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

(0) Obligation:

Clauses:

list([]).
list(.(X, XS)) :- list(XS).
s2l(s(X), .(Y, Xs)) :- s2l(X, Xs).
s2l(0, []).
goal(X) :- ','(s2l(X, XS), list(XS)).

Query: goal(g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="C: goal(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: goal(T1), SCOPE: 1, CLAUSE: 4\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(s2l(T3, X3), list(X3))\n\nT3 is ground\nX3 is free", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: ','(s2l(T3, X3), list(X3)), SCOPE: 2, CLAUSE: 2 | ','(s2l(T3, X3), list(X3)), SCOPE: 2, CLAUSE: 3\n\nT3 is ground\nX3 is free", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ','(s2l(T3, X3), list(X3)), SCOPE: 2, CLAUSE: 2\n\nT3 is ground\nX3 is free", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ','(s2l(T3, X3), list(X3)), SCOPE: 2, CLAUSE: 3\n\nT3 is ground\nX3 is free", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(s2l(T8, X28), list(.(X27, X28)))\n\nT8 is ground\nX27 is free\nX28 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="A: s2l(T8, X28)\n\nT8 is ground\nX28 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: list(.(X27, T10))\n\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: s2l(T8, X28), SCOPE: 3, CLAUSE: 2 | s2l(T8, X28), SCOPE: 3, CLAUSE: 3\n\nT8 is ground\nX28 is free", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: s2l(T8, X28), SCOPE: 3, CLAUSE: 2\n\nT8 is ground\nX28 is free", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: s2l(T8, X28), SCOPE: 3, CLAUSE: 3\n\nT8 is ground\nX28 is free", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: s2l(T16, X63)\n\nT16 is ground\nX63 is free", 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="19: list(.(X27, T10)), SCOPE: 4, CLAUSE: 0 | list(.(X27, T10)), SCOPE: 4, CLAUSE: 1\n\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: list(.(X27, T10)), SCOPE: 4, CLAUSE: 1\n\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="B: list(T22)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: list(T22), SCOPE: 5, CLAUSE: 0 | list(T22), SCOPE: 5, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: list(T22), SCOPE: 5, CLAUSE: 0\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: list(T22), SCOPE: 5, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: list(T29)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: list([])\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: list([]), SCOPE: 6, CLAUSE: 0 | list([]), SCOPE: 6, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: list([]), SCOPE: 6, CLAUSE: 0\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: list([]), SCOPE: 6, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="36: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
37 [label="37: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
38 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 38 [arrowhead = none ];
38 -> 2;

39 [label="ONLY EVAL with clause\ngoal(X2) :- ','(s2l(X2, X3), list(X3)).\nand substitutionT1 -> T3,\nX2 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 39 [arrowhead = none ];
39 -> 3;

40 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 40 [arrowhead = none ];
40 -> 4;

41 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 41 [arrowhead = none ];
41 -> 5;

42 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 42 [arrowhead = none ];
42 -> 6;

43 [label="EVAL with clause\ns2l(s(X24), .(X25, X26)) :- s2l(X24, X26).\nand substitutionX24 -> T8,\nT3 -> s(T8),\nX25 -> X27,\nX26 -> X28,\nX3 -> .(X27, X28)", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 43 [arrowhead = none ];
43 -> 7;

44 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 44 [arrowhead = none ];
44 -> 8;

45 [label="EVAL with clause\ns2l(0, []).\nand substitutionT3 -> 0,\nX3 -> []", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 45 [arrowhead = none ];
45 -> 30;

46 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 46 [arrowhead = none ];
46 -> 31;

47 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
7 -> 47 [arrowhead = none ];
47 -> 9;

48 [label="SPLIT 2\nnew knowledge:\nT8 is ground\nreplacements:X28 -> T10", fontsize=14, style = filled, fillcolor = violet];
7 -> 48 [arrowhead = none ];
48 -> 10;

