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) :- ','(p(X, Y), num(Y)).
p(0, 0).
p(s(X), X).

Query: num(g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: 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: !_1 | 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: 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), num(X3))\n\nT3 is ground\nX3 is free\nnum(T3) !=? num(0)", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(p(T3, X3), num(X3)), SCOPE: 2, CLAUSE: 2 | ','(p(T3, X3), num(X3)), SCOPE: 2, CLAUSE: 3\n\nT3 is ground\nX3 is free\nnum(T3) !=? num(0)", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(p(T3, X3), num(X3)), SCOPE: 2, CLAUSE: 3\n\nT3 is ground\nX3 is free\nnum(T3) !=? num(0)", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: num(T6)\n\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 12 [arrowhead = none ];
12 -> 2;

13 [label="EVAL with clause\nnum(0) :- !_1.\nand substitutionT1 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 13 [arrowhead = none ];
13 -> 3;

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

15 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 15 [arrowhead = none ];
15 -> 5;

16 [label="ONLY EVAL with clause\nnum(X2) :- ','(p(X2, X3), num(X3)).\nand substitutionT1 -> T3,\nX2 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 16 [arrowhead = none ];
16 -> 7;

17 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 17 [arrowhead = none ];
17 -> 6;

18 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 18 [arrowhead = none ];
18 -> 8;

19 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (num(T3), num(0))", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 19 [arrowhead = none ];
19 -> 9;

20 [label="EVAL with clause\np(s(X6), X6).\nand substitutionX6 -> T6,\nT3 -> s(T6),\nX3 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 20 [arrowhead = none ];
20 -> 10;

21 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 21 [arrowhead = none ];
21 -> 11;

22 [label="INSTANCE with matching:\nT1 -> T6", fontsize=14, style = filled, fillcolor = lightgrey];
10 -> 22 [arrowhead = none , style=dashed];
22 -> 1[style=dashed];

}

(2) Obligation:

Triples:

numA(s(X1)) :- numA(X1).

Clauses:

numcA(0).
numcA(s(X1)) :- numcA(X1).

Afs:

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

NUMA_IN_G(s(X1)) → U1_G(X1, numA_in_g(X1))
NUMA_IN_G(s(X1)) → NUMA_IN_G(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:

NUMA_IN_G(s(X1)) → U1_G(X1, numA_in_g(X1))
NUMA_IN_G(s(X1)) → NUMA_IN_G(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:

NUMA_IN_G(s(X1)) → NUMA_IN_G(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:

NUMA_IN_G(s(X1)) → NUMA_IN_G(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:

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

(10) YES