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

(0) Obligation:

Clauses:

less(0, s(X1)).
less(s(X), s(Y)) :- less(X, Y).

Query: less(a,g)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
3 [label="A: less(T1, T2)\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
4 [label="4: less(T1, T2), SCOPE: 1, CLAUSE: 0 | less(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: true | less(T1, s(T4)), SCOPE: 1, CLAUSE: 1\n\nT4 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: less(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT2 is ground\nX3 is free\nless(T1, T2) !=? less(0, s(X3))", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: less(T1, s(T4)), SCOPE: 1, CLAUSE: 1\n\nT4 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: less(T9, T8)\n\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: less(T9, T8), SCOPE: 2, CLAUSE: 0 | less(T9, T8), SCOPE: 2, CLAUSE: 1\n\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: less(T9, T8), SCOPE: 2, CLAUSE: 0\n\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: less(T9, T8), SCOPE: 2, CLAUSE: 1\n\nT8 is ground", 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="38: less(T21, T20)\n\nT20 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
39 [label="39: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
40 [label="40: less(T28, T27)\n\nT27 is ground\nX3 is free\nless(T1, s(T27)) !=? less(0, s(X3))", fontsize=16, style=filled, fillcolor=lightyellow];
42 [label="42: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
44 [label="44: less(T28, T27), SCOPE: 3, CLAUSE: 0 | less(T28, T27), SCOPE: 3, CLAUSE: 1\n\nT27 is ground\nX3 is free\nless(T1, s(T27)) !=? less(0, s(X3))", fontsize=16, style=filled, fillcolor=lightyellow];
46 [label="46: less(T28, T27), SCOPE: 3, CLAUSE: 0\n\nT27 is ground\nX3 is free\nless(T1, s(T27)) !=? less(0, s(X3))", fontsize=16, style=filled, fillcolor=lightyellow];
48 [label="48: less(T28, T27), SCOPE: 3, CLAUSE: 1\n\nT27 is ground\nX3 is free\nless(T1, s(T27)) !=? less(0, s(X3))", fontsize=16, style=filled, fillcolor=lightyellow];
50 [label="50: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
52 [label="52: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
53 [label="53: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
54 [label="54: less(T40, T39)\n\nT39 is ground\nX3 is free\nless(T1, s(s(T39))) !=? less(0, s(X3))", fontsize=16, style=filled, fillcolor=lightyellow];
55 [label="55: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
56 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 56 [arrowhead = none ];
56 -> 4;

57 [label="EVAL with clause\nless(0, s(X3)).\nand substitutionT1 -> 0,\nX3 -> T4,\nT2 -> s(T4)", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 57 [arrowhead = none ];
57 -> 7;

58 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 58 [arrowhead = none ];
58 -> 8;

59 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 59 [arrowhead = none ];
59 -> 9;

60 [label="EVAL with clause\nless(s(X23), s(X24)) :- less(X23, X24).\nand substitutionX23 -> T28,\nT1 -> s(T28),\nX24 -> T27,\nT2 -> s(T27),\nT26 -> T28", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 60 [arrowhead = none ];
60 -> 40;

61 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 61 [arrowhead = none ];
61 -> 42;

62 [label="EVAL with clause\nless(s(X6), s(X7)) :- less(X6, X7).\nand substitutionX6 -> T9,\nT1 -> s(T9),\nT4 -> T8,\nX7 -> T8,\nT7 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 62 [arrowhead = none ];
62 -> 12;

63 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 63 [arrowhead = none ];
63 -> 13;

64 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 64 [arrowhead = none ];
64 -> 14;

65 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 65 [arrowhead = none ];
65 -> 16;

66 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 66 [arrowhead = none ];
66 -> 17;

67 [label="EVAL with clause\nless(0, s(X12)).\nand substitutionT9 -> 0,\nX12 -> T14,\nT8 -> s(T14)", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 67 [arrowhead = none ];
67 -> 35;

68 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 68 [arrowhead = none ];
68 -> 36;

69 [label="EVAL with clause\nless(s(X17), s(X18)) :- less(X17, X18).\nand substitutionX17 -> T21,\nT9 -> s(T21),\nX18 -> T20,\nT8 -> s(T20),\nT19 -> T21", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 69 [arrowhead = none ];
69 -> 38;

