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

(0) Obligation:

Clauses:

ordered([]).
ordered(.(X1, [])).
ordered(.(X, .(Y, Xs))) :- ','(less(X, s(Y)), ordered(.(Y, Xs))).
less(0, s(X2)).
less(s(X), s(Y)) :- less(X, Y).

Query: ordered(g)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: 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\nX8 is free\nordered(T1) !=? ordered([])\nordered(T1) !=? ordered(.(X8, []))", 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: ','(less(T8, s(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: ','(less(T8, s(T9)), ordered(.(T9, T10))), SCOPE: 2, CLAUSE: 3 | ','(less(T8, s(T9)), ordered(.(T9, T10))), SCOPE: 2, CLAUSE: 4\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(less(T8, s(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: ','(less(T8, s(T9)), ordered(.(T9, T10))), SCOPE: 2, CLAUSE: 4\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ordered(.(T15, T10))\n\nT10 is ground\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: ','(less(T20, T21), ordered(.(T21, T10)))\n\nT10 is ground\nT20 is ground\nT21 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: less(T20, T21)\n\nT20 is ground\nT21 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: ordered(.(T21, T10))\n\nT10 is ground\nT21 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: less(T20, T21), SCOPE: 3, CLAUSE: 3 | less(T20, T21), SCOPE: 3, CLAUSE: 4\n\nT20 is ground\nT21 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: less(T20, T21), SCOPE: 3, CLAUSE: 3\n\nT20 is ground\nT21 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: less(T20, T21), SCOPE: 3, CLAUSE: 4\n\nT20 is ground\nT21 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: less(T35, T36)\n\nT35 is ground\nT36 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 31 [arrowhead = none ];
31 -> 2;

32 [label="EVAL with clause\nordered([]).\nand substitutionT1 -> []", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 32 [arrowhead = none ];
32 -> 3;

33 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 33 [arrowhead = none ];
33 -> 4;

34 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 34 [arrowhead = none ];
34 -> 5;

35 [label="EVAL with clause\nordered(.(X8, [])).\nand substitutionX8 -> T3,\nT1 -> .(T3, [])", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 35 [arrowhead = none ];
35 -> 8;

36 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 36 [arrowhead = none ];
36 -> 9;

37 [label="BACKTRACK\nfor clause: ordered(.(X1, []))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 37 [arrowhead = none ];
37 -> 6;

38 [label="BACKTRACK\nfor clause: ordered(.(X, .(Y, Xs))) :- ','(less(X, s(Y)), ordered(.(Y, Xs)))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 38 [arrowhead = none ];
38 -> 7;

39 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 39 [arrowhead = none ];
39 -> 10;

40 [label="EVAL with clause\nordered(.(X15, .(X16, X17))) :- ','(less(X15, s(X16)), ordered(.(X16, X17))).\nand substitutionX15 -> T8,\nX16 -> T9,\nX17 -> T10,\nT1 -> .(T8, .(T9, T10))", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 40 [arrowhead = none ];
40 -> 12;

41 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 41 [arrowhead = none ];
41 -> 13;

42 [label="BACKTRACK\nfor clause: ordered(.(X, .(Y, Xs))) :- ','(less(X, s(Y)), ordered(.(Y, Xs)))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 42 [arrowhead = none ];
42 -> 11;

43 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 43 [arrowhead = none ];
43 -> 14;

44 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 44 [arrowhead = none ];
44 -> 15;

45 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 45 [arrowhead = none ];
45 -> 16;

46 [label="EVAL with clause\nless(0, s(X22)).\nand substitutionT8 -> 0,\nT9 -> T15,\nX22 -> T15", fontsize=14, style = filled, fillcolor = lightblue];
15 -> 46 [arrowhead = none ];
46 -> 17;

47 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 47 [arrowhead = none ];
47 -> 18;

48 [label="EVAL with clause\nless(s(X27), s(X28)) :- less(X27, X28).\nand substitutionX27 -> T20,\nT8 -> s(T20),\nT9 -> T21,\nX28 -> T21", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 48 [arrowhead = none ];
48 -> 19;

49 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 49 [arrowhead = none ];
49 -> 20;

50 [label="INSTANCE with matching:\nT1 -> .(T15, T10)", fontsize=14, style = filled, fillcolor = lightgrey];
17 -> 50 [arrowhead = none , style=dashed];
50 -> 1[style=dashed];

51 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
19 -> 51 [arrowhead = none ];
51 -> 21;

