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

(0) Obligation:

Clauses:

sublist(Xs, Ys) :- ','(app(X1, Zs, Ys), app(Xs, X2, Zs)).
app([], X, X).
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).

Query: sublist(a,g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="C: sublist(T1, T2)\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: sublist(T1, T2), SCOPE: 1, CLAUSE: 0\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="B: ','(app(X7, X8, T6), app(T7, X9, X8))\n\nT6 is ground\nX7 is free\nX8 is free\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: ','(app(X7, X8, T6), app(T7, X9, X8)), SCOPE: 2, CLAUSE: 1 | ','(app(X7, X8, T6), app(T7, X9, X8)), SCOPE: 2, CLAUSE: 2\n\nT6 is ground\nX7 is free\nX8 is free\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ','(app(X7, X8, T6), app(T7, X9, X8)), SCOPE: 2, CLAUSE: 1\n\nT6 is ground\nX7 is free\nX8 is free\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ','(app(X7, X8, T6), app(T7, X9, X8)), SCOPE: 2, CLAUSE: 2\n\nT6 is ground\nX7 is free\nX8 is free\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="A: app(T7, X9, T16)\n\nT16 is ground\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: app(T7, X9, T16), SCOPE: 3, CLAUSE: 1 | app(T7, X9, T16), SCOPE: 3, CLAUSE: 2\n\nT16 is ground\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: app(T7, X9, T16), SCOPE: 3, CLAUSE: 1\n\nT16 is ground\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: app(T7, X9, T16), SCOPE: 3, CLAUSE: 2\n\nT16 is ground\nX9 is free", 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: app(T33, X56, T32)\n\nT32 is ground\nX56 is free", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(app(X80, X81, T41), app(T7, X9, X81))\n\nT41 is ground\nX9 is free\nX80 is free\nX81 is free", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 18 [arrowhead = none ];
18 -> 2;

19 [label="ONLY EVAL with clause\nsublist(X5, X6) :- ','(app(X7, X8, X6), app(X5, X9, X8)).\nand substitutionT1 -> T7,\nX5 -> T7,\nT2 -> T6,\nX6 -> T6,\nT5 -> T7", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 19 [arrowhead = none ];
19 -> 3;

20 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 20 [arrowhead = none ];
20 -> 4;

21 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 21 [arrowhead = none ];
21 -> 5;

22 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 22 [arrowhead = none ];
22 -> 6;

23 [label="ONLY EVAL with clause\napp([], X26, X26).\nand substitutionX7 -> [],\nX8 -> T16,\nX26 -> T16,\nT6 -> T16,\nX27 -> T16", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 23 [arrowhead = none ];
23 -> 7;

24 [label="EVAL with clause\napp(.(X75, X76), X77, .(X75, X78)) :- app(X76, X77, X78).\nand substitutionX75 -> T40,\nX76 -> X80,\nX7 -> .(T40, X80),\nX8 -> X81,\nX77 -> X81,\nX79 -> T40,\nX78 -> T41,\nT6 -> .(T40, T41)", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 24 [arrowhead = none ];
24 -> 16;

25 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 25 [arrowhead = none ];
25 -> 17;

26 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 26 [arrowhead = none ];
26 -> 8;

27 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
8 -> 27 [arrowhead = none ];
27 -> 9;

28 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
8 -> 28 [arrowhead = none ];
28 -> 10;

29 [label="EVAL with clause\napp([], X40, X40).\nand substitutionT7 -> [],\nX9 -> T23,\nX40 -> T23,\nT16 -> T23,\nX41 -> T23", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 29 [arrowhead = none ];
29 -> 11;

30 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 30 [arrowhead = none ];
30 -> 12;

31 [label="EVAL with clause\napp(.(X52, X53), X54, .(X52, X55)) :- app(X53, X54, X55).\nand substitutionX52 -> T30,\nX53 -> T33,\nT7 -> .(T30, T33),\nX9 -> X56,\nX54 -> X56,\nX55 -> T32,\nT16 -> .(T30, T32),\nT31 -> T33", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 31 [arrowhead = none ];
31 -> 14;

32 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 32 [arrowhead = none ];
32 -> 15;

33 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 33 [arrowhead = none ];
33 -> 13;

34 [label="INSTANCE with matching:\nT7 -> T33\nX9 -> X56\nT16 -> T32", fontsize=14, style = filled, fillcolor = lightgrey];
14 -> 34 [arrowhead = none , style=dashed];
34 -> 7[style=dashed];

35 [label="INSTANCE with matching:\nX7 -> X80\nX8 -> X81\nT6 -> T41", fontsize=14, style = filled, fillcolor = lightgrey];
16 -> 35 [arrowhead = none , style=dashed];
35 -> 3[style=dashed];

}

(2) Obligation:

Triples:

appA(.(X1, X2), X3, .(X1, X4)) :- appA(X2, X3, X4).
pB([], X1, X1, X2, X3) :- appA(X2, X3, X1).
pB(.(X1, X2), X3, .(X1, X4), X5, X6) :- pB(X2, X3, X4, X5, X6).
sublistC(X1, X2) :- pB(X3, X4, X2, X1, X5).

Clauses:

appcA([], X1, X1).
appcA(.(X1, X2), X3, .(X1, X4)) :- appcA(X2, X3, X4).
qcB([], X1, X1, X2, X3) :- appcA(X2, X3, X1).
qcB(.(X1, X2), X3, .(X1, X4), X5, X6) :- qcB(X2, X3, X4, X5, X6).

