Left Termination of the query pattern
p(g)
w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

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

Query: p(g)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
2 [label="A: p(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
5 [label="5: p(T1), SCOPE: 1, CLAUSE: 0 | p(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: true | p(0), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: p(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\np(T1) !=? p(0)", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: p(0), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="B: ','(geq(T3, X4), p(X4))\n\nT3 is ground\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: ','(geq(T3, X4), p(X4)), SCOPE: 2, CLAUSE: 2 | ','(geq(T3, X4), p(X4)), SCOPE: 2, CLAUSE: 3\n\nT3 is ground\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: ','(geq(T3, X4), p(X4)), SCOPE: 2, CLAUSE: 2\n\nT3 is ground\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: ','(geq(T3, X4), p(X4)), SCOPE: 2, CLAUSE: 3\n\nT3 is ground\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: p(T8)\n\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
37 [label="37: ','(geq(T11, X18), p(X18))\n\nT11 is ground\nX18 is free", fontsize=16, style=filled, fillcolor=lightyellow];
40 [label="40: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
41 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
2 -> 41 [arrowhead = none ];
41 -> 5;

42 [label="EVAL with clause\np(0).\nand substitutionT1 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 42 [arrowhead = none ];
42 -> 9;

43 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 43 [arrowhead = none ];
43 -> 11;

44 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 44 [arrowhead = none ];
44 -> 16;

45 [label="EVAL with clause\np(s(X3)) :- ','(geq(X3, X4), p(X4)).\nand substitutionX3 -> T3,\nT1 -> s(T3)", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 45 [arrowhead = none ];
45 -> 22;

46 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 46 [arrowhead = none ];
46 -> 24;

47 [label="BACKTRACK\nfor clause: p(s(X)) :- ','(geq(X, Y), p(Y))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 47 [arrowhead = none ];
47 -> 18;

48 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
22 -> 48 [arrowhead = none ];
48 -> 26;

49 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
26 -> 49 [arrowhead = none ];
49 -> 28;

50 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
26 -> 50 [arrowhead = none ];
50 -> 30;

51 [label="ONLY EVAL with clause\ngeq(X9, X9).\nand substitutionT3 -> T8,\nX9 -> T8,\nX4 -> T8", fontsize=14, style = filled, fillcolor = lightblue];
28 -> 51 [arrowhead = none ];
51 -> 32;

52 [label="EVAL with clause\ngeq(s(X16), X17) :- geq(X16, X17).\nand substitutionX16 -> T11,\nT3 -> s(T11),\nX4 -> X18,\nX17 -> X18", fontsize=14, style = filled, fillcolor = lightblue];
30 -> 52 [arrowhead = none ];
52 -> 37;

53 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
30 -> 53 [arrowhead = none ];
53 -> 40;

54 [label="INSTANCE with matching:\nT1 -> T8", fontsize=14, style = filled, fillcolor = lightgrey];
32 -> 54 [arrowhead = none , style=dashed];
54 -> 2[style=dashed];

55 [label="INSTANCE with matching:\nT3 -> T11\nX4 -> X18", fontsize=14, style = filled, fillcolor = lightgrey];
37 -> 55 [arrowhead = none , style=dashed];
55 -> 22[style=dashed];

}

(2) Obligation:

Pi-finite rewrite system:
The TRS R consists of the following rules:

pA_in_g(0) → pA_out_g(0)
pA_in_g(s(T3)) → U1_g(T3, pB_in_ga(T3, X4))
pB_in_ga(T8, T8) → U2_ga(T8, pA_in_g(T8))
U2_ga(T8, pA_out_g(T8)) → pB_out_ga(T8, T8)
pB_in_ga(s(T11), X18) → U3_ga(T11, X18, pB_in_ga(T11, X18))
U3_ga(T11, X18, pB_out_ga(T11, X18)) → pB_out_ga(s(T11), X18)
U1_g(T3, pB_out_ga(T3, X4)) → pA_out_g(s(T3))

The argument filtering Pi contains the following mapping:
pA_in_g(x1)  =  pA_in_g(x1)
0  =  0
pA_out_g(x1)  =  pA_out_g(x1)
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
pB_in_ga(x1, x2)  =  pB_in_ga(x1)
U2_ga(x1, x2)  =  U2_ga(x1, x2)
pB_out_ga(x1, x2)  =  pB_out_ga(x1, x2)
U3_ga(x1, x2, x3)  =  U3_ga(x1, x3)

(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:

PA_IN_G(s(T3)) → U1_G(T3, pB_in_ga(T3, X4))
PA_IN_G(s(T3)) → PB_IN_GA(T3, X4)
PB_IN_GA(T8, T8) → U2_GA(T8, pA_in_g(T8))
PB_IN_GA(T8, T8) → PA_IN_G(T8)
PB_IN_GA(s(T11), X18) → U3_GA(T11, X18, pB_in_ga(T11, X18))
PB_IN_GA(s(T11), X18) → PB_IN_GA(T11, X18)

The TRS R consists of the following rules:

pA_in_g(0) → pA_out_g(0)
pA_in_g(s(T3)) → U1_g(T3, pB_in_ga(T3, X4))
pB_in_ga(T8, T8) → U2_ga(T8, pA_in_g(T8))
U2_ga(T8, pA_out_g(T8)) → pB_out_ga(T8, T8)
pB_in_ga(s(T11), X18) → U3_ga(T11, X18, pB_in_ga(T11, X18))
U3_ga(T11, X18, pB_out_ga(T11, X18)) → pB_out_ga(s(T11), X18)
U1_g(T3, pB_out_ga(T3, X4)) → pA_out_g(s(T3))

