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(N) :- ','(p(N, P), odd(P)).
odd(s(0)) :- !.
odd(s(N)) :- even(P).
p(0, 0).
p(s(X), X).

Query: even(g)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: 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\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: !_1 | even(0), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: even(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\neven(T1) !=? even(0)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(p(T3, X3), odd(X3))\n\nT3 is ground\nX3 is free\neven(T3) !=? even(0)", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(p(T3, X3), odd(X3)), SCOPE: 2, CLAUSE: 4 | ','(p(T3, X3), odd(X3)), SCOPE: 2, CLAUSE: 5\n\nT3 is ground\nX3 is free\neven(T3) !=? even(0)", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(p(T3, X3), odd(X3)), SCOPE: 2, CLAUSE: 5\n\nT3 is ground\nX3 is free\neven(T3) !=? even(0)", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: odd(T6)\n\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: odd(T6), SCOPE: 3, CLAUSE: 2 | odd(T6), SCOPE: 3, CLAUSE: 3\n\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: !_3 | odd(s(0)), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: odd(T6), SCOPE: 3, CLAUSE: 3\n\nT6 is ground\nodd(T6) !=? odd(s(0))", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: even(X11)\n\nX11 is free", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: even(X11), SCOPE: 4, CLAUSE: 0 | even(X11), SCOPE: 4, CLAUSE: 1\n\nX11 is free", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: !_4 | even(X11), SCOPE: 4, CLAUSE: 1\n\nX11 is free", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 23 [arrowhead = none ];
23 -> 2;

24 [label="EVAL with clause\neven(0) :- !_1.\nand substitutionT1 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 24 [arrowhead = none ];
24 -> 3;

25 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 25 [arrowhead = none ];
25 -> 4;

26 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 26 [arrowhead = none ];
26 -> 5;

27 [label="ONLY EVAL with clause\neven(X2) :- ','(p(X2, X3), odd(X3)).\nand substitutionT1 -> T3,\nX2 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 27 [arrowhead = none ];
27 -> 7;

28 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 28 [arrowhead = none ];
28 -> 6;

29 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 29 [arrowhead = none ];
29 -> 8;

30 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (even(T3), even(0))", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 30 [arrowhead = none ];
30 -> 9;

31 [label="EVAL with clause\np(s(X6), X6).\nand substitutionX6 -> T6,\nT3 -> s(T6),\nX3 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 31 [arrowhead = none ];
31 -> 10;

32 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 32 [arrowhead = none ];
32 -> 11;

33 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
10 -> 33 [arrowhead = none ];
33 -> 12;

34 [label="EVAL with clause\nodd(s(0)) :- !_3.\nand substitutionT6 -> s(0)", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 34 [arrowhead = none ];
34 -> 13;

35 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 35 [arrowhead = none ];
35 -> 14;

36 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 36 [arrowhead = none ];
36 -> 15;

37 [label="EVAL with clause\nodd(s(X10)) :- even(X11).\nand substitutionX10 -> T9,\nT6 -> s(T9)", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 37 [arrowhead = none ];
37 -> 17;

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

39 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 39 [arrowhead = none ];
39 -> 16;

40 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
17 -> 40 [arrowhead = none ];
40 -> 19;

41 [label="ONLY EVAL with clause\neven(0) :- !_4.\nand substitutionX11 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
19 -> 41 [arrowhead = none ];
41 -> 20;

42 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
20 -> 42 [arrowhead = none ];
42 -> 21;

43 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
21 -> 43 [arrowhead = none ];
43 -> 22;

}

(2) Obligation:

Clauses:

evenA(0).
evenA(s(s(0))).
evenA(s(s(T9))).

Query: evenA(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:
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

evenA_in_g(0) → evenA_out_g(0)
evenA_in_g(s(s(0))) → evenA_out_g(s(s(0)))
evenA_in_g(s(s(T9))) → evenA_out_g(s(s(T9)))

The argument filtering Pi contains the following mapping:
evenA_in_g(x1)  =  evenA_in_g(x1)
0  =  0
evenA_out_g(x1)  =  evenA_out_g
s(x1)  =  s(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

evenA_in_g(0) → evenA_out_g(0)
evenA_in_g(s(s(0))) → evenA_out_g(s(s(0)))
evenA_in_g(s(s(T9))) → evenA_out_g(s(s(T9)))

The argument filtering Pi contains the following mapping:
evenA_in_g(x1)  =  evenA_in_g(x1)
0  =  0
evenA_out_g(x1)  =  evenA_out_g
s(x1)  =  s(x1)

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
P is empty.
The TRS R consists of the following rules:

evenA_in_g(0) → evenA_out_g(0)
evenA_in_g(s(s(0))) → evenA_out_g(s(s(0)))
evenA_in_g(s(s(T9))) → evenA_out_g(s(s(T9)))

The argument filtering Pi contains the following mapping:
evenA_in_g(x1)  =  evenA_in_g(x1)
0  =  0
evenA_out_g(x1)  =  evenA_out_g
s(x1)  =  s(x1)

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

(6) Obligation:

Pi DP problem:
P is empty.
The TRS R consists of the following rules:

evenA_in_g(0) → evenA_out_g(0)
evenA_in_g(s(s(0))) → evenA_out_g(s(s(0)))
evenA_in_g(s(s(T9))) → evenA_out_g(s(s(T9)))

The argument filtering Pi contains the following mapping:
evenA_in_g(x1)  =  evenA_in_g(x1)
0  =  0
evenA_out_g(x1)  =  evenA_out_g
s(x1)  =  s(x1)

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

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) YES