49 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
9 -> 49 [arrowhead = none ];
49 -> 11;

50 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
10 -> 50 [arrowhead = none ];
50 -> 19;

51 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
11 -> 51 [arrowhead = none ];
51 -> 12;

52 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
11 -> 52 [arrowhead = none ];
52 -> 13;

53 [label="EVAL with clause\ns2l(s(X59), .(X60, X61)) :- s2l(X59, X61).\nand substitutionX59 -> T16,\nT8 -> s(T16),\nX60 -> X62,\nX61 -> X63,\nX28 -> .(X62, X63)", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 53 [arrowhead = none ];
53 -> 14;

54 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 54 [arrowhead = none ];
54 -> 15;

55 [label="EVAL with clause\ns2l(0, []).\nand substitutionT8 -> 0,\nX28 -> []", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 55 [arrowhead = none ];
55 -> 16;

56 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 56 [arrowhead = none ];
56 -> 17;

57 [label="INSTANCE with matching:\nT8 -> T16\nX28 -> X63", fontsize=14, style = filled, fillcolor = lightgrey];
14 -> 57 [arrowhead = none , style=dashed];
57 -> 9[style=dashed];

58 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 58 [arrowhead = none ];
58 -> 18;

59 [label="BACKTRACK\nfor clause: list([])because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
19 -> 59 [arrowhead = none ];
59 -> 20;

60 [label="ONLY EVAL with clause\nlist(.(X75, X76)) :- list(X76).\nand substitutionX27 -> X77,\nX75 -> X77,\nT10 -> T22,\nX76 -> T22,\nT21 -> T22", fontsize=14, style = filled, fillcolor = lightblue];
20 -> 60 [arrowhead = none ];
60 -> 21;

61 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
21 -> 61 [arrowhead = none ];
61 -> 22;

62 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
22 -> 62 [arrowhead = none ];
62 -> 23;

63 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
22 -> 63 [arrowhead = none ];
63 -> 24;

64 [label="EVAL with clause\nlist([]).\nand substitutionT22 -> []", fontsize=14, style = filled, fillcolor = lightblue];
23 -> 64 [arrowhead = none ];
64 -> 25;

65 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
23 -> 65 [arrowhead = none ];
65 -> 26;

66 [label="EVAL with clause\nlist(.(X82, X83)) :- list(X83).\nand substitutionX82 -> T27,\nX83 -> T29,\nT22 -> .(T27, T29),\nT28 -> T29", fontsize=14, style = filled, fillcolor = lightblue];
24 -> 66 [arrowhead = none ];
66 -> 28;

67 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
24 -> 67 [arrowhead = none ];
67 -> 29;

68 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 68 [arrowhead = none ];
68 -> 27;

69 [label="INSTANCE with matching:\nT22 -> T29", fontsize=14, style = filled, fillcolor = lightgrey];
28 -> 69 [arrowhead = none , style=dashed];
69 -> 21[style=dashed];

70 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
30 -> 70 [arrowhead = none ];
70 -> 32;

71 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
32 -> 71 [arrowhead = none ];
71 -> 33;

72 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
32 -> 72 [arrowhead = none ];
72 -> 34;

73 [label="ONLY EVAL with clause\nlist([]).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
33 -> 73 [arrowhead = none ];
73 -> 35;

74 [label="BACKTRACK\nfor clause: list(.(X, XS)) :- list(XS)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
34 -> 74 [arrowhead = none ];
74 -> 37;

75 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
35 -> 75 [arrowhead = none ];
75 -> 36;

}

(2) Obligation:

Triples:

s2lA(s(X1), .(X2, X3)) :- s2lA(X1, X3).
listB(.(X1, X2)) :- listB(X2).
goalC(s(X1)) :- s2lA(X1, X2).
goalC(s(X1)) :- ','(s2lcA(X1, X2), listB(X2)).

Clauses:

s2lcA(s(X1), .(X2, X3)) :- s2lcA(X1, X3).
s2lcA(0, []).
listcB([]).
listcB(.(X1, X2)) :- listcB(X2).

