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

(0) Obligation:

Clauses:

ordered([]).
ordered(.(X, [])).
ordered(.(X, .(Y, Xs))) :- ','(le(X, Y), ordered(.(Y, Xs))).
le(s(X), s(Y)) :- le(X, Y).
le(0, s(0)).
le(0, 0).

Query: ordered(g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: ordered(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: ordered(T1), SCOPE: 1, CLAUSE: 0 | ordered(T1), SCOPE: 1, CLAUSE: 1 | ordered(T1), SCOPE: 1, CLAUSE: 2\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | ordered([]), SCOPE: 1, CLAUSE: 1 | ordered([]), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: ordered(T1), SCOPE: 1, CLAUSE: 1 | ordered(T1), SCOPE: 1, CLAUSE: 2\n\nT1 is ground\nordered(T1) !=? ordered([])", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ordered([]), SCOPE: 1, CLAUSE: 1 | ordered([]), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ordered([]), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: true | ordered(.(T3, [])), SCOPE: 1, CLAUSE: 2\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ordered(T1), SCOPE: 1, CLAUSE: 2\n\nT1 is ground\nX6 is free\nordered(T1) !=? ordered([])\nordered(T1) !=? ordered(.(X6, []))", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ordered(.(T3, [])), SCOPE: 1, CLAUSE: 2\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(le(T8, T9), ordered(.(T9, T10)))\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(le(T8, T9), ordered(.(T9, T10))), SCOPE: 2, CLAUSE: 3 | ','(le(T8, T9), ordered(.(T9, T10))), SCOPE: 2, CLAUSE: 4 | ','(le(T8, T9), ordered(.(T9, T10))), SCOPE: 2, CLAUSE: 5\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(le(T8, T9), ordered(.(T9, T10))), SCOPE: 2, CLAUSE: 3\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(le(T8, T9), ordered(.(T9, T10))), SCOPE: 2, CLAUSE: 4 | ','(le(T8, T9), ordered(.(T9, T10))), SCOPE: 2, CLAUSE: 5\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(le(T19, T20), ordered(.(s(T20), T10)))\n\nT10 is ground\nT19 is ground\nT20 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="B: le(T19, T20)\n\nT19 is ground\nT20 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: ordered(.(s(T20), T10))\n\nT10 is ground\nT20 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: le(T19, T20), SCOPE: 3, CLAUSE: 3 | le(T19, T20), SCOPE: 3, CLAUSE: 4 | le(T19, T20), SCOPE: 3, CLAUSE: 5\n\nT19 is ground\nT20 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: le(T19, T20), SCOPE: 3, CLAUSE: 3\n\nT19 is ground\nT20 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: le(T19, T20), SCOPE: 3, CLAUSE: 4 | le(T19, T20), SCOPE: 3, CLAUSE: 5\n\nT19 is ground\nT20 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: le(T33, T34)\n\nT33 is ground\nT34 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: le(T19, T20), SCOPE: 3, CLAUSE: 4\n\nT19 is ground\nT20 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: le(T19, T20), SCOPE: 3, CLAUSE: 5\n\nT19 is ground\nT20 is ground", 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: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: ','(le(T8, T9), ordered(.(T9, T10))), SCOPE: 2, CLAUSE: 4\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: ','(le(T8, T9), ordered(.(T9, T10))), SCOPE: 2, CLAUSE: 5\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="36: ordered(.(s(0), T10))\n\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
37 [label="37: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
38 [label="38: ordered(.(0, T10))\n\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
39 [label="39: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
40 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 40 [arrowhead = none ];
40 -> 2;

41 [label="EVAL with clause\nordered([]).\nand substitutionT1 -> []", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 41 [arrowhead = none ];
41 -> 3;

42 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 42 [arrowhead = none ];
42 -> 4;

43 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 43 [arrowhead = none ];
43 -> 5;

44 [label="EVAL with clause\nordered(.(X6, [])).\nand substitutionX6 -> T3,\nT1 -> .(T3, [])", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 44 [arrowhead = none ];
44 -> 8;

45 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 45 [arrowhead = none ];
45 -> 9;

46 [label="BACKTRACK\nfor clause: ordered(.(X, []))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 46 [arrowhead = none ];
46 -> 6;

