Left Termination proof of /tmp/tmpt2uKKX/p.pl

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

0 Prolog

↳1 PrologToDTProblemTransformerProof (⇒, 0 ms)

↳2 TRIPLES

↳3 TriplesToPiDPProof (⇒, 9 ms)

↳4 PiDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 PiDP

↳7 PiDPToQDPProof (⇒, 0 ms)

↳8 QDP

↳9 QDPSizeChangeProof (⇔, 0 ms)

↳10 YES

(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) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: p(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: p(T1), SCOPE: 1, CLAUSE: 0 | p(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | p(0), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: p(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\np(T1) !=? p(0)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: p(0), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="B: ','(geq(T3, X4), p(X4))\n\nT3 is ground\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(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];
10 [label="10: ','(geq(T3, X4), p(X4)), SCOPE: 2, CLAUSE: 2\n\nT3 is ground\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(geq(T3, X4), p(X4)), SCOPE: 2, CLAUSE: 3\n\nT3 is ground\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: p(T8)\n\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(geq(T11, X18), p(X18))\n\nT11 is ground\nX18 is free", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 15 [arrowhead = none ];
15 -> 2;

16 [label="EVAL with clause\np(0).\nand substitutionT1 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 16 [arrowhead = none ];
16 -> 3;

17 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 17 [arrowhead = none ];
17 -> 4;

18 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 18 [arrowhead = none ];
18 -> 5;

19 [label="EVAL with clause\np(s(X3)) :- ','(geq(X3, X4), p(X4)).\nand substitutionX3 -> T3,\nT1 -> s(T3)", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 19 [arrowhead = none ];
19 -> 7;

20 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 20 [arrowhead = none ];
20 -> 8;

21 [label="BACKTRACK\nfor clause: p(s(X)) :- ','(geq(X, Y), p(Y))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 21 [arrowhead = none ];
21 -> 6;

22 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 22 [arrowhead = none ];
22 -> 9;

23 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 23 [arrowhead = none ];
23 -> 10;

24 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 24 [arrowhead = none ];
24 -> 11;

25 [label="ONLY EVAL with clause\ngeq(X9, X9).\nand substitutionT3 -> T8,\nX9 -> T8,\nX4 -> T8", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 25 [arrowhead = none ];
25 -> 12;

26 [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];
11 -> 26 [arrowhead = none ];
26 -> 13;

27 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 27 [arrowhead = none ];
27 -> 14;

28 [label="INSTANCE with matching:\nT1 -> T8", fontsize=14, style = filled, fillcolor = lightgrey];
12 -> 28 [arrowhead = none , style=dashed];
28 -> 1[style=dashed];

29 [label="INSTANCE with matching:\nT3 -> T11\nX4 -> X18", fontsize=14, style = filled, fillcolor = lightgrey];
13 -> 29 [arrowhead = none , style=dashed];
29 -> 7[style=dashed];

}

(2) Obligation:

Triples:

pB(X1, X1) :- pA(X1).
pB(s(X1), X2) :- pB(X1, X2).
pA(s(X1)) :- pB(X1, X2).

Clauses:

pcA(0).
pcA(s(X1)) :- qcB(X1, X2).
qcB(X1, X1) :- pcA(X1).
qcB(s(X1), X2) :- qcB(X1, X2).

Afs:

pA(x1) = pA(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:
pA_in: (b)
pB_in: (b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

PA_IN_G(s(X1)) → U3_G(X1, pB_in_ga(X1, X2))
PA_IN_G(s(X1)) → PB_IN_GA(X1, X2)
PB_IN_GA(X1, X1) → U1_GA(X1, pA_in_g(X1))
PB_IN_GA(X1, X1) → PA_IN_G(X1)
PB_IN_GA(s(X1), X2) → U2_GA(X1, X2, pB_in_ga(X1, X2))
PB_IN_GA(s(X1), X2) → PB_IN_GA(X1, X2)

R is empty.
The argument filtering Pi contains the following mapping:
pA_in_g(x1) = pA_in_g(x1)
s(x1) = s(x1)
pB_in_ga(x1, x2) = pB_in_ga(x1)
PA_IN_G(x1) = PA_IN_G(x1)
U3_G(x1, x2) = U3_G(x1, x2)
PB_IN_GA(x1, x2) = PB_IN_GA(x1)
U1_GA(x1, x2) = U1_GA(x1, x2)
U2_GA(x1, x2, x3) = U2_GA(x1, x3)

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:

PA_IN_G(s(X1)) → U3_G(X1, pB_in_ga(X1, X2))
PA_IN_G(s(X1)) → PB_IN_GA(X1, X2)
PB_IN_GA(X1, X1) → U1_GA(X1, pA_in_g(X1))
PB_IN_GA(X1, X1) → PA_IN_G(X1)
PB_IN_GA(s(X1), X2) → U2_GA(X1, X2, pB_in_ga(X1, X2))
PB_IN_GA(s(X1), X2) → PB_IN_GA(X1, X2)

(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(X1)) → PB_IN_GA(X1, X2)
PB_IN_GA(X1, X1) → PA_IN_G(X1)
PB_IN_GA(s(X1), X2) → PB_IN_GA(X1, X2)

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

(7) PiDPToQDPProof (SOUND 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:

PA_IN_G(s(X1)) → PB_IN_GA(X1)
PB_IN_GA(X1) → PA_IN_G(X1)
PB_IN_GA(s(X1)) → PB_IN_GA(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:

PB_IN_GA(X1) → PA_IN_G(X1)
The graph contains the following edges 1 >= 1
PB_IN_GA(s(X1)) → PB_IN_GA(X1)
The graph contains the following edges 1 > 1
PA_IN_G(s(X1)) → PB_IN_GA(X1)
The graph contains the following edges 1 > 1