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

(0) Obligation:

Clauses:

less(0, Y) :- ','(!, =(Y, s(X2))).
less(X, Y) :- ','(p(X, X1), ','(p(Y, Y1), less(X1, Y1))).
p(0, 0).
p(s(X), X).
=(X, X).

Query: less(g,a)

(1) BuiltinConflictTransformerProof (EQUIVALENT transformation)

Renamed defined predicates conflicting with built-in predicates [PROLOG].

(2) Obligation:

Clauses:

less(0, Y) :- ','(!, user_defined_=(Y, s(X2))).
less(X, Y) :- ','(p(X, X1), ','(p(Y, Y1), less(X1, Y1))).
p(0, 0).
p(s(X), X).
user_defined_=(X, X).

Query: less(g,a)

(3) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: less(T1, T2)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: less(T1, T2), SCOPE: 1, CLAUSE: 0 | less(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(!_1, user_defined_=(T5, s(X5))) | less(0, T2), SCOPE: 1, CLAUSE: 1\n\nX5 is free", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: less(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nX4 is free\nless(T1, T2) !=? less(0, X4)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: user_defined_=(T5, s(X5))\n\nX5 is free", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: user_defined_=(T5, s(X5)), SCOPE: 2, CLAUSE: 4\n\nX5 is free", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(p(T14, X13), ','(p(T16, X14), less(X13, X14)))\n\nT14 is ground\nX4 is free\nX13 is free\nX14 is free\nless(T14, T2) !=? less(0, X4)", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(p(T14, X13), ','(p(T16, X14), less(X13, X14))), SCOPE: 3, CLAUSE: 2 | ','(p(T14, X13), ','(p(T16, X14), less(X13, X14))), SCOPE: 3, CLAUSE: 3\n\nT14 is ground\nX4 is free\nX13 is free\nX14 is free\nless(T14, T2) !=? less(0, X4)", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(p(T14, X13), ','(p(T16, X14), less(X13, X14))), SCOPE: 3, CLAUSE: 3\n\nT14 is ground\nX4 is free\nX13 is free\nX14 is free\nless(T14, T2) !=? less(0, X4)", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(p(T16, X14), less(T19, X14))\n\nT19 is ground\nX14 is free", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(p(T16, X14), less(T19, X14)), SCOPE: 4, CLAUSE: 2 | ','(p(T16, X14), less(T19, X14)), SCOPE: 4, CLAUSE: 3\n\nT19 is ground\nX14 is free", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(p(T16, X14), less(T19, X14)), SCOPE: 4, CLAUSE: 2\n\nT19 is ground\nX14 is free", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(p(T16, X14), less(T19, X14)), SCOPE: 4, CLAUSE: 3\n\nT19 is ground\nX14 is free", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: less(T19, 0)\n\nT19 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: less(T19, 0), SCOPE: 5, CLAUSE: 0 | less(T19, 0), SCOPE: 5, CLAUSE: 1\n\nT19 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: ','(!_5, user_defined_=(0, s(X27))) | less(0, 0), SCOPE: 5, CLAUSE: 1\n\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: less(T19, 0), SCOPE: 5, CLAUSE: 1\n\nT19 is ground\nX26 is free\nless(T19, 0) !=? less(0, X26)", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: user_defined_=(0, s(X27))\n\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: user_defined_=(0, s(X27)), SCOPE: 6, CLAUSE: 4\n\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: ','(p(T22, X40), ','(p(0, X41), less(X40, X41)))\n\nT22 is ground\nX26 is free\nX40 is free\nX41 is free\nless(T22, 0) !=? less(0, X26)", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: ','(p(T22, X40), ','(p(0, X41), less(X40, X41))), SCOPE: 7, CLAUSE: 2 | ','(p(T22, X40), ','(p(0, X41), less(X40, X41))), SCOPE: 7, CLAUSE: 3\n\nT22 is ground\nX26 is free\nX40 is free\nX41 is free\nless(T22, 0) !=? less(0, X26)", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: ','(p(T22, X40), ','(p(0, X41), less(X40, X41))), SCOPE: 7, CLAUSE: 3\n\nT22 is ground\nX26 is free\nX40 is free\nX41 is free\nless(T22, 0) !=? less(0, X26)", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: ','(p(0, X41), less(T25, X41))\n\nT25 is ground\nX41 is free", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: ','(p(0, X41), less(T25, X41)), SCOPE: 8, CLAUSE: 2 | ','(p(0, X41), less(T25, X41)), SCOPE: 8, CLAUSE: 3\n\nT25 is ground\nX41 is free", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: ','(p(0, X41), less(T25, X41)), SCOPE: 8, CLAUSE: 2\n\nT25 is ground\nX41 is free", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: ','(p(0, X41), less(T25, X41)), SCOPE: 8, CLAUSE: 3\n\nT25 is ground\nX41 is free", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: less(T25, 0)\n\nT25 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="36: less(T19, T29)\n\nT19 is ground", 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="EVAL with clause\nless(0, X4) :- ','(!_1, user_defined_=(X4, s(X5))).\nand substitutionT1 -> 0,\nT2 -> T5,\nX4 -> T5,\nT4 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 39 [arrowhead = none ];
39 -> 3;