47 [label="BACKTRACK\nfor clause: ordered(.(X, .(Y, Xs))) :- ','(le(X, Y), ordered(.(Y, Xs)))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 47 [arrowhead = none ];
47 -> 7;

48 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 48 [arrowhead = none ];
48 -> 10;

49 [label="EVAL with clause\nordered(.(X13, .(X14, X15))) :- ','(le(X13, X14), ordered(.(X14, X15))).\nand substitutionX13 -> T8,\nX14 -> T9,\nX15 -> T10,\nT1 -> .(T8, .(T9, T10))", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 49 [arrowhead = none ];
49 -> 12;

50 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 50 [arrowhead = none ];
50 -> 13;

51 [label="BACKTRACK\nfor clause: ordered(.(X, .(Y, Xs))) :- ','(le(X, Y), ordered(.(Y, Xs)))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 51 [arrowhead = none ];
51 -> 11;

52 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 52 [arrowhead = none ];
52 -> 14;

53 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 53 [arrowhead = none ];
53 -> 15;

54 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 54 [arrowhead = none ];
54 -> 16;

55 [label="EVAL with clause\nle(s(X24), s(X25)) :- le(X24, X25).\nand substitutionX24 -> T19,\nT8 -> s(T19),\nX25 -> T20,\nT9 -> s(T20)", fontsize=14, style = filled, fillcolor = lightblue];
15 -> 55 [arrowhead = none ];
55 -> 17;

56 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 56 [arrowhead = none ];
56 -> 18;

57 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
16 -> 57 [arrowhead = none ];
57 -> 34;

58 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
16 -> 58 [arrowhead = none ];
58 -> 35;

59 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
17 -> 59 [arrowhead = none ];
59 -> 19;

60 [label="SPLIT 2\nnew knowledge:\nT19 is ground\nT20 is ground", fontsize=14, style = filled, fillcolor = violet];
17 -> 60 [arrowhead = none ];
60 -> 20;

61 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
19 -> 61 [arrowhead = none ];
61 -> 21;

62 [label="INSTANCE with matching:\nT1 -> .(s(T20), T10)", fontsize=14, style = filled, fillcolor = lightgrey];
20 -> 62 [arrowhead = none , style=dashed];
62 -> 1[style=dashed];

63 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
21 -> 63 [arrowhead = none ];
63 -> 22;

64 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
21 -> 64 [arrowhead = none ];
64 -> 23;

65 [label="EVAL with clause\nle(s(X38), s(X39)) :- le(X38, X39).\nand substitutionX38 -> T33,\nT19 -> s(T33),\nX39 -> T34,\nT20 -> s(T34)", fontsize=14, style = filled, fillcolor = lightblue];
22 -> 65 [arrowhead = none ];
65 -> 24;

66 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
22 -> 66 [arrowhead = none ];
66 -> 25;

67 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
23 -> 67 [arrowhead = none ];
67 -> 26;

68 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
23 -> 68 [arrowhead = none ];
68 -> 27;

69 [label="INSTANCE with matching:\nT19 -> T33\nT20 -> T34", fontsize=14, style = filled, fillcolor = lightgrey];
24 -> 69 [arrowhead = none , style=dashed];
69 -> 19[style=dashed];

70 [label="EVAL with clause\nle(0, s(0)).\nand substitutionT19 -> 0,\nT20 -> s(0)", fontsize=14, style = filled, fillcolor = lightblue];
26 -> 70 [arrowhead = none ];
70 -> 28;

71 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
26 -> 71 [arrowhead = none ];
71 -> 29;

72 [label="EVAL with clause\nle(0, 0).\nand substitutionT19 -> 0,\nT20 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
27 -> 72 [arrowhead = none ];
72 -> 31;

73 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
27 -> 73 [arrowhead = none ];
73 -> 32;

74 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
28 -> 74 [arrowhead = none ];
74 -> 30;

75 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
31 -> 75 [arrowhead = none ];
75 -> 33;

76 [label="EVAL with clause\nle(0, s(0)).\nand substitutionT8 -> 0,\nT9 -> s(0)", fontsize=14, style = filled, fillcolor = lightblue];
34 -> 76 [arrowhead = none ];
76 -> 36;

