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

(0) Obligation:

Clauses:

num(0).
num(X) :- ','(no(zero(0)), ','(p(X, Y), num(Y))).
p(0, 0).
p(s(X), X).
zero(0).
no(X) :- ','(X, ','(!, failure(a))).
no(X1).
failure(b).

Query: num(g)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: num(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: num(T1), SCOPE: 1, CLAUSE: 0 | num(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | num(0), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: num(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nnum(T1) !=? num(0)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: num(0), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ','(no(zero(0)), ','(p(0, X4), num(X4)))\n\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(no(zero(0)), ','(p(0, X4), num(X4))), SCOPE: 2, CLAUSE: 5 | ','(no(zero(0)), ','(p(0, X4), num(X4))), SCOPE: 2, CLAUSE: 6\n\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(','(call(zero(0)), ','(!_2, failure(a))), ','(p(0, X4), num(X4))) | ','(no(zero(0)), ','(p(0, X4), num(X4))), SCOPE: 2, CLAUSE: 6\n\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(zero(0), ','(','(!_2, failure(a)), ','(p(0, X4), num(X4)))) | ?_3 | ','(no(zero(0)), ','(p(0, X4), num(X4))), SCOPE: 2, CLAUSE: 6\n\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(zero(0), ','(','(!_2, failure(a)), ','(p(0, X4), num(X4)))), SCOPE: 4, CLAUSE: 4 | ?_4 | ?_3 | ','(no(zero(0)), ','(p(0, X4), num(X4))), SCOPE: 2, CLAUSE: 6\n\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(','(!_2, failure(a)), ','(p(0, X4), num(X4))) | ?_4 | ?_3 | ','(no(zero(0)), ','(p(0, X4), num(X4))), SCOPE: 2, CLAUSE: 6\n\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(failure(a), ','(p(0, X4), num(X4)))\n\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(failure(a), ','(p(0, X4), num(X4))), SCOPE: 5, CLAUSE: 7\n\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(no(zero(0)), ','(p(T3, X10), num(X10)))\n\nT3 is ground\nX10 is free\nnum(T3) !=? num(0)", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(no(zero(0)), ','(p(T3, X10), num(X10))), SCOPE: 6, CLAUSE: 5 | ','(no(zero(0)), ','(p(T3, X10), num(X10))), SCOPE: 6, CLAUSE: 6\n\nT3 is ground\nX10 is free\nnum(T3) !=? num(0)", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(','(call(zero(0)), ','(!_6, failure(a))), ','(p(T3, X10), num(X10))) | ','(no(zero(0)), ','(p(T3, X10), num(X10))), SCOPE: 6, CLAUSE: 6\n\nT3 is ground\nX10 is free\nnum(T3) !=? num(0)", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(zero(0), ','(','(!_6, failure(a)), ','(p(T3, X10), num(X10)))) | ?_7 | ','(no(zero(0)), ','(p(T3, X10), num(X10))), SCOPE: 6, CLAUSE: 6\n\nT3 is ground\nX10 is free\nnum(T3) !=? num(0)", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: ','(zero(0), ','(','(!_6, failure(a)), ','(p(T3, X10), num(X10)))), SCOPE: 8, CLAUSE: 4 | ?_8 | ?_7 | ','(no(zero(0)), ','(p(T3, X10), num(X10))), SCOPE: 6, CLAUSE: 6\n\nT3 is ground\nX10 is free\nnum(T3) !=? num(0)", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: ','(','(!_6, failure(a)), ','(p(T3, X10), num(X10))) | ?_8 | ?_7 | ','(no(zero(0)), ','(p(T3, X10), num(X10))), SCOPE: 6, CLAUSE: 6\n\nT3 is ground\nX10 is free\nnum(T3) !=? num(0)", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: ','(failure(a), ','(p(T3, X10), num(X10)))\n\nT3 is ground\nX10 is free\nnum(T3) !=? num(0)", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: ','(failure(a), ','(p(T3, X10), num(X10))), SCOPE: 9, CLAUSE: 7\n\nT3 is ground\nX10 is free\nnum(T3) !=? num(0)", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 24 [arrowhead = none ];
24 -> 2;

25 [label="EVAL with clause\nnum(0).\nand substitutionT1 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 25 [arrowhead = none ];
25 -> 3;

26 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 26 [arrowhead = none ];
26 -> 4;

27 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 27 [arrowhead = none ];
27 -> 5;

28 [label="ONLY EVAL with clause\nnum(X9) :- ','(no(zero(0)), ','(p(X9, X10), num(X10))).\nand substitutionT1 -> T3,\nX9 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 28 [arrowhead = none ];
28 -> 15;

29 [label="ONLY EVAL with clause\nnum(X3) :- ','(no(zero(0)), ','(p(X3, X4), num(X4))).\nand substitutionX3 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 29 [arrowhead = none ];
29 -> 6;

30 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
6 -> 30 [arrowhead = none ];
30 -> 7;

31 [label="ONLY EVAL with clause\nno(X7) :- ','(call(X7), ','(!_2, failure(a))).\nand substitutionX7 -> zero(0)", fontsize=14, style = filled, fillcolor = lightblue];
7 -> 31 [arrowhead = none ];
31 -> 8;

32 [label="CALL", fontsize=14, style = filled, fillcolor = lightcyan];
8 -> 32 [arrowhead = none ];
32 -> 9;

33 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
9 -> 33 [arrowhead = none ];
33 -> 10;

34 [label="ONLY EVAL with clause\nzero(0).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 34 [arrowhead = none ];
34 -> 11;

35 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 35 [arrowhead = none ];
35 -> 12;

36 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 36 [arrowhead = none ];
36 -> 13;

37 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 37 [arrowhead = none ];
37 -> 14;

38 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
15 -> 38 [arrowhead = none ];
38 -> 16;

39 [label="ONLY EVAL with clause\nno(X13) :- ','(call(X13), ','(!_6, failure(a))).\nand substitutionX13 -> zero(0)", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 39 [arrowhead = none ];
39 -> 17;

40 [label="CALL", fontsize=14, style = filled, fillcolor = lightcyan];
17 -> 40 [arrowhead = none ];
40 -> 18;

41 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
18 -> 41 [arrowhead = none ];
41 -> 19;

42 [label="ONLY EVAL with clause\nzero(0).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
19 -> 42 [arrowhead = none ];
42 -> 20;

43 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
20 -> 43 [arrowhead = none ];
43 -> 21;

44 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
21 -> 44 [arrowhead = none ];
44 -> 22;

45 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
22 -> 45 [arrowhead = none ];
45 -> 23;

}

(2) Obligation:

Clauses:

numA(0).

Query: numA(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:

numA_in_g(0) → numA_out_g(0)

The argument filtering Pi contains the following mapping:
numA_in_g(x1)  =  numA_in_g(x1)
0  =  0
numA_out_g(x1)  =  numA_out_g

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

numA_in_g(0) → numA_out_g(0)

The argument filtering Pi contains the following mapping:
numA_in_g(x1)  =  numA_in_g(x1)
0  =  0
numA_out_g(x1)  =  numA_out_g

(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:

numA_in_g(0) → numA_out_g(0)

The argument filtering Pi contains the following mapping:
numA_in_g(x1)  =  numA_in_g(x1)
0  =  0
numA_out_g(x1)  =  numA_out_g

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:

numA_in_g(0) → numA_out_g(0)

The argument filtering Pi contains the following mapping:
numA_in_g(x1)  =  numA_in_g(x1)
0  =  0
numA_out_g(x1)  =  numA_out_g

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