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) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
2 [label="A: num(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
4 [label="4: num(T1), SCOPE: 1, CLAUSE: 0 | num(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: !_1 | num(0), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: num(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nnum(T1) !=? num(0)", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: ','(p(T3, X3), num(X3))\n\nT3 is ground\nX3 is free\nnum(T3) !=? num(0)", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: ','(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];
26 [label="26: ','(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];
27 [label="27: num(T6)\n\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
2 -> 31 [arrowhead = none ];
31 -> 4;

32 [label="EVAL with clause\nnum(0) :- !_1.\nand substitutionT1 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 32 [arrowhead = none ];
32 -> 8;

33 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 33 [arrowhead = none ];
33 -> 11;

34 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 34 [arrowhead = none ];
34 -> 15;

35 [label="ONLY EVAL with clause\nnum(X2) :- ','(p(X2, X3), num(X3)).\nand substitutionT1 -> T3,\nX2 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 35 [arrowhead = none ];
35 -> 20;

36 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 36 [arrowhead = none ];
36 -> 18;

37 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
20 -> 37 [arrowhead = none ];
37 -> 24;

38 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (num(T3), num(0))", fontsize=14, style = filled, fillcolor = lightgreen];
24 -> 38 [arrowhead = none ];
38 -> 26;

39 [label="EVAL with clause\np(s(X6), X6).\nand substitutionX6 -> T6,\nT3 -> s(T6),\nX3 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
26 -> 39 [arrowhead = none ];
39 -> 27;

40 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
26 -> 40 [arrowhead = none ];
40 -> 30;

41 [label="INSTANCE with matching:\nT1 -> T6", fontsize=14, style = filled, fillcolor = lightgrey];
27 -> 41 [arrowhead = none , style=dashed];
41 -> 2[style=dashed];

}

(2) 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))

Pi is empty.

(3) 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))

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

(4) 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))

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(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))

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

(7) UsableRulesProof (EQUIVALENT transformation)

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

(8) 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

(9) PiDPToQDPProof (EQUIVALENT transformation)

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

(10) 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.

(11) 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

(12) YES