Afs:

goalC(x1)  =  goalC(x1)

(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:
goalC_in: (b)
s2lA_in: (b,f)
s2lcA_in: (b,f)
listB_in: (b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

GOALC_IN_G(s(X1)) → U3_G(X1, s2lA_in_ga(X1, X2))
GOALC_IN_G(s(X1)) → S2LA_IN_GA(X1, X2)
S2LA_IN_GA(s(X1), .(X2, X3)) → U1_GA(X1, X2, X3, s2lA_in_ga(X1, X3))
S2LA_IN_GA(s(X1), .(X2, X3)) → S2LA_IN_GA(X1, X3)
GOALC_IN_G(s(X1)) → U4_G(X1, s2lcA_in_ga(X1, X2))
U4_G(X1, s2lcA_out_ga(X1, X2)) → U5_G(X1, listB_in_g(X2))
U4_G(X1, s2lcA_out_ga(X1, X2)) → LISTB_IN_G(X2)
LISTB_IN_G(.(X1, X2)) → U2_G(X1, X2, listB_in_g(X2))
LISTB_IN_G(.(X1, X2)) → LISTB_IN_G(X2)

The TRS R consists of the following rules:

s2lcA_in_ga(s(X1), .(X2, X3)) → U7_ga(X1, X2, X3, s2lcA_in_ga(X1, X3))
s2lcA_in_ga(0, []) → s2lcA_out_ga(0, [])
U7_ga(X1, X2, X3, s2lcA_out_ga(X1, X3)) → s2lcA_out_ga(s(X1), .(X2, X3))

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
s2lA_in_ga(x1, x2)  =  s2lA_in_ga(x1)
.(x1, x2)  =  .(x2)
s2lcA_in_ga(x1, x2)  =  s2lcA_in_ga(x1)
U7_ga(x1, x2, x3, x4)  =  U7_ga(x1, x4)
0  =  0
s2lcA_out_ga(x1, x2)  =  s2lcA_out_ga(x1, x2)
listB_in_g(x1)  =  listB_in_g(x1)
GOALC_IN_G(x1)  =  GOALC_IN_G(x1)
U3_G(x1, x2)  =  U3_G(x1, x2)
S2LA_IN_GA(x1, x2)  =  S2LA_IN_GA(x1)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x1, x4)
U4_G(x1, x2)  =  U4_G(x1, x2)
U5_G(x1, x2)  =  U5_G(x1, x2)
LISTB_IN_G(x1)  =  LISTB_IN_G(x1)
U2_G(x1, x2, x3)  =  U2_G(x2, x3)

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:

GOALC_IN_G(s(X1)) → U3_G(X1, s2lA_in_ga(X1, X2))
GOALC_IN_G(s(X1)) → S2LA_IN_GA(X1, X2)
S2LA_IN_GA(s(X1), .(X2, X3)) → U1_GA(X1, X2, X3, s2lA_in_ga(X1, X3))
S2LA_IN_GA(s(X1), .(X2, X3)) → S2LA_IN_GA(X1, X3)
GOALC_IN_G(s(X1)) → U4_G(X1, s2lcA_in_ga(X1, X2))
U4_G(X1, s2lcA_out_ga(X1, X2)) → U5_G(X1, listB_in_g(X2))
U4_G(X1, s2lcA_out_ga(X1, X2)) → LISTB_IN_G(X2)
LISTB_IN_G(.(X1, X2)) → U2_G(X1, X2, listB_in_g(X2))
LISTB_IN_G(.(X1, X2)) → LISTB_IN_G(X2)

The TRS R consists of the following rules:

s2lcA_in_ga(s(X1), .(X2, X3)) → U7_ga(X1, X2, X3, s2lcA_in_ga(X1, X3))
s2lcA_in_ga(0, []) → s2lcA_out_ga(0, [])
U7_ga(X1, X2, X3, s2lcA_out_ga(X1, X3)) → s2lcA_out_ga(s(X1), .(X2, X3))