70 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 70 [arrowhead = none ];
70 -> 39;

71 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
35 -> 71 [arrowhead = none ];
71 -> 37;

72 [label="INSTANCE with matching:\nT1 -> T21\nT2 -> T20", fontsize=14, style = filled, fillcolor = lightgrey];
38 -> 72 [arrowhead = none , style=dashed];
72 -> 3[style=dashed];

73 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
40 -> 73 [arrowhead = none ];
73 -> 44;

74 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
44 -> 74 [arrowhead = none ];
74 -> 46;

75 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
44 -> 75 [arrowhead = none ];
75 -> 48;

76 [label="EVAL with clause\nless(0, s(X29)).\nand substitutionT28 -> 0,\nX29 -> T33,\nT27 -> s(T33)", fontsize=14, style = filled, fillcolor = lightblue];
46 -> 76 [arrowhead = none ];
76 -> 50;

77 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
46 -> 77 [arrowhead = none ];
77 -> 52;

78 [label="EVAL with clause\nless(s(X34), s(X35)) :- less(X34, X35).\nand substitutionX34 -> T40,\nT28 -> s(T40),\nX35 -> T39,\nT27 -> s(T39),\nT38 -> T40", fontsize=14, style = filled, fillcolor = lightblue];
48 -> 78 [arrowhead = none ];
78 -> 54;

79 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
48 -> 79 [arrowhead = none ];
79 -> 55;

80 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
50 -> 80 [arrowhead = none ];
80 -> 53;

81 [label="INSTANCE with matching:\nT1 -> T40\nT2 -> T39", fontsize=14, style = filled, fillcolor = lightgrey];
54 -> 81 [arrowhead = none , style=dashed];
81 -> 3[style=dashed];

}

(2) Obligation:

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

lessA_in_ag(0, s(T4)) → lessA_out_ag(0, s(T4))
lessA_in_ag(s(0), s(s(T14))) → lessA_out_ag(s(0), s(s(T14)))
lessA_in_ag(s(s(T21)), s(s(T20))) → U1_ag(T21, T20, lessA_in_ag(T21, T20))
lessA_in_ag(s(s(T40)), s(s(T39))) → U2_ag(T40, T39, lessA_in_ag(T40, T39))
U2_ag(T40, T39, lessA_out_ag(T40, T39)) → lessA_out_ag(s(s(T40)), s(s(T39)))
U1_ag(T21, T20, lessA_out_ag(T21, T20)) → lessA_out_ag(s(s(T21)), s(s(T20)))

The argument filtering Pi contains the following mapping:
lessA_in_ag(x1, x2)  =  lessA_in_ag(x2)
s(x1)  =  s(x1)
lessA_out_ag(x1, x2)  =  lessA_out_ag(x1, x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
U2_ag(x1, x2, x3)  =  U2_ag(x2, x3)

(3) 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:

LESSA_IN_AG(s(s(T21)), s(s(T20))) → U1_AG(T21, T20, lessA_in_ag(T21, T20))
LESSA_IN_AG(s(s(T21)), s(s(T20))) → LESSA_IN_AG(T21, T20)
LESSA_IN_AG(s(s(T40)), s(s(T39))) → U2_AG(T40, T39, lessA_in_ag(T40, T39))

The TRS R consists of the following rules:

lessA_in_ag(0, s(T4)) → lessA_out_ag(0, s(T4))
lessA_in_ag(s(0), s(s(T14))) → lessA_out_ag(s(0), s(s(T14)))
lessA_in_ag(s(s(T21)), s(s(T20))) → U1_ag(T21, T20, lessA_in_ag(T21, T20))
lessA_in_ag(s(s(T40)), s(s(T39))) → U2_ag(T40, T39, lessA_in_ag(T40, T39))
U2_ag(T40, T39, lessA_out_ag(T40, T39)) → lessA_out_ag(s(s(T40)), s(s(T39)))
U1_ag(T21, T20, lessA_out_ag(T21, T20)) → lessA_out_ag(s(s(T21)), s(s(T20)))

