_out_gg(T23, T6)) → pB_out_agg(T23, .(T23, T24), T6)
U2_agg(X25, T15, T16, T6, pB_out_agg(X25, T16, T6)) → pB_out_agg(X25, .(T15, T16), T6)
U1_gg(T5, T6, pB_out_agg(X5, T5, T6)) → overlapA_out_gg(T5, T6)

The argument filtering Pi contains the following mapping:
overlapA_in_gg(x1, x2)  =  overlapA_in_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x2, x3)
pB_in_agg(x1, x2, x3)  =  pB_in_agg(x2, x3)
.(x1, x2)  =  .(x1, x2)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x2, x3, x4, x5)
U3_agg(x1, x2, x3, x4)  =  U3_agg(x1, x2, x3, x4)
member1C_in_gg(x1, x2)  =  member1C_in_gg(x1, x2)
U4_gg(x1, x2, x3, x4)  =  U4_gg(x1, x2, x3, x4)
member1C_out_gg(x1, x2)  =  member1C_out_gg(x1, x2)
pB_out_agg(x1, x2, x3)  =  pB_out_agg(x1, x2, x3)
overlapA_out_gg(x1, x2)  =  overlapA_out_gg(x1, x2)
MEMBER1C_IN_GG(x1, x2)  =  MEMBER1C_IN_GG(x1, x2)

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

(8) UsableRulesProof (EQUIVALENT transformation)

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

(9) Obligation:

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

MEMBER1C_IN_GG(T40, .(T41, T42)) → MEMBER1C_IN_GG(T40, T42)

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

(10) PiDPToQDPProof (EQUIVALENT transformation)

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

(11) Obligation:

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

MEMBER1C_IN_GG(T40, .(T41, T42)) → MEMBER1C_IN_GG(T40, T42)

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

(12) 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:

• MEMBER1C_IN_GG(T40, .(T41, T42)) → MEMBER1C_IN_GG(T40, T42)
  The graph contains the following edges 1 >= 1, 2 > 2

(13) YES

(14) Obligation:

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

PB_IN_AGG(X25, .(T15, T16), T6) → PB_IN_AGG(X25, T16, T6)

The TRS R consists of the following rules:

overlapA_in_gg(T5, T6) → U1_gg(T5, T6, pB_in_agg(X5, T5, T6))
pB_in_agg(X25, .(T15, T16), T6) → U2_agg(X25, T15, T16, T6, pB_in_agg(X25, T16, T6))
pB_in_agg(T23, .(T23, T24), T6) → U3_agg(T23, T24, T6, member1C_in_gg(T23, T6))
member1C_in_gg(T40, .(T41, T42)) → U4_gg(T40, T41, T42, member1C_in_gg(T40, T42))
member1C_in_gg(T50, .(T50, T51)) → member1C_out_gg(T50, .(T50, T51))
U4_gg(T40, T41, T42, member1C_out_gg(T40, T42)) → member1C_out_gg(T40, .(T41, T42))
U3_agg(T23, T24, T6, member1C_out_gg(T23, T6)) → pB_out_agg(T23, .(T23, T24), T6)
U2_agg(X25, T15, T16, T6, pB_out_agg(X25, T16, T6)) → pB_out_agg(X25, .(T15, T16), T6)
U1_gg(T5, T6, pB_out_agg(X5, T5, T6)) → overlapA_out_gg(T5, T6)

The argument filtering Pi contains the following mapping:
overlapA_in_gg(x1, x2)  =  overlapA_in_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x2, x3)
pB_in_agg(x1, x2, x3)  =  pB_in_agg(x2, x3)
.(x1, x2)  =  .(x1, x2)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x2, x3, x4, x5)
U3_agg(x1, x2, x3, x4)  =  U3_agg(x1, x2, x3, x4)
member1C_in_gg(x1, x2)  =  member1C_in_gg(x1, x2)
U4_gg(x1, x2, x3, x4)  =  U4_gg(x1, x2, x3, x4)
member1C_out_gg(x1, x2)  =  member1C_out_gg(x1, x2)
pB_out_agg(x1, x2, x3)  =  pB_out_agg(x1, x2, x3)
overlapA_out_gg(x1, x2)  =  overlapA_out_gg(x1, x2)
PB_IN_AGG(x1, x2, x3)  =  PB_IN_AGG(x2, x3)

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

(15) UsableRulesProof (EQUIVALENT transformation)

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

(16) Obligation:

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

PB_IN_AGG(X25, .(T15, T16), T6) → PB_IN_AGG(X25, T16, 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

(17) PiDPToQDPProof (SOUND transformation)

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

(18) Obligation:

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

PB_IN_AGG(.(T15, T16), T6) → PB_IN_AGG(T16, T6)

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

(19) 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(.(T15, T16), T6) → PB_IN_AGG(T16, T6)
  The graph contains the following edges 1 > 1, 2 >= 2

(20) YES