77 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
34 -> 77 [arrowhead = none ];
77 -> 37;

78 [label="EVAL with clause\nle(0, 0).\nand substitutionT8 -> 0,\nT9 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
35 -> 78 [arrowhead = none ];
78 -> 38;

79 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
35 -> 79 [arrowhead = none ];
79 -> 39;

80 [label="INSTANCE with matching:\nT1 -> .(s(0), T10)", fontsize=14, style = filled, fillcolor = lightgrey];
36 -> 80 [arrowhead = none , style=dashed];
80 -> 1[style=dashed];

81 [label="INSTANCE with matching:\nT1 -> .(0, T10)", fontsize=14, style = filled, fillcolor = lightgrey];
38 -> 81 [arrowhead = none , style=dashed];
81 -> 1[style=dashed];

}

(2) Obligation:

Triples:

leB(s(X1), s(X2)) :- leB(X1, X2).
orderedA(.(s(X1), .(s(X2), X3))) :- leB(X1, X2).
orderedA(.(s(X1), .(s(X2), X3))) :- ','(lecB(X1, X2), orderedA(.(s(X2), X3))).
orderedA(.(0, .(s(0), X1))) :- orderedA(.(s(0), X1)).
orderedA(.(0, .(0, X1))) :- orderedA(.(0, X1)).

Clauses:

orderedcA([]).
orderedcA(.(X1, [])).
orderedcA(.(s(X1), .(s(X2), X3))) :- ','(lecB(X1, X2), orderedcA(.(s(X2), X3))).
orderedcA(.(0, .(s(0), X1))) :- orderedcA(.(s(0), X1)).
orderedcA(.(0, .(0, X1))) :- orderedcA(.(0, X1)).
lecB(s(X1), s(X2)) :- lecB(X1, X2).
lecB(0, s(0)).
lecB(0, 0).

Afs:

orderedA(x1)  =  orderedA(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:
orderedA_in: (b)
leB_in: (b,b)
lecB_in: (b,b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

ORDEREDA_IN_G(.(s(X1), .(s(X2), X3))) → U2_G(X1, X2, X3, leB_in_gg(X1, X2))
ORDEREDA_IN_G(.(s(X1), .(s(X2), X3))) → LEB_IN_GG(X1, X2)
LEB_IN_GG(s(X1), s(X2)) → U1_GG(X1, X2, leB_in_gg(X1, X2))
LEB_IN_GG(s(X1), s(X2)) → LEB_IN_GG(X1, X2)
ORDEREDA_IN_G(.(s(X1), .(s(X2), X3))) → U3_G(X1, X2, X3, lecB_in_gg(X1, X2))
U3_G(X1, X2, X3, lecB_out_gg(X1, X2)) → U4_G(X1, X2, X3, orderedA_in_g(.(s(X2), X3)))
U3_G(X1, X2, X3, lecB_out_gg(X1, X2)) → ORDEREDA_IN_G(.(s(X2), X3))
ORDEREDA_IN_G(.(0, .(s(0), X1))) → U5_G(X1, orderedA_in_g(.(s(0), X1)))
ORDEREDA_IN_G(.(0, .(s(0), X1))) → ORDEREDA_IN_G(.(s(0), X1))
ORDEREDA_IN_G(.(0, .(0, X1))) → U6_G(X1, orderedA_in_g(.(0, X1)))
ORDEREDA_IN_G(.(0, .(0, X1))) → ORDEREDA_IN_G(.(0, X1))

The TRS R consists of the following rules:

lecB_in_gg(s(X1), s(X2)) → U12_gg(X1, X2, lecB_in_gg(X1, X2))
lecB_in_gg(0, s(0)) → lecB_out_gg(0, s(0))
lecB_in_gg(0, 0) → lecB_out_gg(0, 0)
U12_gg(X1, X2, lecB_out_gg(X1, X2)) → lecB_out_gg(s(X1), s(X2))

Pi is empty.
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:

ORDEREDA_IN_G(.(s(X1), .(s(X2), X3))) → U2_G(X1, X2, X3, leB_in_gg(X1, X2))
ORDEREDA_IN_G(.(s(X1), .(s(X2), X3))) → LEB_IN_GG(X1, X2)
LEB_IN_GG(s(X1), s(X2)) → U1_GG(X1, X2, leB_in_gg(X1, X2))
LEB_IN_GG(s(X1), s(X2)) → LEB_IN_GG(X1, X2)
ORDEREDA_IN_G(.(s(X1), .(s(X2), X3))) → U3_G(X1, X2, X3, lecB_in_gg(X1, X2))
U3_G(X1, X2, X3, lecB_out_gg(X1, X2)) → U4_G(X1, X2, X3, orderedA_in_g(.(s(X2), X3)))
U3_G(X1, X2, X3, lecB_out_gg(X1, X2)) → ORDEREDA_IN_G(.(s(X2), X3))
ORDEREDA_IN_G(.(0, .(s(0), X1))) → U5_G(X1, orderedA_in_g(.(s(0), X1)))
ORDEREDA_IN_G(.(0, .(s(0), X1))) → ORDEREDA_IN_G(.(s(0), X1))
ORDEREDA_IN_G(.(0, .(0, X1))) → U6_G(X1, orderedA_in_g(.(0, X1)))
ORDEREDA_IN_G(.(0, .(0, X1))) → ORDEREDA_IN_G(.(0, X1))

The TRS R consists of the following rules:

lecB_in_gg(s(X1), s(X2)) → U12_gg(X1, X2, lecB_in_gg(X1, X2))
lecB_in_gg(0, s(0)) → lecB_out_gg(0, s(0))
lecB_in_gg(0, 0) → lecB_out_gg(0, 0)
U12_gg(X1, X2, lecB_out_gg(X1, X2)) → lecB_out_gg(s(X1), s(X2))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 7 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

LEB_IN_GG(s(X1), s(X2)) → LEB_IN_GG(X1, X2)

The TRS R consists of the following rules:

lecB_in_gg(s(X1), s(X2)) → U12_gg(X1, X2, lecB_in_gg(X1, X2))
lecB_in_gg(0, s(0)) → lecB_out_gg(0, s(0))
lecB_in_gg(0, 0) → lecB_out_gg(0, 0)
U12_gg(X1, X2, lecB_out_gg(X1, X2)) → lecB_out_gg(s(X1), s(X2))

Pi is empty.
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:

LEB_IN_GG(s(X1), s(X2)) → LEB_IN_GG(X1, X2)

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:

LEB_IN_GG(s(X1), s(X2)) → LEB_IN_GG(X1, 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:

• LEB_IN_GG(s(X1), s(X2)) → LEB_IN_GG(X1, X2)
  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:

ORDEREDA_IN_G(.(s(X1), .(s(X2), X3))) → U3_G(X1, X2, X3, lecB_in_gg(X1, X2))
U3_G(X1, X2, X3, lecB_out_gg(X1, X2)) → ORDEREDA_IN_G(.(s(X2), X3))

The TRS R consists of the following rules:

lecB_in_gg(s(X1), s(X2)) → U12_gg(X1, X2, lecB_in_gg(X1, X2))
lecB_in_gg(0, s(0)) → lecB_out_gg(0, s(0))
lecB_in_gg(0, 0) → lecB_out_gg(0, 0)
U12_gg(X1, X2, lecB_out_gg(X1, X2)) → lecB_out_gg(s(X1), s(X2))

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

(15) PiDPToQDPProof (EQUIVALENT transformation)

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

(16) Obligation:

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

ORDEREDA_IN_G(.(s(X1), .(s(X2), X3))) → U3_G(X1, X2, X3, lecB_in_gg(X1, X2))
U3_G(X1, X2, X3, lecB_out_gg(X1, X2)) → ORDEREDA_IN_G(.(s(X2), X3))

The TRS R consists of the following rules:

lecB_in_gg(s(X1), s(X2)) → U12_gg(X1, X2, lecB_in_gg(X1, X2))
lecB_in_gg(0, s(0)) → lecB_out_gg(0, s(0))
lecB_in_gg(0, 0) → lecB_out_gg(0, 0)
U12_gg(X1, X2, lecB_out_gg(X1, X2)) → lecB_out_gg(s(X1), s(X2))

The set Q consists of the following terms:

lecB_in_gg(x0, x1)
U12_gg(x0, x1, x2)