40 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 40 [arrowhead = none ];
40 -> 4;

41 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 41 [arrowhead = none ];
41 -> 5;

42 [label="ONLY EVAL with clause\nless(X11, X12) :- ','(p(X11, X13), ','(p(X12, X14), less(X13, X14))).\nand substitutionT1 -> T14,\nX11 -> T14,\nT2 -> T16,\nX12 -> T16,\nT15 -> T16", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 42 [arrowhead = none ];
42 -> 10;

43 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
5 -> 43 [arrowhead = none ];
43 -> 6;

44 [label="EVAL with clause\nuser_defined_=(X8, X8).\nand substitutionT5 -> s(T11),\nX8 -> s(T11),\nX5 -> T11,\nT10 -> s(T11)", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 44 [arrowhead = none ];
44 -> 7;

45 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 45 [arrowhead = none ];
45 -> 8;

46 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 46 [arrowhead = none ];
46 -> 9;

47 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
10 -> 47 [arrowhead = none ];
47 -> 11;

48 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (less(T14, T2), less(0, X4))", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 48 [arrowhead = none ];
48 -> 12;

49 [label="EVAL with clause\np(s(X17), X17).\nand substitutionX17 -> T19,\nT14 -> s(T19),\nX13 -> T19", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 49 [arrowhead = none ];
49 -> 13;

50 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 50 [arrowhead = none ];
50 -> 14;

51 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
13 -> 51 [arrowhead = none ];
51 -> 15;

52 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
15 -> 52 [arrowhead = none ];
52 -> 16;

53 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
15 -> 53 [arrowhead = none ];
53 -> 17;

54 [label="EVAL with clause\np(0, 0).\nand substitutionT16 -> 0,\nX14 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 54 [arrowhead = none ];
54 -> 18;

55 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 55 [arrowhead = none ];
55 -> 19;

56 [label="EVAL with clause\np(s(X60), X60).\nand substitutionX60 -> T29,\nT16 -> s(T29),\nX14 -> T29,\nT28 -> T29", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 56 [arrowhead = none ];
56 -> 36;

57 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 57 [arrowhead = none ];
57 -> 37;

58 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
18 -> 58 [arrowhead = none ];
58 -> 20;

59 [label="EVAL with clause\nless(0, X26) :- ','(!_5, user_defined_=(X26, s(X27))).\nand substitutionT19 -> 0,\nX26 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
20 -> 59 [arrowhead = none ];
59 -> 21;

60 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
20 -> 60 [arrowhead = none ];
60 -> 22;

61 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
21 -> 61 [arrowhead = none ];
61 -> 23;

62 [label="ONLY EVAL with clause\nless(X38, X39) :- ','(p(X38, X40), ','(p(X39, X41), less(X40, X41))).\nand substitutionT19 -> T22,\nX38 -> T22,\nX39 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
22 -> 62 [arrowhead = none ];
62 -> 26;

63 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
23 -> 63 [arrowhead = none ];
63 -> 24;

64 [label="BACKTRACK\nfor clause: user_defined_=(X, X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
24 -> 64 [arrowhead = none ];
64 -> 25;

65 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
26 -> 65 [arrowhead = none ];
65 -> 27;

