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(N) :- ','(p(N, P), 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: ','(p(T9, X11), even(X11))\n\nT9 is ground\nX11 is free\nodd(T9) !=? odd(s(0))", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(p(T9, X11), even(X11)), SCOPE: 4, CLAUSE: 4 | ','(p(T9, X11), even(X11)), SCOPE: 4, CLAUSE: 5\n\nT9 is ground\nX11 is free\nodd(T9) !=? odd(s(0))", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: ','(p(T9, X11), even(X11)), SCOPE: 4, CLAUSE: 4\n\nT9 is ground\nX11 is free\nodd(T9) !=? odd(s(0))", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: ','(p(T9, X11), even(X11)), SCOPE: 4, CLAUSE: 5\n\nT9 is ground\nX11 is free\nodd(T9) !=? odd(s(0))", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: even(0)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: even(0), SCOPE: 5, CLAUSE: 0 | even(0), SCOPE: 5, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: !_5 | even(0), SCOPE: 5, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: even(T12)\n\nT12 is ground\nodd(s(T12)) !=? odd(s(0))", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 29 [arrowhead = none ];
29 -> 2;

30 [label="EVAL with clause\neven(0) :- !_1.\nand substitutionT1 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 30 [arrowhead = none ];
30 -> 3;

31 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 31 [arrowhead = none ];
31 -> 4;

32 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 32 [arrowhead = none ];
32 -> 5;

33 [label="ONLY EVAL with clause\neven(X2) :- ','(p(X2, X3), odd(X3)).\nand substitutionT1 -> T3,\nX2 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 33 [arrowhead = none ];
33 -> 7;

34 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 34 [arrowhead = none ];
34 -> 6;

35 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 35 [arrowhead = none ];
35 -> 8;

36 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (even(T3), even(0))", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 36 [arrowhead = none ];
36 -> 9;

37 [label="EVAL with clause\np(s(X6), X6).\nand substitutionX6 -> T6,\nT3 -> s(T6),\nX3 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 37 [arrowhead = none ];
37 -> 10;

38 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 38 [arrowhead = none ];
38 -> 11;

39 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
10 -> 39 [arrowhead = none ];
39 -> 12;

40 [label="EVAL with clause\nodd(s(0)) :- !_3.\nand substitutionT6 -> s(0)", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 40 [arrowhead = none ];
40 -> 13;

41 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 41 [arrowhead = none ];
41 -> 14;

42 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 42 [arrowhead = none ];
42 -> 15;

43 [label="ONLY EVAL with clause\nodd(X10) :- ','(p(X10, X11), even(X11)).\nand substitutionT6 -> T9,\nX10 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 43 [arrowhead = none ];
43 -> 17;

44 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 44 [arrowhead = none ];
44 -> 16;

45 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
17 -> 45 [arrowhead = none ];
45 -> 18;

46 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
18 -> 46 [arrowhead = none ];
46 -> 19;

47 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
18 -> 47 [arrowhead = none ];
47 -> 20;

48 [label="EVAL with clause\np(0, 0).\nand substitutionT9 -> 0,\nX11 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
19 -> 48 [arrowhead = none ];
48 -> 21;

49 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
19 -> 49 [arrowhead = none ];
49 -> 22;

50 [label="EVAL with clause\np(s(X14), X14).\nand substitutionX14 -> T12,\nT9 -> s(T12),\nX11 -> T12", fontsize=14, style = filled, fillcolor = lightblue];
20 -> 50 [arrowhead = none ];
50 -> 27;

51 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
20 -> 51 [arrowhead = none ];
51 -> 28;

52 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
21 -> 52 [arrowhead = none ];
52 -> 23;

53 [label="ONLY EVAL with clause\neven(0) :- !_5.\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
23 -> 53 [arrowhead = none ];
53 -> 24;

54 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
24 -> 54 [arrowhead = none ];
54 -> 25;

55 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 55 [arrowhead = none ];
55 -> 26;

56 [label="INSTANCE with matching:\nT1 -> T12", fontsize=14, style = filled, fillcolor = lightgrey];
27 -> 56 [arrowhead = none , style=dashed];
56 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

evenA(0).
evenA(s(s(0))).
evenA(s(0)).
evenA(s(s(T12))) :- evenA(T12).

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:
evenA_in: (b)
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(0)) → evenA_out_g(s(0))
evenA_in_g(s(s(T12))) → U1_g(T12, evenA_in_g(T12))
U1_g(T12, evenA_out_g(T12)) → evenA_out_g(s(s(T12)))

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)
U1_g(x1, x2)  =  U1_g(x2)

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(0)) → evenA_out_g(s(0))
evenA_in_g(s(s(T12))) → U1_g(T12, evenA_in_g(T12))
U1_g(T12, evenA_out_g(T12)) → evenA_out_g(s(s(T12)))

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)
U1_g(x1, x2)  =  U1_g(x2)

(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:

EVENA_IN_G(s(s(T12))) → U1_G(T12, evenA_in_g(T12))
EVENA_IN_G(s(s(T12))) → EVENA_IN_G(T12)

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(0)) → evenA_out_g(s(0))
evenA_in_g(s(s(T12))) → U1_g(T12, evenA_in_g(T12))
U1_g(T12, evenA_out_g(T12)) → evenA_out_g(s(s(T12)))

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)
U1_g(x1, x2)  =  U1_g(x2)
EVENA_IN_G(x1)  =  EVENA_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)

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

(6) Obligation:

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

EVENA_IN_G(s(s(T12))) → U1_G(T12, evenA_in_g(T12))
EVENA_IN_G(s(s(T12))) → EVENA_IN_G(T12)

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(0)) → evenA_out_g(s(0))
evenA_in_g(s(s(T12))) → U1_g(T12, evenA_in_g(T12))
U1_g(T12, evenA_out_g(T12)) → evenA_out_g(s(s(T12)))

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)
U1_g(x1, x2)  =  U1_g(x2)
EVENA_IN_G(x1)  =  EVENA_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

EVENA_IN_G(s(s(T12))) → EVENA_IN_G(T12)

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(0)) → evenA_out_g(s(0))
evenA_in_g(s(s(T12))) → U1_g(T12, evenA_in_g(T12))
U1_g(T12, evenA_out_g(T12)) → evenA_out_g(s(s(T12)))

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)
U1_g(x1, x2)  =  U1_g(x2)
EVENA_IN_G(x1)  =  EVENA_IN_G(x1)

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

(9) UsableRulesProof (EQUIVALENT transformation)

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

(10) Obligation:

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

EVENA_IN_G(s(s(T12))) → EVENA_IN_G(T12)

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

(11) PiDPToQDPProof (EQUIVALENT transformation)

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

(12) Obligation:

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

EVENA_IN_G(s(s(T12))) → EVENA_IN_G(T12)

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

(13) 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(T12))) → EVENA_IN_G(T12)
  The graph contains the following edges 1 > 1

(14) YES