Afs:

sublistC(x1, x2)  =  sublistC(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:
sublistC_in: (f,b)
pB_in: (f,f,b,f,f)
appA_in: (f,f,b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

SUBLISTC_IN_AG(X1, X2) → U4_AG(X1, X2, pB_in_aagaa(X3, X4, X2, X1, X5))
SUBLISTC_IN_AG(X1, X2) → PB_IN_AAGAA(X3, X4, X2, X1, X5)
PB_IN_AAGAA([], X1, X1, X2, X3) → U2_AAGAA(X1, X2, X3, appA_in_aag(X2, X3, X1))
PB_IN_AAGAA([], X1, X1, X2, X3) → APPA_IN_AAG(X2, X3, X1)
APPA_IN_AAG(.(X1, X2), X3, .(X1, X4)) → U1_AAG(X1, X2, X3, X4, appA_in_aag(X2, X3, X4))
APPA_IN_AAG(.(X1, X2), X3, .(X1, X4)) → APPA_IN_AAG(X2, X3, X4)
PB_IN_AAGAA(.(X1, X2), X3, .(X1, X4), X5, X6) → U3_AAGAA(X1, X2, X3, X4, X5, X6, pB_in_aagaa(X2, X3, X4, X5, X6))
PB_IN_AAGAA(.(X1, X2), X3, .(X1, X4), X5, X6) → PB_IN_AAGAA(X2, X3, X4, X5, X6)

R is empty.
The argument filtering Pi contains the following mapping:
pB_in_aagaa(x1, x2, x3, x4, x5)  =  pB_in_aagaa(x3)
appA_in_aag(x1, x2, x3)  =  appA_in_aag(x3)
.(x1, x2)  =  .(x1, x2)
[]  =  []
SUBLISTC_IN_AG(x1, x2)  =  SUBLISTC_IN_AG(x2)
U4_AG(x1, x2, x3)  =  U4_AG(x2, x3)
PB_IN_AAGAA(x1, x2, x3, x4, x5)  =  PB_IN_AAGAA(x3)
U2_AAGAA(x1, x2, x3, x4)  =  U2_AAGAA(x1, x4)
APPA_IN_AAG(x1, x2, x3)  =  APPA_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x1, x4, x5)
U3_AAGAA(x1, x2, x3, x4, x5, x6, x7)  =  U3_AAGAA(x1, x4, x7)

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:

SUBLISTC_IN_AG(X1, X2) → U4_AG(X1, X2, pB_in_aagaa(X3, X4, X2, X1, X5))
SUBLISTC_IN_AG(X1, X2) → PB_IN_AAGAA(X3, X4, X2, X1, X5)
PB_IN_AAGAA([], X1, X1, X2, X3) → U2_AAGAA(X1, X2, X3, appA_in_aag(X2, X3, X1))
PB_IN_AAGAA([], X1, X1, X2, X3) → APPA_IN_AAG(X2, X3, X1)
APPA_IN_AAG(.(X1, X2), X3, .(X1, X4)) → U1_AAG(X1, X2, X3, X4, appA_in_aag(X2, X3, X4))
APPA_IN_AAG(.(X1, X2), X3, .(X1, X4)) → APPA_IN_AAG(X2, X3, X4)
PB_IN_AAGAA(.(X1, X2), X3, .(X1, X4), X5, X6) → U3_AAGAA(X1, X2, X3, X4, X5, X6, pB_in_aagaa(X2, X3, X4, X5, X6))
PB_IN_AAGAA(.(X1, X2), X3, .(X1, X4), X5, X6) → PB_IN_AAGAA(X2, X3, X4, X5, X6)

R is empty.
The argument filtering Pi contains the following mapping:
pB_in_aagaa(x1, x2, x3, x4, x5)  =  pB_in_aagaa(x3)
appA_in_aag(x1, x2, x3)  =  appA_in_aag(x3)
.(x1, x2)  =  .(x1, x2)
[]  =  []
SUBLISTC_IN_AG(x1, x2)  =  SUBLISTC_IN_AG(x2)
U4_AG(x1, x2, x3)  =  U4_AG(x2, x3)
PB_IN_AAGAA(x1, x2, x3, x4, x5)  =  PB_IN_AAGAA(x3)
U2_AAGAA(x1, x2, x3, x4)  =  U2_AAGAA(x1, x4)
APPA_IN_AAG(x1, x2, x3)  =  APPA_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x1, x4, x5)
U3_AAGAA(x1, x2, x3, x4, x5, x6, x7)  =  U3_AAGAA(x1, x4, x7)

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:

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

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

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

(8) PiDPToQDPProof (SOUND 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:

APPA_IN_AAG(.(X1, X4)) → APPA_IN_AAG(X4)

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:

• APPA_IN_AAG(.(X1, X4)) → APPA_IN_AAG(X4)
  The graph contains the following edges 1 > 1

(11) YES

(12) Obligation:

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

PB_IN_AAGAA(.(X1, X2), X3, .(X1, X4), X5, X6) → PB_IN_AAGAA(X2, X3, X4, X5, X6)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
PB_IN_AAGAA(x1, x2, x3, x4, x5)  =  PB_IN_AAGAA(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:

PB_IN_AAGAA(.(X1, X4)) → PB_IN_AAGAA(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:

• PB_IN_AAGAA(.(X1, X4)) → PB_IN_AAGAA(X4)
  The graph contains the following edges 1 > 1

(16) YES