66 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (less(T22, 0), less(0, X26))", fontsize=14, style = filled, fillcolor = lightgreen];
27 -> 66 [arrowhead = none ];
66 -> 28;

67 [label="EVAL with clause\np(s(X47), X47).\nand substitutionX47 -> T25,\nT22 -> s(T25),\nX40 -> T25", fontsize=14, style = filled, fillcolor = lightblue];
28 -> 67 [arrowhead = none ];
67 -> 29;

68 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
28 -> 68 [arrowhead = none ];
68 -> 30;

69 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
29 -> 69 [arrowhead = none ];
69 -> 31;

70 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
31 -> 70 [arrowhead = none ];
70 -> 32;

71 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
31 -> 71 [arrowhead = none ];
71 -> 33;

72 [label="ONLY EVAL with clause\np(0, 0).\nand substitutionX41 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
32 -> 72 [arrowhead = none ];
72 -> 34;

73 [label="BACKTRACK\nfor clause: p(s(X), X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
33 -> 73 [arrowhead = none ];
73 -> 35;

74 [label="INSTANCE with matching:\nT19 -> T25", fontsize=14, style = filled, fillcolor = lightgrey];
34 -> 74 [arrowhead = none , style=dashed];
74 -> 18[style=dashed];

75 [label="INSTANCE with matching:\nT1 -> T19\nT2 -> T29", fontsize=14, style = filled, fillcolor = lightgrey];
36 -> 75 [arrowhead = none , style=dashed];
75 -> 1[style=dashed];

}

(4) Obligation:

Clauses:

lessA(s(T25)) :- lessA(T25).
lessB(0, s(T11)).
lessB(s(T19), 0) :- lessA(T19).
lessB(s(T19), s(T29)) :- lessB(T19, T29).

Query: lessB(g,a)

(5) PrologToPiTRSProof (SOUND transformation)

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

lessB_in_ga(0, s(T11)) → lessB_out_ga(0, s(T11))
lessB_in_ga(s(T19), 0) → U2_ga(T19, lessA_in_g(T19))
lessA_in_g(s(T25)) → U1_g(T25, lessA_in_g(T25))
U1_g(T25, lessA_out_g(T25)) → lessA_out_g(s(T25))
U2_ga(T19, lessA_out_g(T19)) → lessB_out_ga(s(T19), 0)
lessB_in_ga(s(T19), s(T29)) → U3_ga(T19, T29, lessB_in_ga(T19, T29))
U3_ga(T19, T29, lessB_out_ga(T19, T29)) → lessB_out_ga(s(T19), s(T29))

The argument filtering Pi contains the following mapping:
lessB_in_ga(x1, x2)  =  lessB_in_ga(x1)
0  =  0
lessB_out_ga(x1, x2)  =  lessB_out_ga
s(x1)  =  s(x1)
U2_ga(x1, x2)  =  U2_ga(x2)
lessA_in_g(x1)  =  lessA_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
lessA_out_g(x1)  =  lessA_out_g
U3_ga(x1, x2, x3)  =  U3_ga(x3)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(6) Obligation:

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

lessB_in_ga(0, s(T11)) → lessB_out_ga(0, s(T11))
lessB_in_ga(s(T19), 0) → U2_ga(T19, lessA_in_g(T19))
lessA_in_g(s(T25)) → U1_g(T25, lessA_in_g(T25))
U1_g(T25, lessA_out_g(T25)) → lessA_out_g(s(T25))
U2_ga(T19, lessA_out_g(T19)) → lessB_out_ga(s(T19), 0)
lessB_in_ga(s(T19), s(T29)) → U3_ga(T19, T29, lessB_in_ga(T19, T29))
U3_ga(T19, T29, lessB_out_ga(T19, T29)) → lessB_out_ga(s(T19), s(T29))

The argument filtering Pi contains the following mapping:
lessB_in_ga(x1, x2)  =  lessB_in_ga(x1)
0  =  0
lessB_out_ga(x1, x2)  =  lessB_out_ga
s(x1)  =  s(x1)
U2_ga(x1, x2)  =  U2_ga(x2)
lessA_in_g(x1)  =  lessA_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
lessA_out_g(x1)  =  lessA_out_g
U3_ga(x1, x2, x3)  =  U3_ga(x3)