The argument filtering Pi contains the following mapping:
lessA_in_ag(x1, x2)  =  lessA_in_ag(x2)
s(x1)  =  s(x1)
lessA_out_ag(x1, x2)  =  lessA_out_ag(x1, x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
U2_ag(x1, x2, x3)  =  U2_ag(x2, x3)
LESSA_IN_AG(x1, x2)  =  LESSA_IN_AG(x2)
U1_AG(x1, x2, x3)  =  U1_AG(x2, x3)
U2_AG(x1, x2, x3)  =  U2_AG(x2, x3)

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

(4) Obligation:

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

LESSA_IN_AG(s(s(T21)), s(s(T20))) → U1_AG(T21, T20, lessA_in_ag(T21, T20))
LESSA_IN_AG(s(s(T21)), s(s(T20))) → LESSA_IN_AG(T21, T20)
LESSA_IN_AG(s(s(T40)), s(s(T39))) → U2_AG(T40, T39, lessA_in_ag(T40, T39))

The TRS R consists of the following rules:

lessA_in_ag(0, s(T4)) → lessA_out_ag(0, s(T4))
lessA_in_ag(s(0), s(s(T14))) → lessA_out_ag(s(0), s(s(T14)))
lessA_in_ag(s(s(T21)), s(s(T20))) → U1_ag(T21, T20, lessA_in_ag(T21, T20))
lessA_in_ag(s(s(T40)), s(s(T39))) → U2_ag(T40, T39, lessA_in_ag(T40, T39))
U2_ag(T40, T39, lessA_out_ag(T40, T39)) → lessA_out_ag(s(s(T40)), s(s(T39)))
U1_ag(T21, T20, lessA_out_ag(T21, T20)) → lessA_out_ag(s(s(T21)), s(s(T20)))

The argument filtering Pi contains the following mapping:
lessA_in_ag(x1, x2)  =  lessA_in_ag(x2)
s(x1)  =  s(x1)
lessA_out_ag(x1, x2)  =  lessA_out_ag(x1, x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
U2_ag(x1, x2, x3)  =  U2_ag(x2, x3)
LESSA_IN_AG(x1, x2)  =  LESSA_IN_AG(x2)
U1_AG(x1, x2, x3)  =  U1_AG(x2, x3)
U2_AG(x1, x2, x3)  =  U2_AG(x2, x3)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes.

(6) Obligation:

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

LESSA_IN_AG(s(s(T21)), s(s(T20))) → LESSA_IN_AG(T21, T20)

The TRS R consists of the following rules:

lessA_in_ag(0, s(T4)) → lessA_out_ag(0, s(T4))
lessA_in_ag(s(0), s(s(T14))) → lessA_out_ag(s(0), s(s(T14)))
lessA_in_ag(s(s(T21)), s(s(T20))) → U1_ag(T21, T20, lessA_in_ag(T21, T20))
lessA_in_ag(s(s(T40)), s(s(T39))) → U2_ag(T40, T39, lessA_in_ag(T40, T39))
U2_ag(T40, T39, lessA_out_ag(T40, T39)) → lessA_out_ag(s(s(T40)), s(s(T39)))
U1_ag(T21, T20, lessA_out_ag(T21, T20)) → lessA_out_ag(s(s(T21)), s(s(T20)))

The argument filtering Pi contains the following mapping:
lessA_in_ag(x1, x2)  =  lessA_in_ag(x2)
s(x1)  =  s(x1)
lessA_out_ag(x1, x2)  =  lessA_out_ag(x1, x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
U2_ag(x1, x2, x3)  =  U2_ag(x2, x3)
LESSA_IN_AG(x1, x2)  =  LESSA_IN_AG(x2)

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

(7) UsableRulesProof (EQUIVALENT transformation)

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

(8) Obligation:

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

LESSA_IN_AG(s(s(T21)), s(s(T20))) → LESSA_IN_AG(T21, T20)

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

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

(9) PiDPToQDPProof (SOUND transformation)

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

(10) Obligation:

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

LESSA_IN_AG(s(s(T20))) → LESSA_IN_AG(T20)

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

(11) 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_AG(s(s(T20))) → LESSA_IN_AG(T20)
  The graph contains the following edges 1 > 1

(12) YES