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) 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: !_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:

Clauses:

numA(0).
numA(s(T6)) :- numA(T6).

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:
numA_in: (b)
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)
numA_in_g(s(T6)) → U1_g(T6, numA_in_g(T6))
U1_g(T6, numA_out_g(T6)) → numA_out_g(s(T6))

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
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:

numA_in_g(0) → numA_out_g(0)
numA_in_g(s(T6)) → U1_g(T6, numA_in_g(T6))
U1_g(T6, numA_out_g(T6)) → numA_out_g(s(T6))

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
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:

NUMA_IN_G(s(T6)) → U1_G(T6, numA_in_g(T6))
NUMA_IN_G(s(T6)) → NUMA_IN_G(T6)

The TRS R consists of the following rules:

numA_in_g(0) → numA_out_g(0)
numA_in_g(s(T6)) → U1_g(T6, numA_in_g(T6))
U1_g(T6, numA_out_g(T6)) → numA_out_g(s(T6))

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
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
NUMA_IN_G(x1)  =  NUMA_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:

NUMA_IN_G(s(T6)) → U1_G(T6, numA_in_g(T6))
NUMA_IN_G(s(T6)) → NUMA_IN_G(T6)

The TRS R consists of the following rules:

numA_in_g(0) → numA_out_g(0)
numA_in_g(s(T6)) → U1_g(T6, numA_in_g(T6))
U1_g(T6, numA_out_g(T6)) → numA_out_g(s(T6))

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
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
NUMA_IN_G(x1)  =  NUMA_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:

NUMA_IN_G(s(T6)) → NUMA_IN_G(T6)

The TRS R consists of the following rules:

numA_in_g(0) → numA_out_g(0)
numA_in_g(s(T6)) → U1_g(T6, numA_in_g(T6))
U1_g(T6, numA_out_g(T6)) → numA_out_g(s(T6))

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
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
NUMA_IN_G(x1)  =  NUMA_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:

NUMA_IN_G(s(T6)) → NUMA_IN_G(T6)

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:

NUMA_IN_G(s(T6)) → NUMA_IN_G(T6)

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:

• NUMA_IN_G(s(T6)) → NUMA_IN_G(T6)
  The graph contains the following edges 1 > 1

(14) YES