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

(0) Obligation:

Clauses:

even(0).
even(s(s(0))).
even(s(s(s(X)))) :- odd(X).
odd(s(0)).
odd(s(X)) :- even(s(s(X))).

Query: even(g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: even(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: even(T1), SCOPE: 1, CLAUSE: 0 | even(T1), SCOPE: 1, CLAUSE: 1 | even(T1), SCOPE: 1, CLAUSE: 2\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | even(0), SCOPE: 1, CLAUSE: 1 | even(0), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: even(T1), SCOPE: 1, CLAUSE: 1 | even(T1), SCOPE: 1, CLAUSE: 2\n\nT1 is ground\neven(T1) !=? even(0)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: even(0), SCOPE: 1, CLAUSE: 1 | even(0), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: even(0), 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 | even(s(s(0))), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: even(T1), SCOPE: 1, CLAUSE: 2\n\nT1 is ground\neven(T1) !=? even(0)\neven(T1) !=? even(s(s(0)))", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: even(s(s(0))), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: odd(T3)\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: odd(T3), SCOPE: 2, CLAUSE: 3 | odd(T3), SCOPE: 2, CLAUSE: 4\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: odd(T3), SCOPE: 2, CLAUSE: 3\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: odd(T3), SCOPE: 2, CLAUSE: 4\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: even(s(s(T6)))\n\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 22 [arrowhead = none ];
22 -> 2;

23 [label="EVAL with clause\neven(0).\nand substitutionT1 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 23 [arrowhead = none ];
23 -> 3;

24 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 24 [arrowhead = none ];
24 -> 4;

25 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 25 [arrowhead = none ];
25 -> 5;

26 [label="EVAL with clause\neven(s(s(0))).\nand substitutionT1 -> s(s(0))", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 26 [arrowhead = none ];
26 -> 8;

27 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 27 [arrowhead = none ];
27 -> 9;

28 [label="BACKTRACK\nfor clause: even(s(s(0)))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 28 [arrowhead = none ];
28 -> 6;

29 [label="BACKTRACK\nfor clause: even(s(s(s(X)))) :- odd(X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 29 [arrowhead = none ];
29 -> 7;

30 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 30 [arrowhead = none ];
30 -> 10;

31 [label="EVAL with clause\neven(s(s(s(X4)))) :- odd(X4).\nand substitutionX4 -> T3,\nT1 -> s(s(s(T3)))", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 31 [arrowhead = none ];
31 -> 12;

32 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 32 [arrowhead = none ];
32 -> 13;

33 [label="BACKTRACK\nfor clause: even(s(s(s(X)))) :- odd(X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 33 [arrowhead = none ];
33 -> 11;

34 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 34 [arrowhead = none ];
34 -> 14;

35 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 35 [arrowhead = none ];
35 -> 15;

36 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 36 [arrowhead = none ];
36 -> 16;

37 [label="EVAL with clause\nodd(s(0)).\nand substitutionT3 -> s(0)", fontsize=14, style = filled, fillcolor = lightblue];
15 -> 37 [arrowhead = none ];
37 -> 17;

38 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 38 [arrowhead = none ];
38 -> 18;

39 [label="EVAL with clause\nodd(s(X7)) :- even(s(s(X7))).\nand substitutionX7 -> T6,\nT3 -> s(T6)", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 39 [arrowhead = none ];
39 -> 20;

40 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 40 [arrowhead = none ];
40 -> 21;

41 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 41 [arrowhead = none ];
41 -> 19;

42 [label="INSTANCE with matching:\nT1 -> s(s(T6))", fontsize=14, style = filled, fillcolor = lightgrey];
20 -> 42 [arrowhead = none , style=dashed];
42 -> 1[style=dashed];

}

(2) Obligation:

Triples:

evenA(s(s(s(s(X1))))) :- evenA(s(s(X1))).

Clauses:

evencA(0).
evencA(s(s(0))).
evencA(s(s(s(s(0))))).
evencA(s(s(s(s(X1))))) :- evencA(s(s(X1))).

Afs:

evenA(x1)  =  evenA(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:
evenA_in: (b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

EVENA_IN_G(s(s(s(s(X1))))) → U1_G(X1, evenA_in_g(s(s(X1))))
EVENA_IN_G(s(s(s(s(X1))))) → EVENA_IN_G(s(s(X1)))

R is empty.
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:

EVENA_IN_G(s(s(s(s(X1))))) → U1_G(X1, evenA_in_g(s(s(X1))))
EVENA_IN_G(s(s(s(s(X1))))) → EVENA_IN_G(s(s(X1)))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Obligation:

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

EVENA_IN_G(s(s(s(s(X1))))) → EVENA_IN_G(s(s(X1)))

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

(7) PiDPToQDPProof (EQUIVALENT transformation)

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

(8) Obligation:

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

EVENA_IN_G(s(s(s(s(X1))))) → EVENA_IN_G(s(s(X1)))

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

(9) 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:

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

(10) YES