Left Termination proof of /tmp/tmpBtas0b/ts01.pl

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

0 Prolog

↳1 PrologToPrologProblemTransformerProof (⇒, 0 ms)

↳2 Prolog

↳3 PrologToPiTRSProof (⇒, 0 ms)

↳4 PiTRS

↳5 DependencyPairsProof (⇔, 1 ms)

↳6 PiDP

↳7 PisEmptyProof (⇔, 0 ms)

↳8 YES

(0) Obligation:

Clauses:

p(X) :- q(X).
p(X) :- ','(no(q(X)), p(X)).
q(X).
no(X) :- ','(X, ','(!, failure(a))).
no(X1).
failure(b).

Query: p(a)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: p(T1)\n\n", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: p(T1), SCOPE: 1, CLAUSE: 0 | p(T1), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: q(T4) | p(T1), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: q(T4), SCOPE: 2, CLAUSE: 2 | ?_2 | p(T1), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: q(T4), SCOPE: 2, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ?_2 | p(T1), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: p(T1), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(no(q(T15)), p(T15))\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(no(q(T15)), p(T15)), SCOPE: 3, CLAUSE: 3 | ','(no(q(T15)), p(T15)), SCOPE: 3, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(','(call(q(T21)), ','(!_3, failure(a))), p(T21)) | ','(no(q(T15)), p(T15)), SCOPE: 3, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(q(T21), ','(','(!_3, failure(a)), p(T21))) | ?_4 | ','(no(q(T15)), p(T15)), SCOPE: 3, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(q(T21), ','(','(!_3, failure(a)), p(T21))), SCOPE: 5, CLAUSE: 2 | ?_5 | ?_4 | ','(no(q(T15)), p(T15)), SCOPE: 3, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(','(!_3, failure(a)), p(T26)) | ?_5 | ?_4 | ','(no(q(T15)), p(T15)), SCOPE: 3, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(failure(a), p(T26))\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(failure(a), p(T26)), SCOPE: 6, CLAUSE: 5\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 19 [arrowhead = none ];
19 -> 2;

20 [label="ONLY EVAL with clause\np(X3) :- q(X3).\nand substitutionT1 -> T4,\nX3 -> T4,\nT3 -> T4", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 20 [arrowhead = none ];
20 -> 3;

21 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 21 [arrowhead = none ];
21 -> 4;

22 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 22 [arrowhead = none ];
22 -> 5;

23 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 23 [arrowhead = none ];
23 -> 6;

24 [label="ONLY EVAL with clause\nq(X8).\nand substitutionT4 -> T9,\nX8 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 24 [arrowhead = none ];
24 -> 7;

25 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 25 [arrowhead = none ];
25 -> 9;

26 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 26 [arrowhead = none ];
26 -> 8;

27 [label="ONLY EVAL with clause\np(X12) :- ','(no(q(X12)), p(X12)).\nand substitutionT1 -> T15,\nX12 -> T15,\nT14 -> T15", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 27 [arrowhead = none ];
27 -> 10;

28 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
10 -> 28 [arrowhead = none ];
28 -> 11;

29 [label="ONLY EVAL with clause\nno(X16) :- ','(call(X16), ','(!_3, failure(a))).\nand substitutionT15 -> T21,\nX16 -> q(T21),\nT20 -> T21", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 29 [arrowhead = none ];
29 -> 12;

30 [label="CALL", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 30 [arrowhead = none ];
30 -> 13;

31 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
13 -> 31 [arrowhead = none ];
31 -> 14;

32 [label="ONLY EVAL with clause\nq(X19).\nand substitutionT21 -> T26,\nX19 -> T26,\nT25 -> T26", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 32 [arrowhead = none ];
32 -> 15;

33 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 33 [arrowhead = none ];
33 -> 16;

34 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
16 -> 34 [arrowhead = none ];
34 -> 17;

35 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 35 [arrowhead = none ];
35 -> 18;

}

(2) Obligation:

Clauses:

pA(T9).

Query: pA(a)

(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:
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

pA_in_a(T9) → pA_out_a(T9)

The argument filtering Pi contains the following mapping:
pA_in_a(x1) = pA_in_a
pA_out_a(x1) = pA_out_a

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

pA_in_a(T9) → pA_out_a(T9)

The argument filtering Pi contains the following mapping:
pA_in_a(x1) = pA_in_a
pA_out_a(x1) = pA_out_a

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
P is empty.
The TRS R consists of the following rules:

pA_in_a(T9) → pA_out_a(T9)

The argument filtering Pi contains the following mapping:
pA_in_a(x1) = pA_in_a
pA_out_a(x1) = pA_out_a

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

(6) Obligation:

Pi DP problem:
P is empty.
The TRS R consists of the following rules:

pA_in_a(T9) → pA_out_a(T9)

The argument filtering Pi contains the following mapping:
pA_in_a(x1) = pA_in_a
pA_out_a(x1) = pA_out_a

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

(7) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,R,Pi) chain.

(0) Obligation:

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

(2) Obligation:

(3) PrologToPiTRSProof (SOUND transformation)

(4) Obligation:

(5) DependencyPairsProof (EQUIVALENT transformation)

(6) Obligation:

(7) PisEmptyProof (EQUIVALENT transformation)

(8) YES