The argument filtering Pi contains the following mapping:
pA_in_g(x1)  =  pA_in_g(x1)
0  =  0
pA_out_g(x1)  =  pA_out_g(x1)
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
pB_in_ga(x1, x2)  =  pB_in_ga(x1)
U2_ga(x1, x2)  =  U2_ga(x1, x2)
pB_out_ga(x1, x2)  =  pB_out_ga(x1, x2)
U3_ga(x1, x2, x3)  =  U3_ga(x1, x3)
PA_IN_G(x1)  =  PA_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x1, x2)
PB_IN_GA(x1, x2)  =  PB_IN_GA(x1)
U2_GA(x1, x2)  =  U2_GA(x1, x2)
U3_GA(x1, x2, x3)  =  U3_GA(x1, x3)

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

(4) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

PA_IN_G(s(T3)) → U1_G(T3, pB_in_ga(T3, X4))
PA_IN_G(s(T3)) → PB_IN_GA(T3, X4)
PB_IN_GA(T8, T8) → U2_GA(T8, pA_in_g(T8))
PB_IN_GA(T8, T8) → PA_IN_G(T8)
PB_IN_GA(s(T11), X18) → U3_GA(T11, X18, pB_in_ga(T11, X18))
PB_IN_GA(s(T11), X18) → PB_IN_GA(T11, X18)

The TRS R consists of the following rules:

pA_in_g(0) → pA_out_g(0)
pA_in_g(s(T3)) → U1_g(T3, pB_in_ga(T3, X4))
pB_in_ga(T8, T8) → U2_ga(T8, pA_in_g(T8))
U2_ga(T8, pA_out_g(T8)) → pB_out_ga(T8, T8)
pB_in_ga(s(T11), X18) → U3_ga(T11, X18, pB_in_ga(T11, X18))
U3_ga(T11, X18, pB_out_ga(T11, X18)) → pB_out_ga(s(T11), X18)
U1_g(T3, pB_out_ga(T3, X4)) → pA_out_g(s(T3))

The argument filtering Pi contains the following mapping:
pA_in_g(x1)  =  pA_in_g(x1)
0  =  0
pA_out_g(x1)  =  pA_out_g(x1)
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
pB_in_ga(x1, x2)  =  pB_in_ga(x1)
U2_ga(x1, x2)  =  U2_ga(x1, x2)
pB_out_ga(x1, x2)  =  pB_out_ga(x1, x2)
U3_ga(x1, x2, x3)  =  U3_ga(x1, x3)
PA_IN_G(x1)  =  PA_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x1, x2)
PB_IN_GA(x1, x2)  =  PB_IN_GA(x1)
U2_GA(x1, x2)  =  U2_GA(x1, x2)
U3_GA(x1, x2, x3)  =  U3_GA(x1, x3)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(6) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

PA_IN_G(s(T3)) → PB_IN_GA(T3, X4)
PB_IN_GA(T8, T8) → PA_IN_G(T8)
PB_IN_GA(s(T11), X18) → PB_IN_GA(T11, X18)

The TRS R consists of the following rules:

pA_in_g(0) → pA_out_g(0)
pA_in_g(s(T3)) → U1_g(T3, pB_in_ga(T3, X4))
pB_in_ga(T8, T8) → U2_ga(T8, pA_in_g(T8))
U2_ga(T8, pA_out_g(T8)) → pB_out_ga(T8, T8)
pB_in_ga(s(T11), X18) → U3_ga(T11, X18, pB_in_ga(T11, X18))
U3_ga(T11, X18, pB_out_ga(T11, X18)) → pB_out_ga(s(T11), X18)
U1_g(T3, pB_out_ga(T3, X4)) → pA_out_g(s(T3))

The argument filtering Pi contains the following mapping:
pA_in_g(x1)  =  pA_in_g(x1)
0  =  0
pA_out_g(x1)  =  pA_out_g(x1)
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
pB_in_ga(x1, x2)  =  pB_in_ga(x1)
U2_ga(x1, x2)  =  U2_ga(x1, x2)
pB_out_ga(x1, x2)  =  pB_out_ga(x1, x2)
U3_ga(x1, x2, x3)  =  U3_ga(x1, x3)
PA_IN_G(x1)  =  PA_IN_G(x1)
PB_IN_GA(x1, x2)  =  PB_IN_GA(x1)

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:

PA_IN_G(s(T3)) → PB_IN_GA(T3, X4)
PB_IN_GA(T8, T8) → PA_IN_G(T8)
PB_IN_GA(s(T11), X18) → PB_IN_GA(T11, X18)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
PA_IN_G(x1)  =  PA_IN_G(x1)
PB_IN_GA(x1, x2)  =  PB_IN_GA(x1)

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

(9) PiDPToQDPProof (SOUND 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:

PA_IN_G(s(T3)) → PB_IN_GA(T3)
PB_IN_GA(T8) → PA_IN_G(T8)
PB_IN_GA(s(T11)) → PB_IN_GA(T11)

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:

• PB_IN_GA(T8) → PA_IN_G(T8)
  The graph contains the following edges 1 >= 1

• PB_IN_GA(s(T11)) → PB_IN_GA(T11)
  The graph contains the following edges 1 > 1

• PA_IN_G(s(T3)) → PB_IN_GA(T3)
  The graph contains the following edges 1 > 1

(12) YES