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(s(X)) :- num(X).

Query: num(g)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

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];
5 [label="5: num(T1), SCOPE: 1, CLAUSE: 0 | num(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: true | num(0), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: num(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nnum(T1) !=? num(0)", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: num(0), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: num(T3)\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: num(T3), SCOPE: 2, CLAUSE: 0 | num(T3), SCOPE: 2, CLAUSE: 1\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: num(T3), SCOPE: 2, CLAUSE: 0\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: num(T3), SCOPE: 2, CLAUSE: 1\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
37 [label="37: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
39 [label="39: num(T6)\n\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
41 [label="41: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
42 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 42 [arrowhead = none ];
42 -> 5;

43 [label="EVAL with clause\nnum(0).\nand substitutionT1 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 43 [arrowhead = none ];
43 -> 9;

44 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 44 [arrowhead = none ];
44 -> 12;

45 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 45 [arrowhead = none ];
45 -> 17;

46 [label="EVAL with clause\nnum(s(X3)) :- num(X3).\nand substitutionX3 -> T3,\nT1 -> s(T3)", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 46 [arrowhead = none ];
46 -> 22;

47 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 47 [arrowhead = none ];
47 -> 24;

48 [label="BACKTRACK\nfor clause: num(s(X)) :- num(X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 48 [arrowhead = none ];
48 -> 19;

49 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
22 -> 49 [arrowhead = none ];
49 -> 27;

50 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
27 -> 50 [arrowhead = none ];
50 -> 29;

51 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
27 -> 51 [arrowhead = none ];
51 -> 31;

52 [label="EVAL with clause\nnum(0).\nand substitutionT3 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
29 -> 52 [arrowhead = none ];
52 -> 33;

53 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
29 -> 53 [arrowhead = none ];
53 -> 35;

54 [label="EVAL with clause\nnum(s(X6)) :- num(X6).\nand substitutionX6 -> T6,\nT3 -> s(T6)", fontsize=14, style = filled, fillcolor = lightblue];
31 -> 54 [arrowhead = none ];
54 -> 39;

55 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
31 -> 55 [arrowhead = none ];
55 -> 41;

56 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
33 -> 56 [arrowhead = none ];
56 -> 37;

57 [label="INSTANCE with matching:\nT1 -> T6", fontsize=14, style = filled, fillcolor = lightgrey];
39 -> 57 [arrowhead = none , style=dashed];
57 -> 1[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(0)) → numA_out_g(s(0))
numA_in_g(s(s(T6))) → U1_g(T6, numA_in_g(T6))
U1_g(T6, numA_out_g(T6)) → numA_out_g(s(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(s(T6))) → U1_G(T6, numA_in_g(T6))
NUMA_IN_G(s(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(0)) → numA_out_g(s(0))
numA_in_g(s(s(T6))) → U1_g(T6, numA_in_g(T6))
U1_g(T6, numA_out_g(T6)) → numA_out_g(s(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(s(T6))) → U1_G(T6, numA_in_g(T6))
NUMA_IN_G(s(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(0)) → numA_out_g(s(0))
numA_in_g(s(s(T6))) → U1_g(T6, numA_in_g(T6))
U1_g(T6, numA_out_g(T6)) → numA_out_g(s(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(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(0)) → numA_out_g(s(0))
numA_in_g(s(s(T6))) → U1_g(T6, numA_in_g(T6))
U1_g(T6, numA_out_g(T6)) → numA_out_g(s(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(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(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(s(T6))) → NUMA_IN_G(T6)
  The graph contains the following edges 1 > 1

(12) YES