Left Termination of the query pattern num_in_1(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).

Queries:

num(g).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

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: num(T1), SCOPE: 1, CLAUSE: 0\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: num(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", 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: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(no(zero(0)), ','(p(T2, X4), num(X4)))\n\nT2 is ground\nX4 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(no(zero(0)), ','(p(T2, X4), num(X4))), SCOPE: 2, CLAUSE: 5 | ','(no(zero(0)), ','(p(T2, X4), num(X4))), SCOPE: 2, CLAUSE: 6 | ?_2\n\nT2 is ground\nX4 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(zero(0), ','(!_2, ','(failure(a), ','(p(T2, X4), num(X4))))) | ','(no(zero(0)), ','(p(T2, X4), num(X4))), SCOPE: 2, CLAUSE: 6 | ?_2\n\nT2 is ground\nX4 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(zero(0), ','(!_2, ','(failure(a), ','(p(T2, X4), num(X4))))), SCOPE: 3, CLAUSE: 4 | ','(no(zero(0)), ','(p(T2, X4), num(X4))), SCOPE: 2, CLAUSE: 6 | ?_2\n\nT2 is ground\nX4 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(!_2, ','(failure(a), ','(p(T2, X4), num(X4)))) | ','(no(zero(0)), ','(p(T2, X4), num(X4))), SCOPE: 2, CLAUSE: 6 | ?_2\n\nT2 is ground\nX4 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(failure(a), ','(p(T2, X4), num(X4)))\n\nT2 is ground\nX4 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(failure(a), ','(p(T2, X4), num(X4))), SCOPE: 4, CLAUSE: 7\n\nT2 is ground\nX4 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 16 [arrowhead = none ];
16 -> 2;

17 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
2 -> 17 [arrowhead = none ];
17 -> 3;

18 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
2 -> 18 [arrowhead = none ];
18 -> 4;

19 [label="EVAL with clause\nnum(0).\nand substitution\nT1 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 19 [arrowhead = none ];
19 -> 5;

20 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 20 [arrowhead = none ];
20 -> 6;

21 [label="EVAL with clause\nnum(X3) :- ','(no(zero(0)), ','(p(X3, X4), num(X4))).\nand substitution\nT1 -> T2\nX3 -> T2", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 21 [arrowhead = none ];
21 -> 8;

22 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 22 [arrowhead = none ];
22 -> 7;

23 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
8 -> 23 [arrowhead = none ];
23 -> 9;

24 [label="EVAL with clause\nno(X6) :- ','(X6, ','(!, failure(a))).\nand substitution\nX6 -> zero(0)", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 24 [arrowhead = none ];
24 -> 10;

25 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
10 -> 25 [arrowhead = none ];
25 -> 11;

26 [label="EVAL with clause\nzero(0).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 26 [arrowhead = none ];
26 -> 12;

27 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 27 [arrowhead = none ];
27 -> 13;

28 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
13 -> 28 [arrowhead = none ];
28 -> 14;

29 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = white];
14 -> 29 [arrowhead = none ];
29 -> 15;

}

(2) Obligation:

Clauses:

num1(0).

Queries:

num1(g).

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. 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:

num1_in_g(0) → num1_out_g(0)

The argument filtering Pi contains the following mapping:
num1_in_g(x1)  =  num1_in_g(x1)
0  =  0
num1_out_g(x1)  =  num1_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:

num1_in_g(0) → num1_out_g(0)

The argument filtering Pi contains the following mapping:
num1_in_g(x1)  =  num1_in_g(x1)
0  =  0
num1_out_g(x1)  =  num1_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:

num1_in_g(0) → num1_out_g(0)

The argument filtering Pi contains the following mapping:
num1_in_g(x1)  =  num1_in_g(x1)
0  =  0
num1_out_g(x1)  =  num1_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:

num1_in_g(0) → num1_out_g(0)

The argument filtering Pi contains the following mapping:
num1_in_g(x1)  =  num1_in_g(x1)
0  =  0
num1_out_g(x1)  =  num1_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) TRUE