Left Termination of the query pattern p_in_1(a) w.r.t. the given Prolog program could not be shown:

(0) Obligation:

Clauses:

p(X) :- q(X).
p(X) :- p(X).
q(X) :- !.

Queries:

p(a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

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: p(T1), SCOPE: 1, CLAUSE: 0\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: p(T1), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: q(T3)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: q(T3), SCOPE: 2, CLAUSE: 2 | ?_2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: !_2 | ?_2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: p(T6)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: p(T6), SCOPE: 3, CLAUSE: 0 | p(T6), SCOPE: 3, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: p(T6), SCOPE: 3, CLAUSE: 0\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: p(T6), SCOPE: 3, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: q(T8)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: q(T8), SCOPE: 4, CLAUSE: 2 | ?_4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: !_4 | ?_4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: p(T11)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 20 [arrowhead = none ];
20 -> 2;

21 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
2 -> 21 [arrowhead = none ];
21 -> 3;

22 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
2 -> 22 [arrowhead = none ];
22 -> 4;

23 [label="EVAL with clause\np(X3) :- q(X3).\nand substitution\nT1 -> T3\nX3 -> T3\nT2 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 23 [arrowhead = none ];
23 -> 5;

24 [label="EVAL with clause\np(X7) :- p(X7).\nand substitution\nT1 -> T6\nX7 -> T6\nT5 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 24 [arrowhead = none ];
24 -> 10;

25 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
5 -> 25 [arrowhead = none ];
25 -> 6;

26 [label="EVAL with clause\nq(X5) :- !.\nand substitution\nT3 -> T4\nX5 -> T4", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 26 [arrowhead = none ];
26 -> 7;

27 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 27 [arrowhead = none ];
27 -> 8;

28 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 28 [arrowhead = none ];
28 -> 9;

29 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
10 -> 29 [arrowhead = none ];
29 -> 11;

30 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
11 -> 30 [arrowhead = none ];
30 -> 12;

31 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
11 -> 31 [arrowhead = none ];
31 -> 13;

32 [label="EVAL with clause\np(X10) :- q(X10).\nand substitution\nT6 -> T8\nX10 -> T8\nT7 -> T8", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 32 [arrowhead = none ];
32 -> 14;

33 [label="EVAL with clause\np(X14) :- p(X14).\nand substitution\nT6 -> T11\nX14 -> T11\nT10 -> T11", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 33 [arrowhead = none ];
33 -> 19;

34 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
14 -> 34 [arrowhead = none ];
34 -> 15;

35 [label="EVAL with clause\nq(X12) :- !.\nand substitution\nT8 -> T9\nX12 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
15 -> 35 [arrowhead = none ];
35 -> 16;

36 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 36 [arrowhead = none ];
36 -> 17;

37 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 37 [arrowhead = none ];
37 -> 18;

38 [label="INSTANCE with matching:\nT1 -> T11", fontsize=14, style = filled, fillcolor = lightgrey];
19 -> 38 [arrowhead = none , style=dashed];
38 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

p1(T4).
p1(T9).
p1(T11) :- p1(T11).

Queries:

p1(a).

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
p1_in: (f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(x2)

(5) 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:

P1_IN_A(T11) → U1_A(T11, p1_in_a(T11))
P1_IN_A(T11) → P1_IN_A(T11)

The TRS R consists of the following rules:

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(x2)
P1_IN_A(x1)  =  P1_IN_A
U1_A(x1, x2)  =  U1_A(x2)

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

(6) Obligation:

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

P1_IN_A(T11) → U1_A(T11, p1_in_a(T11))
P1_IN_A(T11) → P1_IN_A(T11)

The TRS R consists of the following rules:

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(x2)
P1_IN_A(x1)  =  P1_IN_A
U1_A(x1, x2)  =  U1_A(x2)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

P1_IN_A(T11) → P1_IN_A(T11)

The TRS R consists of the following rules:

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(x2)
P1_IN_A(x1)  =  P1_IN_A

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

(9) UsableRulesProof (EQUIVALENT transformation)

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

(10) Obligation:

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

P1_IN_A(T11) → P1_IN_A(T11)

R is empty.
The argument filtering Pi contains the following mapping:
P1_IN_A(x1)  =  P1_IN_A

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

(11) PiDPToQDPProof (SOUND transformation)

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

(12) Obligation:

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

P1_IN_A → P1_IN_A

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

(13) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = P1_IN_A evaluates to t =P1_IN_A

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

• Semiunifier: [ ]
• Matcher: [ ]

---

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from P1_IN_A to P1_IN_A.

(14) FALSE

(15) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
p1_in: (f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(16) Obligation:

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

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(x2)

(17) 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:

P1_IN_A(T11) → U1_A(T11, p1_in_a(T11))
P1_IN_A(T11) → P1_IN_A(T11)

The TRS R consists of the following rules:

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(x2)
P1_IN_A(x1)  =  P1_IN_A
U1_A(x1, x2)  =  U1_A(x2)

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

(18) Obligation:

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

P1_IN_A(T11) → U1_A(T11, p1_in_a(T11))
P1_IN_A(T11) → P1_IN_A(T11)

The TRS R consists of the following rules:

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(x2)
P1_IN_A(x1)  =  P1_IN_A
U1_A(x1, x2)  =  U1_A(x2)

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

(19) DependencyGraphProof (EQUIVALENT transformation)

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

(20) Obligation:

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

P1_IN_A(T11) → P1_IN_A(T11)

The TRS R consists of the following rules:

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(x2)
P1_IN_A(x1)  =  P1_IN_A

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

(21) UsableRulesProof (EQUIVALENT transformation)

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

(22) Obligation:

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

P1_IN_A(T11) → P1_IN_A(T11)

R is empty.
The argument filtering Pi contains the following mapping:
P1_IN_A(x1)  =  P1_IN_A

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

(23) PiDPToQDPProof (SOUND transformation)

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

(24) Obligation:

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

P1_IN_A → P1_IN_A

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

(25) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = P1_IN_A evaluates to t =P1_IN_A

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

• Semiunifier: [ ]
• Matcher: [ ]

---

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from P1_IN_A to P1_IN_A.

(26) FALSE