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

(17) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

lecB_in_gg(0, s(0)) → lecB_out_gg(0, s(0))
lecB_in_gg(0, 0) → lecB_out_gg(0, 0)

Used ordering: Polynomial interpretation [POLO]:

POL(.(x₁, x₂)) = 1 + 2·x₁ + 2·x₂   
POL(0) = 0   
POL(ORDEREDA_IN_G(x₁)) = x₁   
POL(U12_gg(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(U3_G(x₁, x₂, x₃, x₄)) = 1 + x₁ + 2·x₂ + 2·x₃ + 2·x₄   
POL(lecB_in_gg(x₁, x₂)) = 1 + x₁ + x₂   
POL(lecB_out_gg(x₁, x₂)) = x₁ + x₂   
POL(s(x₁)) = 2·x₁   

(18) Obligation:

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

ORDEREDA_IN_G(.(s(X1), .(s(X2), X3))) → U3_G(X1, X2, X3, lecB_in_gg(X1, X2))
U3_G(X1, X2, X3, lecB_out_gg(X1, X2)) → ORDEREDA_IN_G(.(s(X2), X3))

The TRS R consists of the following rules:

lecB_in_gg(s(X1), s(X2)) → U12_gg(X1, X2, lecB_in_gg(X1, X2))
U12_gg(X1, X2, lecB_out_gg(X1, X2)) → lecB_out_gg(s(X1), s(X2))

The set Q consists of the following terms:

lecB_in_gg(x0, x1)
U12_gg(x0, x1, x2)

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

(19) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

U3_G(X1, X2, X3, lecB_out_gg(X1, X2)) → ORDEREDA_IN_G(.(s(X2), X3))

Used ordering: Polynomial interpretation [POLO]:

POL(.(x₁, x₂)) = x₁ + x₂   
POL(ORDEREDA_IN_G(x₁)) = 2·x₁   
POL(U12_gg(x₁, x₂, x₃)) = x₁ + 2·x₂ + x₃   
POL(U3_G(x₁, x₂, x₃, x₄)) = 2·x₁ + 2·x₂ + 2·x₃ + x₄   
POL(lecB_in_gg(x₁, x₂)) = 2·x₁ + 2·x₂   
POL(lecB_out_gg(x₁, x₂)) = 2 + x₁ + 2·x₂   
POL(s(x₁)) = 2·x₁   

(20) Obligation:

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

ORDEREDA_IN_G(.(s(X1), .(s(X2), X3))) → U3_G(X1, X2, X3, lecB_in_gg(X1, X2))

The TRS R consists of the following rules:

lecB_in_gg(s(X1), s(X2)) → U12_gg(X1, X2, lecB_in_gg(X1, X2))
U12_gg(X1, X2, lecB_out_gg(X1, X2)) → lecB_out_gg(s(X1), s(X2))

The set Q consists of the following terms:

lecB_in_gg(x0, x1)
U12_gg(x0, x1, x2)

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

(21) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(22) TRUE

(23) Obligation:

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

ORDEREDA_IN_G(.(0, .(0, X1))) → ORDEREDA_IN_G(.(0, X1))

The TRS R consists of the following rules:

lecB_in_gg(s(X1), s(X2)) → U12_gg(X1, X2, lecB_in_gg(X1, X2))
lecB_in_gg(0, s(0)) → lecB_out_gg(0, s(0))
lecB_in_gg(0, 0) → lecB_out_gg(0, 0)
U12_gg(X1, X2, lecB_out_gg(X1, X2)) → lecB_out_gg(s(X1), s(X2))

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

(24) UsableRulesProof (EQUIVALENT transformation)

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

(25) Obligation:

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

ORDEREDA_IN_G(.(0, .(0, X1))) → ORDEREDA_IN_G(.(0, X1))

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

(26) PiDPToQDPProof (EQUIVALENT transformation)

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

(27) Obligation:

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

ORDEREDA_IN_G(.(0, .(0, X1))) → ORDEREDA_IN_G(.(0, X1))

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

(28) 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:

• ORDEREDA_IN_G(.(0, .(0, X1))) → ORDEREDA_IN_G(.(0, X1))
  The graph contains the following edges 1 > 1

(29) YES