(7) 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:

LESSB_IN_GA(s(T19), 0) → U2_GA(T19, lessA_in_g(T19))
LESSB_IN_GA(s(T19), 0) → LESSA_IN_G(T19)
LESSA_IN_G(s(T25)) → U1_G(T25, lessA_in_g(T25))
LESSA_IN_G(s(T25)) → LESSA_IN_G(T25)
LESSB_IN_GA(s(T19), s(T29)) → U3_GA(T19, T29, lessB_in_ga(T19, T29))
LESSB_IN_GA(s(T19), s(T29)) → LESSB_IN_GA(T19, T29)

The TRS R consists of the following rules:

lessB_in_ga(0, s(T11)) → lessB_out_ga(0, s(T11))
lessB_in_ga(s(T19), 0) → U2_ga(T19, lessA_in_g(T19))
lessA_in_g(s(T25)) → U1_g(T25, lessA_in_g(T25))
U1_g(T25, lessA_out_g(T25)) → lessA_out_g(s(T25))
U2_ga(T19, lessA_out_g(T19)) → lessB_out_ga(s(T19), 0)
lessB_in_ga(s(T19), s(T29)) → U3_ga(T19, T29, lessB_in_ga(T19, T29))
U3_ga(T19, T29, lessB_out_ga(T19, T29)) → lessB_out_ga(s(T19), s(T29))

The argument filtering Pi contains the following mapping:
lessB_in_ga(x1, x2)  =  lessB_in_ga(x1)
0  =  0
lessB_out_ga(x1, x2)  =  lessB_out_ga
s(x1)  =  s(x1)
U2_ga(x1, x2)  =  U2_ga(x2)
lessA_in_g(x1)  =  lessA_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
lessA_out_g(x1)  =  lessA_out_g
U3_ga(x1, x2, x3)  =  U3_ga(x3)
LESSB_IN_GA(x1, x2)  =  LESSB_IN_GA(x1)
U2_GA(x1, x2)  =  U2_GA(x2)
LESSA_IN_G(x1)  =  LESSA_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)
U3_GA(x1, x2, x3)  =  U3_GA(x3)

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

(8) Obligation:

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

LESSB_IN_GA(s(T19), 0) → U2_GA(T19, lessA_in_g(T19))
LESSB_IN_GA(s(T19), 0) → LESSA_IN_G(T19)
LESSA_IN_G(s(T25)) → U1_G(T25, lessA_in_g(T25))
LESSA_IN_G(s(T25)) → LESSA_IN_G(T25)
LESSB_IN_GA(s(T19), s(T29)) → U3_GA(T19, T29, lessB_in_ga(T19, T29))
LESSB_IN_GA(s(T19), s(T29)) → LESSB_IN_GA(T19, T29)

The TRS R consists of the following rules:

lessB_in_ga(0, s(T11)) → lessB_out_ga(0, s(T11))
lessB_in_ga(s(T19), 0) → U2_ga(T19, lessA_in_g(T19))
lessA_in_g(s(T25)) → U1_g(T25, lessA_in_g(T25))
U1_g(T25, lessA_out_g(T25)) → lessA_out_g(s(T25))
U2_ga(T19, lessA_out_g(T19)) → lessB_out_ga(s(T19), 0)
lessB_in_ga(s(T19), s(T29)) → U3_ga(T19, T29, lessB_in_ga(T19, T29))
U3_ga(T19, T29, lessB_out_ga(T19, T29)) → lessB_out_ga(s(T19), s(T29))

The argument filtering Pi contains the following mapping:
lessB_in_ga(x1, x2)  =  lessB_in_ga(x1)
0  =  0
lessB_out_ga(x1, x2)  =  lessB_out_ga
s(x1)  =  s(x1)
U2_ga(x1, x2)  =  U2_ga(x2)
lessA_in_g(x1)  =  lessA_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
lessA_out_g(x1)  =  lessA_out_g
U3_ga(x1, x2, x3)  =  U3_ga(x3)
LESSB_IN_GA(x1, x2)  =  LESSB_IN_GA(x1)
U2_GA(x1, x2)  =  U2_GA(x2)
LESSA_IN_G(x1)  =  LESSA_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)
U3_GA(x1, x2, x3)  =  U3_GA(x3)