52 [label="SPLIT 2\nnew knowledge:\nT20 is ground\nT21 is ground", fontsize=14, style = filled, fillcolor = violet];
19 -> 52 [arrowhead = none ];
52 -> 22;

53 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
21 -> 53 [arrowhead = none ];
53 -> 23;

54 [label="INSTANCE with matching:\nT1 -> .(T21, T10)", fontsize=14, style = filled, fillcolor = lightgrey];
22 -> 54 [arrowhead = none , style=dashed];
54 -> 1[style=dashed];

55 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
23 -> 55 [arrowhead = none ];
55 -> 24;

56 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
23 -> 56 [arrowhead = none ];
56 -> 25;

57 [label="EVAL with clause\nless(0, s(X37)).\nand substitutionT20 -> 0,\nX37 -> T30,\nT21 -> s(T30)", fontsize=14, style = filled, fillcolor = lightblue];
24 -> 57 [arrowhead = none ];
57 -> 26;

58 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
24 -> 58 [arrowhead = none ];
58 -> 27;

59 [label="EVAL with clause\nless(s(X42), s(X43)) :- less(X42, X43).\nand substitutionX42 -> T35,\nT20 -> s(T35),\nX43 -> T36,\nT21 -> s(T36)", fontsize=14, style = filled, fillcolor = lightblue];
25 -> 59 [arrowhead = none ];
59 -> 29;

60 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 60 [arrowhead = none ];
60 -> 30;

61 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
26 -> 61 [arrowhead = none ];
61 -> 28;

62 [label="INSTANCE with matching:\nT20 -> T35\nT21 -> T36", fontsize=14, style = filled, fillcolor = lightgrey];
29 -> 62 [arrowhead = none , style=dashed];
62 -> 21[style=dashed];

}

(2) Obligation:

Clauses:

lessA(0, s(T30)).
lessA(s(T35), s(T36)) :- lessA(T35, T36).
orderedB([]).
orderedB(.(T3, [])).
orderedB(.(0, .(T15, T10))) :- orderedB(.(T15, T10)).
orderedB(.(s(T20), .(T21, T10))) :- lessA(T20, T21).
orderedB(.(s(T20), .(T21, T10))) :- ','(lessA(T20, T21), orderedB(.(T21, T10))).

Query: orderedB(g)

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
orderedB_in: (b)
lessA_in: (b,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

orderedB_in_g([]) → orderedB_out_g([])
orderedB_in_g(.(T3, [])) → orderedB_out_g(.(T3, []))
orderedB_in_g(.(0, .(T15, T10))) → U2_g(T15, T10, orderedB_in_g(.(T15, T10)))
orderedB_in_g(.(s(T20), .(T21, T10))) → U3_g(T20, T21, T10, lessA_in_gg(T20, T21))
lessA_in_gg(0, s(T30)) → lessA_out_gg(0, s(T30))
lessA_in_gg(s(T35), s(T36)) → U1_gg(T35, T36, lessA_in_gg(T35, T36))
U1_gg(T35, T36, lessA_out_gg(T35, T36)) → lessA_out_gg(s(T35), s(T36))
U3_g(T20, T21, T10, lessA_out_gg(T20, T21)) → orderedB_out_g(.(s(T20), .(T21, T10)))
U3_g(T20, T21, T10, lessA_out_gg(T20, T21)) → U4_g(T20, T21, T10, orderedB_in_g(.(T21, T10)))
U4_g(T20, T21, T10, orderedB_out_g(.(T21, T10))) → orderedB_out_g(.(s(T20), .(T21, T10)))
U2_g(T15, T10, orderedB_out_g(.(T15, T10))) → orderedB_out_g(.(0, .(T15, T10)))

The argument filtering Pi contains the following mapping:
orderedB_in_g(x1)  =  orderedB_in_g(x1)
[]  =  []
orderedB_out_g(x1)  =  orderedB_out_g
.(x1, x2)  =  .(x1, x2)
0  =  0
U2_g(x1, x2, x3)  =  U2_g(x3)
s(x1)  =  s(x1)
U3_g(x1, x2, x3, x4)  =  U3_g(x2, x3, x4)
lessA_in_gg(x1, x2)  =  lessA_in_gg(x1, x2)
lessA_out_gg(x1, x2)  =  lessA_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U4_g(x1, x2, x3, x4)  =  U4_g(x4)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