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
s2lA_in_ga(x1, x2)  =  s2lA_in_ga(x1)
.(x1, x2)  =  .(x2)
s2lcA_in_ga(x1, x2)  =  s2lcA_in_ga(x1)
U7_ga(x1, x2, x3, x4)  =  U7_ga(x1, x4)
0  =  0
s2lcA_out_ga(x1, x2)  =  s2lcA_out_ga(x1, x2)
listB_in_g(x1)  =  listB_in_g(x1)
GOALC_IN_G(x1)  =  GOALC_IN_G(x1)
U3_G(x1, x2)  =  U3_G(x1, x2)
S2LA_IN_GA(x1, x2)  =  S2LA_IN_GA(x1)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x1, x4)
U4_G(x1, x2)  =  U4_G(x1, x2)
U5_G(x1, x2)  =  U5_G(x1, x2)
LISTB_IN_G(x1)  =  LISTB_IN_G(x1)
U2_G(x1, x2, x3)  =  U2_G(x2, x3)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 7 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

LISTB_IN_G(.(X1, X2)) → LISTB_IN_G(X2)

The TRS R consists of the following rules:

s2lcA_in_ga(s(X1), .(X2, X3)) → U7_ga(X1, X2, X3, s2lcA_in_ga(X1, X3))
s2lcA_in_ga(0, []) → s2lcA_out_ga(0, [])
U7_ga(X1, X2, X3, s2lcA_out_ga(X1, X3)) → s2lcA_out_ga(s(X1), .(X2, X3))

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
.(x1, x2)  =  .(x2)
s2lcA_in_ga(x1, x2)  =  s2lcA_in_ga(x1)
U7_ga(x1, x2, x3, x4)  =  U7_ga(x1, x4)
0  =  0
s2lcA_out_ga(x1, x2)  =  s2lcA_out_ga(x1, x2)
LISTB_IN_G(x1)  =  LISTB_IN_G(x1)

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:

LISTB_IN_G(.(X1, X2)) → LISTB_IN_G(X2)

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

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

(10) PiDPToQDPProof (SOUND 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:

LISTB_IN_G(.(X2)) → LISTB_IN_G(X2)

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:

• LISTB_IN_G(.(X2)) → LISTB_IN_G(X2)
  The graph contains the following edges 1 > 1

(13) YES

(14) Obligation:

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

S2LA_IN_GA(s(X1), .(X2, X3)) → S2LA_IN_GA(X1, X3)

The TRS R consists of the following rules:

s2lcA_in_ga(s(X1), .(X2, X3)) → U7_ga(X1, X2, X3, s2lcA_in_ga(X1, X3))
s2lcA_in_ga(0, []) → s2lcA_out_ga(0, [])
U7_ga(X1, X2, X3, s2lcA_out_ga(X1, X3)) → s2lcA_out_ga(s(X1), .(X2, X3))

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
.(x1, x2)  =  .(x2)
s2lcA_in_ga(x1, x2)  =  s2lcA_in_ga(x1)
U7_ga(x1, x2, x3, x4)  =  U7_ga(x1, x4)
0  =  0
s2lcA_out_ga(x1, x2)  =  s2lcA_out_ga(x1, x2)
S2LA_IN_GA(x1, x2)  =  S2LA_IN_GA(x1)

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:

S2LA_IN_GA(s(X1), .(X2, X3)) → S2LA_IN_GA(X1, X3)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
.(x1, x2)  =  .(x2)
S2LA_IN_GA(x1, x2)  =  S2LA_IN_GA(x1)

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:

S2LA_IN_GA(s(X1)) → S2LA_IN_GA(X1)

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:

• S2LA_IN_GA(s(X1)) → S2LA_IN_GA(X1)
  The graph contains the following edges 1 > 1

(20) YES