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

(9) DependencyGraphProof (EQUIVALENT transformation)

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

(10) Complex Obligation (AND)

(11) Obligation:

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

LESSA_IN_G(s(T25)) → LESSA_IN_G(T25)

The TRS R consists of the following rules:

lessB_in_ga(0, s(T11)) → lessB_out_ga(0, s(T11))
lessB_in_ga(s(T19), 0) → U2_ga(T19, lessA_in_g(T19))
lessA_in_g(s(T25)) → U1_g(T25, lessA_in_g(T25))
U1_g(T25, lessA_out_g(T25)) → lessA_out_g(s(T25))
U2_ga(T19, lessA_out_g(T19)) → lessB_out_ga(s(T19), 0)
lessB_in_ga(s(T19), s(T29)) → U3_ga(T19, T29, lessB_in_ga(T19, T29))
U3_ga(T19, T29, lessB_out_ga(T19, T29)) → lessB_out_ga(s(T19), s(T29))

The argument filtering Pi contains the following mapping:
lessB_in_ga(x1, x2)  =  lessB_in_ga(x1)
0  =  0
lessB_out_ga(x1, x2)  =  lessB_out_ga
s(x1)  =  s(x1)
U2_ga(x1, x2)  =  U2_ga(x2)
lessA_in_g(x1)  =  lessA_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
lessA_out_g(x1)  =  lessA_out_g
U3_ga(x1, x2, x3)  =  U3_ga(x3)
LESSA_IN_G(x1)  =  LESSA_IN_G(x1)

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

(12) UsableRulesProof (EQUIVALENT transformation)

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

(13) Obligation:

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

LESSA_IN_G(s(T25)) → LESSA_IN_G(T25)

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

(14) PiDPToQDPProof (EQUIVALENT transformation)

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

(15) Obligation:

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

LESSA_IN_G(s(T25)) → LESSA_IN_G(T25)

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

(16) 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_G(s(T25)) → LESSA_IN_G(T25)
  The graph contains the following edges 1 > 1

(17) YES

(18) Obligation:

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

LESSB_IN_GA(s(T19), s(T29)) → LESSB_IN_GA(T19, T29)

The TRS R consists of the following rules:

lessB_in_ga(0, s(T11)) → lessB_out_ga(0, s(T11))
lessB_in_ga(s(T19), 0) → U2_ga(T19, lessA_in_g(T19))
lessA_in_g(s(T25)) → U1_g(T25, lessA_in_g(T25))
U1_g(T25, lessA_out_g(T25)) → lessA_out_g(s(T25))
U2_ga(T19, lessA_out_g(T19)) → lessB_out_ga(s(T19), 0)
lessB_in_ga(s(T19), s(T29)) → U3_ga(T19, T29, lessB_in_ga(T19, T29))
U3_ga(T19, T29, lessB_out_ga(T19, T29)) → lessB_out_ga(s(T19), s(T29))

The argument filtering Pi contains the following mapping:
lessB_in_ga(x1, x2)  =  lessB_in_ga(x1)
0  =  0
lessB_out_ga(x1, x2)  =  lessB_out_ga
s(x1)  =  s(x1)
U2_ga(x1, x2)  =  U2_ga(x2)
lessA_in_g(x1)  =  lessA_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
lessA_out_g(x1)  =  lessA_out_g
U3_ga(x1, x2, x3)  =  U3_ga(x3)
LESSB_IN_GA(x1, x2)  =  LESSB_IN_GA(x1)

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

(19) UsableRulesProof (EQUIVALENT transformation)

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

(20) Obligation:

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

LESSB_IN_GA(s(T19), s(T29)) → LESSB_IN_GA(T19, T29)

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

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

(21) PiDPToQDPProof (SOUND transformation)

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

(22) Obligation:

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

LESSB_IN_GA(s(T19)) → LESSB_IN_GA(T19)

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

(23) 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:

• LESSB_IN_GA(s(T19)) → LESSB_IN_GA(T19)
  The graph contains the following edges 1 > 1

(24) YES