orderedB_in_g([]) → orderedB_out_g([])
orderedB_in_g(.(T3, [])) → orderedB_out_g(.(T3, []))
orderedB_in_g(.(0, .(T15, T10))) → U2_g(T15, T10, orderedB_in_g(.(T15, T10)))
orderedB_in_g(.(s(T20), .(T21, T10))) → U3_g(T20, T21, T10, lessA_in_gg(T20, T21))
lessA_in_gg(0, s(T30)) → lessA_out_gg(0, s(T30))
lessA_in_gg(s(T35), s(T36)) → U1_gg(T35, T36, lessA_in_gg(T35, T36))
U1_gg(T35, T36, lessA_out_gg(T35, T36)) → lessA_out_gg(s(T35), s(T36))
U3_g(T20, T21, T10, lessA_out_gg(T20, T21)) → orderedB_out_g(.(s(T20), .(T21, T10)))
U3_g(T20, T21, T10, lessA_out_gg(T20, T21)) → U4_g(T20, T21, T10, orderedB_in_g(.(T21, T10)))
U4_g(T20, T21, T10, orderedB_out_g(.(T21, T10))) → orderedB_out_g(.(s(T20), .(T21, T10)))
U2_g(T15, T10, orderedB_out_g(.(T15, T10))) → orderedB_out_g(.(0, .(T15, T10)))

The argument filtering Pi contains the following mapping:
orderedB_in_g(x1)  =  orderedB_in_g(x1)
[]  =  []
orderedB_out_g(x1)  =  orderedB_out_g
.(x1, x2)  =  .(x1, x2)
0  =  0
U2_g(x1, x2, x3)  =  U2_g(x3)
s(x1)  =  s(x1)
U3_g(x1, x2, x3, x4)  =  U3_g(x2, x3, x4)
lessA_in_gg(x1, x2)  =  lessA_in_gg(x1, x2)
lessA_out_gg(x1, x2)  =  lessA_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U4_g(x1, x2, x3, x4)  =  U4_g(x4)

(5) 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:

ORDEREDB_IN_G(.(0, .(T15, T10))) → U2_G(T15, T10, orderedB_in_g(.(T15, T10)))
ORDEREDB_IN_G(.(0, .(T15, T10))) → ORDEREDB_IN_G(.(T15, T10))
ORDEREDB_IN_G(.(s(T20), .(T21, T10))) → U3_G(T20, T21, T10, lessA_in_gg(T20, T21))
ORDEREDB_IN_G(.(s(T20), .(T21, T10))) → LESSA_IN_GG(T20, T21)
LESSA_IN_GG(s(T35), s(T36)) → U1_GG(T35, T36, lessA_in_gg(T35, T36))
LESSA_IN_GG(s(T35), s(T36)) → LESSA_IN_GG(T35, T36)
U3_G(T20, T21, T10, lessA_out_gg(T20, T21)) → U4_G(T20, T21, T10, orderedB_in_g(.(T21, T10)))
U3_G(T20, T21, T10, lessA_out_gg(T20, T21)) → ORDEREDB_IN_G(.(T21, T10))

The TRS R consists of the following rules:

orderedB_in_g([]) → orderedB_out_g([])
orderedB_in_g(.(T3, [])) → orderedB_out_g(.(T3, []))
orderedB_in_g(.(0, .(T15, T10))) → U2_g(T15, T10, orderedB_in_g(.(T15, T10)))
orderedB_in_g(.(s(T20), .(T21, T10))) → U3_g(T20, T21, T10, lessA_in_gg(T20, T21))
lessA_in_gg(0, s(T30)) → lessA_out_gg(0, s(T30))
lessA_in_gg(s(T35), s(T36)) → U1_gg(T35, T36, lessA_in_gg(T35, T36))
U1_gg(T35, T36, lessA_out_gg(T35, T36)) → lessA_out_gg(s(T35), s(T36))
U3_g(T20, T21, T10, lessA_out_gg(T20, T21)) → orderedB_out_g(.(s(T20), .(T21, T10)))
U3_g(T20, T21, T10, lessA_out_gg(T20, T21)) → U4_g(T20, T21, T10, orderedB_in_g(.(T21, T10)))
U4_g(T20, T21, T10, orderedB_out_g(.(T21, T10))) → orderedB_out_g(.(s(T20), .(T21, T10)))
U2_g(T15, T10, orderedB_out_g(.(T15, T10))) → orderedB_out_g(.(0, .(T15, T10)))

