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

(0) Obligation:

Clauses:

p :- r.
r :- ','(!, q).
r :- q.
q.
q :- r.

Queries:

p().

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: p\n\n", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: p, SCOPE: 1, CLAUSE: 0\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: r\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: r, SCOPE: 2, CLAUSE: 1 | r, SCOPE: 2, CLAUSE: 2 | ?_2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ','(!_2, q) | r, SCOPE: 2, CLAUSE: 2 | ?_2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: q\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: q, SCOPE: 3, CLAUSE: 3 | q, SCOPE: 3, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: q, SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: q, SCOPE: 3, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: r\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 13 [arrowhead = none ];
13 -> 2;

14 [label="EVAL with clause\np :- r.\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 14 [arrowhead = none ];
14 -> 3;

15 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 15 [arrowhead = none ];
15 -> 4;

16 [label="EVAL with clause\nr :- ','(!, q).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 16 [arrowhead = none ];
16 -> 5;

17 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 17 [arrowhead = none ];
17 -> 6;

18 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
6 -> 18 [arrowhead = none ];
18 -> 7;

19 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
7 -> 19 [arrowhead = none ];
19 -> 8;

20 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
7 -> 20 [arrowhead = none ];
20 -> 9;

21 [label="EVAL with clause\nq.\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 21 [arrowhead = none ];
21 -> 10;

22 [label="EVAL with clause\nq :- r.\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 22 [arrowhead = none ];
22 -> 12;

23 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 23 [arrowhead = none ];
23 -> 11;

24 [label="INSTANCE", fontsize=14, style = filled, fillcolor = lightgrey];
12 -> 24 [arrowhead = none , style=dashed];
24 -> 3[style=dashed];

}

(2) Obligation:

Clauses:

r3.
r3 :- r3.
p1.
p1 :- r3.

Queries:

p1().

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

p1_in_ → p1_out_
p1_in_ → U2_(r3_in_)
r3_in_ → r3_out_
r3_in_ → U1_(r3_in_)
U1_(r3_out_) → r3_out_
U2_(r3_out_) → p1_out_

Pi is empty.

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_ → p1_out_
p1_in_ → U2_(r3_in_)
r3_in_ → r3_out_
r3_in_ → U1_(r3_in_)
U1_(r3_out_) → r3_out_
U2_(r3_out_) → p1_out_

Pi is empty.

(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_ → U2_¹(r3_in_)
P1_IN_ → R3_IN_
R3_IN_ → U1_¹(r3_in_)
R3_IN_ → R3_IN_

The TRS R consists of the following rules:

p1_in_ → p1_out_
p1_in_ → U2_(r3_in_)
r3_in_ → r3_out_
r3_in_ → U1_(r3_in_)
U1_(r3_out_) → r3_out_
U2_(r3_out_) → p1_out_

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

(6) Obligation:

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

P1_IN_ → U2_¹(r3_in_)
P1_IN_ → R3_IN_
R3_IN_ → U1_¹(r3_in_)
R3_IN_ → R3_IN_

The TRS R consists of the following rules:

p1_in_ → p1_out_
p1_in_ → U2_(r3_in_)
r3_in_ → r3_out_
r3_in_ → U1_(r3_in_)
U1_(r3_out_) → r3_out_
U2_(r3_out_) → p1_out_

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

R3_IN_ → R3_IN_

The TRS R consists of the following rules:

p1_in_ → p1_out_
p1_in_ → U2_(r3_in_)
r3_in_ → r3_out_
r3_in_ → U1_(r3_in_)
U1_(r3_out_) → r3_out_
U2_(r3_out_) → p1_out_

Pi is empty.
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:

R3_IN_ → R3_IN_

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

(11) PiDPToQDPProof (EQUIVALENT 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:

R3_IN_ → R3_IN_

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 = R3_IN_ evaluates to t =R3_IN_

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 R3_IN_ to R3_IN_.

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

p1_in_ → p1_out_
p1_in_ → U2_(r3_in_)
r3_in_ → r3_out_
r3_in_ → U1_(r3_in_)
U1_(r3_out_) → r3_out_
U2_(r3_out_) → p1_out_

Pi is empty.

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_ → p1_out_
p1_in_ → U2_(r3_in_)
r3_in_ → r3_out_
r3_in_ → U1_(r3_in_)
U1_(r3_out_) → r3_out_
U2_(r3_out_) → p1_out_

Pi is empty.

(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_ → U2_¹(r3_in_)
P1_IN_ → R3_IN_
R3_IN_ → U1_¹(r3_in_)
R3_IN_ → R3_IN_

The TRS R consists of the following rules:

p1_in_ → p1_out_
p1_in_ → U2_(r3_in_)
r3_in_ → r3_out_
r3_in_ → U1_(r3_in_)
U1_(r3_out_) → r3_out_
U2_(r3_out_) → p1_out_

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

(18) Obligation:

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

P1_IN_ → U2_¹(r3_in_)
P1_IN_ → R3_IN_
R3_IN_ → U1_¹(r3_in_)
R3_IN_ → R3_IN_

The TRS R consists of the following rules:

p1_in_ → p1_out_
p1_in_ → U2_(r3_in_)
r3_in_ → r3_out_
r3_in_ → U1_(r3_in_)
U1_(r3_out_) → r3_out_
U2_(r3_out_) → p1_out_

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

(19) DependencyGraphProof (EQUIVALENT transformation)

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

(20) Obligation:

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

R3_IN_ → R3_IN_

The TRS R consists of the following rules:

p1_in_ → p1_out_
p1_in_ → U2_(r3_in_)
r3_in_ → r3_out_
r3_in_ → U1_(r3_in_)
U1_(r3_out_) → r3_out_
U2_(r3_out_) → p1_out_

Pi is empty.
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:

R3_IN_ → R3_IN_

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

(23) PiDPToQDPProof (EQUIVALENT 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:

R3_IN_ → R3_IN_

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 = R3_IN_ evaluates to t =R3_IN_

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

• Matcher: [ ]
• Semiunifier: [ ]

---

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from R3_IN_ to R3_IN_.

(26) FALSE