The argument filtering Pi contains the following mapping:
orderedB_in_g(x1)  =  orderedB_in_g(x1)
[]  =  []
orderedB_out_g(x1)  =  orderedB_out_g
.(x1, x2)  =  .(x1, x2)
0  =  0
U2_g(x1, x2, x3)  =  U2_g(x3)
s(x1)  =  s(x1)
U3_g(x1, x2, x3, x4)  =  U3_g(x2, x3, x4)
lessA_in_gg(x1, x2)  =  lessA_in_gg(x1, x2)
lessA_out_gg(x1, x2)  =  lessA_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U4_g(x1, x2, x3, x4)  =  U4_g(x4)
ORDEREDB_IN_G(x1)  =  ORDEREDB_IN_G(x1)
U2_G(x1, x2, x3)  =  U2_G(x3)
U3_G(x1, x2, x3, x4)  =  U3_G(x2, x3, x4)
LESSA_IN_GG(x1, x2)  =  LESSA_IN_GG(x1, x2)
U1_GG(x1, x2, x3)  =  U1_GG(x3)
U4_G(x1, x2, x3, x4)  =  U4_G(x4)

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

(6) Obligation:

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

ORDEREDB_IN_G(.(0, .(T15, T10))) → U2_G(T15, T10, orderedB_in_g(.(T15, T10)))
ORDEREDB_IN_G(.(0, .(T15, T10))) → ORDEREDB_IN_G(.(T15, T10))
ORDEREDB_IN_G(.(s(T20), .(T21, T10))) → U3_G(T20, T21, T10, lessA_in_gg(T20, T21))
ORDEREDB_IN_G(.(s(T20), .(T21, T10))) → LESSA_IN_GG(T20, T21)
LESSA_IN_GG(s(T35), s(T36)) → U1_GG(T35, T36, lessA_in_gg(T35, T36))
LESSA_IN_GG(s(T35), s(T36)) → LESSA_IN_GG(T35, T36)
U3_G(T20, T21, T10, lessA_out_gg(T20, T21)) → U4_G(T20, T21, T10, orderedB_in_g(.(T21, T10)))
U3_G(T20, T21, T10, lessA_out_gg(T20, T21)) → ORDEREDB_IN_G(.(T21, T10))

The TRS R consists of the following rules:

orderedB_in_g([]) → orderedB_out_g([])
orderedB_in_g(.(T3, [])) → orderedB_out_g(.(T3, []))
orderedB_in_g(.(0, .(T15, T10))) → U2_g(T15, T10, orderedB_in_g(.(T15, T10)))
orderedB_in_g(.(s(T20), .(T21, T10))) → U3_g(T20, T21, T10, lessA_in_gg(T20, T21))
lessA_in_gg(0, s(T30)) → lessA_out_gg(0, s(T30))
lessA_in_gg(s(T35), s(T36)) → U1_gg(T35, T36, lessA_in_gg(T35, T36))
U1_gg(T35, T36, lessA_out_gg(T35, T36)) → lessA_out_gg(s(T35), s(T36))
U3_g(T20, T21, T10, lessA_out_gg(T20, T21)) → orderedB_out_g(.(s(T20), .(T21, T10)))
U3_g(T20, T21, T10, lessA_out_gg(T20, T21)) → U4_g(T20, T21, T10, orderedB_in_g(.(T21, T10)))
U4_g(T20, T21, T10, orderedB_out_g(.(T21, T10))) → orderedB_out_g(.(s(T20), .(T21, T10)))
U2_g(T15, T10, orderedB_out_g(.(T15, T10))) → orderedB_out_g(.(0, .(T15, T10)))

The argument filtering Pi contains the following mapping:
orderedB_in_g(x1)  =  orderedB_in_g(x1)
[]  =  []
orderedB_out_g(x1)  =  orderedB_out_g
.(x1, x2)  =  .(x1, x2)
0  =  0
U2_g(x1, x2, x3)  =  U2_g(x3)
s(x1)  =  s(x1)
U3_g(x1, x2, x3, x4)  =  U3_g(x2, x3, x4)
lessA_in_gg(x1, x2)  =  lessA_in_gg(x1, x2)
lessA_out_gg(x1, x2)  =  lessA_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U4_g(x1, x2, x3, x4)  =  U4_g(x4)
ORDEREDB_IN_G(x1)  =  ORDEREDB_IN_G(x1)
U2_G(x1, x2, x3)  =  U2_G(x3)
U3_G(x1, x2, x3, x4)  =  U3_G(x2, x3, x4)
LESSA_IN_GG(x1, x2)  =  LESSA_IN_GG(x1, x2)
U1_GG(x1, x2, x3)  =  U1_GG(x3)
U4_G(x1, x2, x3, x4)  =  U4_G(x4)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 4 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

LESSA_IN_GG(s(T35), s(T36)) → LESSA_IN_GG(T35, T36)

The TRS R consists of the following rules:

orderedB_in_g([]) → orderedB_out_g([])
orderedB_in_g(.(T3, [])) → orderedB_out_g(.(T3, []))
orderedB_in_g(.(0, .(T15, T10))) → U2_g(T15, T10, orderedB_in_g(.(T15, T10)))
orderedB_in_g(.(s(T20), .(T21, T10))) → U3_g(T20, T21, T10, lessA_in_gg(T20, T21))
lessA_in_gg(0, s(T30)) → lessA_out_gg(0, s(T30))
lessA_in_gg(s(T35), s(T36)) → U1_gg(T35, T36, lessA_in_gg(T35, T36))
U1_gg(T35, T36, lessA_out_gg(T35, T36)) → lessA_out_gg(s(T35), s(T36))
U3_g(T20, T21, T10, lessA_out_gg(T20, T21)) → orderedB_out_g(.(s(T20), .(T21, T10)))
U3_g(T20, T21, T10, lessA_out_gg(T20, T21)) → U4_g(T20, T21, T10, orderedB_in_g(.(T21, T10)))
U4_g(T20, T21, T10, orderedB_out_g(.(T21, T10))) → orderedB_out_g(.(s(T20), .(T21, T10)))
U2_g(T15, T10, orderedB_out_g(.(T15, T10))) → orderedB_out_g(.(0, .(T15, T10)))

The argument filtering Pi contains the following mapping:
orderedB_in_g(x1)  =  orderedB_in_g(x1)
[]  =  []
orderedB_out_g(x1)  =  orderedB_out_g
.(x1, x2)  =  .(x1, x2)
0  =  0
U2_g(x1, x2, x3)  =  U2_g(x3)
s(x1)  =  s(x1)
U3_g(x1, x2, x3, x4)  =  U3_g(x2, x3, x4)
lessA_in_gg(x1, x2)  =  lessA_in_gg(x1, x2)
lessA_out_gg(x1, x2)  =  lessA_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U4_g(x1, x2, x3, x4)  =  U4_g(x4)
LESSA_IN_GG(x1, x2)  =  LESSA_IN_GG(x1, x2)

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

(10) UsableRulesProof (EQUIVALENT transformation)

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

(11) Obligation:

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

LESSA_IN_GG(s(T35), s(T36)) → LESSA_IN_GG(T35, T36)

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

(12) PiDPToQDPProof (EQUIVALENT transformation)

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

(13) Obligation:

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

LESSA_IN_GG(s(T35), s(T36)) → LESSA_IN_GG(T35, T36)

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

(14) 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:

• LESSA_IN_GG(s(T35), s(T36)) → LESSA_IN_GG(T35, T36)
  The graph contains the following edges 1 > 1, 2 > 2

(15) YES

(16) Obligation:

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

ORDEREDB_IN_G(.(s(T20), .(T21, T10))) → U3_G(T20, T21, T10, lessA_in_gg(T20, T21))
U3_G(T20, T21, T10, lessA_out_gg(T20, T21)) → ORDEREDB_IN_G(.(T21, T10))
ORDEREDB_IN_G(.(0, .(T15, T10))) → ORDEREDB_IN_G(.(T15, T10))

The TRS R consists of the following rules:

orderedB_in_g([]) → orderedB_out_g([])
orderedB_in_g(.(T3, [])) → orderedB_out_g(.(T3, []))
orderedB_in_g(.(0, .(T15, T10))) → U2_g(T15, T10, orderedB_in_g(.(T15, T10)))
orderedB_in_g(.(s(T20), .(T21, T10))) → U3_g(T20, T21, T10, lessA_in_gg(T20, T21))
lessA_in_gg(0, s(T30)) → lessA_out_gg(0, s(T30))
lessA_in_gg(s(T35), s(T36)) → U1_gg(T35, T36, lessA_in_gg(T35, T36))
U1_gg(T35, T36, lessA_out_gg(T35, T36)) → lessA_out_gg(s(T35), s(T36))
U3_g(T20, T21, T10, lessA_out_gg(T20, T21)) → orderedB_out_g(.(s(T20), .(T21, T10)))
U3_g(T20, T21, T10, lessA_out_gg(T20, T21)) → U4_g(T20, T21, T10, orderedB_in_g(.(T21, T10)))
U4_g(T20, T21, T10, orderedB_out_g(.(T21, T10))) → orderedB_out_g(.(s(T20), .(T21, T10)))
U2_g(T15, T10, orderedB_out_g(.(T15, T10))) → orderedB_out_g(.(0, .(T15, T10)))

The argument filtering Pi contains the following mapping:
orderedB_in_g(x1)  =  orderedB_in_g(x1)
[]  =  []
orderedB_out_g(x1)  =  orderedB_out_g
.(x1, x2)  =  .(x1, x2)
0  =  0
U2_g(x1, x2, x3)  =  U2_g(x3)
s(x1)  =  s(x1)
U3_g(x1, x2, x3, x4)  =  U3_g(x2, x3, x4)
lessA_in_gg(x1, x2)  =  lessA_in_gg(x1, x2)
lessA_out_gg(x1, x2)  =  lessA_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U4_g(x1, x2, x3, x4)  =  U4_g(x4)
ORDEREDB_IN_G(x1)  =  ORDEREDB_IN_G(x1)
U3_G(x1, x2, x3, x4)  =  U3_G(x2, x3, x4)

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

(17) UsableRulesProof (EQUIVALENT transformation)

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

(18) Obligation:

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

ORDEREDB_IN_G(.(s(T20), .(T21, T10))) → U3_G(T20, T21, T10, lessA_in_gg(T20, T21))
U3_G(T20, T21, T10, lessA_out_gg(T20, T21)) → ORDEREDB_IN_G(.(T21, T10))
ORDEREDB_IN_G(.(0, .(T15, T10))) → ORDEREDB_IN_G(.(T15, T10))

The TRS R consists of the following rules:

lessA_in_gg(0, s(T30)) → lessA_out_gg(0, s(T30))
lessA_in_gg(s(T35), s(T36)) → U1_gg(T35, T36, lessA_in_gg(T35, T36))
U1_gg(T35, T36, lessA_out_gg(T35, T36)) → lessA_out_gg(s(T35), s(T36))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
0  =  0
s(x1)  =  s(x1)
lessA_in_gg(x1, x2)  =  lessA_in_gg(x1, x2)
lessA_out_gg(x1, x2)  =  lessA_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(x3)
ORDEREDB_IN_G(x1)  =  ORDEREDB_IN_G(x1)
U3_G(x1, x2, x3, x4)  =  U3_G(x2, x3, x4)

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

(19) PiDPToQDPProof (SOUND transformation)

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

(20) Obligation:

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

ORDEREDB_IN_G(.(s(T20), .(T21, T10))) → U3_G(T21, T10, lessA_in_gg(T20, T21))
U3_G(T21, T10, lessA_out_gg) → ORDEREDB_IN_G(.(T21, T10))
ORDEREDB_IN_G(.(0, .(T15, T10))) → ORDEREDB_IN_G(.(T15, T10))

The TRS R consists of the following rules:

lessA_in_gg(0, s(T30)) → lessA_out_gg
lessA_in_gg(s(T35), s(T36)) → U1_gg(lessA_in_gg(T35, T36))
U1_gg(lessA_out_gg) → lessA_out_gg

The set Q consists of the following terms:

lessA_in_gg(x0, x1)
U1_gg(x0)

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

(21) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

ORDEREDB_IN_G(.(s(T20), .(T21, T10))) → U3_G(T21, T10, lessA_in_gg(T20, T21))
U3_G(T21, T10, lessA_out_gg) → ORDEREDB_IN_G(.(T21, T10))
ORDEREDB_IN_G(.(0, .(T15, T10))) → ORDEREDB_IN_G(.(T15, T10))

The following rules are removed from R:

lessA_in_gg(0, s(T30)) → lessA_out_gg
lessA_in_gg(s(T35), s(T36)) → U1_gg(lessA_in_gg(T35, T36))
U1_gg(lessA_out_gg) → lessA_out_gg

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(.(x₁, x₂)) = 2·x₁ + 2·x₂   
POL(0) = 1   
POL(ORDEREDB_IN_G(x₁)) = x₁   
POL(U1_gg(x₁)) = 2·x₁   
POL(U3_G(x₁, x₂, x₃)) = 2·x₁ + 2·x₂ + x₃   
POL(lessA_in_gg(x₁, x₂)) = x₁ + 2·x₂   
POL(lessA_out_gg) = 1   
POL(s(x₁)) = 2·x₁   

(22) Obligation:

Q DP problem:
P is empty.
R is empty.
The set Q consists of the following terms:

lessA_in_gg(x0, x1)
U1_gg(x0)

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

(23) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(24) YES