Left Termination of the query pattern q_in_1(a) w.r.t. the given Prolog program could not be shown:

(0) Obligation:

Clauses:

q(X) :- ','(p(X), !).
p(0).
p(s(X)) :- p(X).

Queries:

q(a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: q(T1)\n\n", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: q(T1), SCOPE: 1, CLAUSE: 0 | ?_1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(p(T3), !_1) | ?_1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: ','(p(T3), !_1), SCOPE: 2, CLAUSE: 1 | ','(p(T3), !_1), SCOPE: 2, CLAUSE: 2 | ?_1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: !_1 | ','(p(T3), !_1), SCOPE: 2, CLAUSE: 2 | ?_1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ','(p(T3), !_1), SCOPE: 2, CLAUSE: 2 | ?_1\n\np(T3) !=? p(0)", 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(T5), !_1) | ?_1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: \n\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 11 [arrowhead = none ];
11 -> 2;

12 [label="EVAL with clause\nq(X2) :- ','(p(X2), !).\nand substitution\nT1 -> T3\nX2 -> T3\nT2 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 12 [arrowhead = none ];
12 -> 3;

13 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 13 [arrowhead = none ];
13 -> 4;

14 [label="EVAL with clause\np(0).\nand substitution\nT3 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 14 [arrowhead = none ];
14 -> 5;

15 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 15 [arrowhead = none ];
15 -> 6;

16 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 16 [arrowhead = none ];
16 -> 7;

17 [label="EVAL with clause\np(s(X4)) :- p(X4).\nand substitution\nX4 -> T5\nT3 -> s(T5)\nT4 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 17 [arrowhead = none ];
17 -> 9;

18 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 18 [arrowhead = none ];
18 -> 10;

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

20 [label="INSTANCE with matching:\nT3 -> T5", fontsize=14, style = filled, fillcolor = lightgrey];
9 -> 20 [arrowhead = none , style=dashed];
20 -> 3[style=dashed];

}

(2) Obligation:

Clauses:

p3(0).
p3(s(T5)) :- p3(T5).
q1(0).
q1(s(T5)) :- p3(T5).

Queries:

q1(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:
q1_in: (f)
p3_in: (f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

q1_in_a(0) → q1_out_a(0)
q1_in_a(s(T5)) → U2_a(T5, p3_in_a(T5))
p3_in_a(0) → p3_out_a(0)
p3_in_a(s(T5)) → U1_a(T5, p3_in_a(T5))
U1_a(T5, p3_out_a(T5)) → p3_out_a(s(T5))
U2_a(T5, p3_out_a(T5)) → q1_out_a(s(T5))

The argument filtering Pi contains the following mapping:
q1_in_a(x1)  =  q1_in_a
q1_out_a(x1)  =  q1_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
p3_in_a(x1)  =  p3_in_a
p3_out_a(x1)  =  p3_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
s(x1)  =  s(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

q1_in_a(0) → q1_out_a(0)
q1_in_a(s(T5)) → U2_a(T5, p3_in_a(T5))
p3_in_a(0) → p3_out_a(0)
p3_in_a(s(T5)) → U1_a(T5, p3_in_a(T5))
U1_a(T5, p3_out_a(T5)) → p3_out_a(s(T5))
U2_a(T5, p3_out_a(T5)) → q1_out_a(s(T5))

The argument filtering Pi contains the following mapping:
q1_in_a(x1)  =  q1_in_a
q1_out_a(x1)  =  q1_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
p3_in_a(x1)  =  p3_in_a
p3_out_a(x1)  =  p3_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
s(x1)  =  s(x1)

(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:

Q1_IN_A(s(T5)) → U2_A(T5, p3_in_a(T5))
Q1_IN_A(s(T5)) → P3_IN_A(T5)
P3_IN_A(s(T5)) → U1_A(T5, p3_in_a(T5))
P3_IN_A(s(T5)) → P3_IN_A(T5)

The TRS R consists of the following rules:

q1_in_a(0) → q1_out_a(0)
q1_in_a(s(T5)) → U2_a(T5, p3_in_a(T5))
p3_in_a(0) → p3_out_a(0)
p3_in_a(s(T5)) → U1_a(T5, p3_in_a(T5))
U1_a(T5, p3_out_a(T5)) → p3_out_a(s(T5))
U2_a(T5, p3_out_a(T5)) → q1_out_a(s(T5))

The argument filtering Pi contains the following mapping:
q1_in_a(x1)  =  q1_in_a
q1_out_a(x1)  =  q1_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
p3_in_a(x1)  =  p3_in_a
p3_out_a(x1)  =  p3_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
s(x1)  =  s(x1)
Q1_IN_A(x1)  =  Q1_IN_A
U2_A(x1, x2)  =  U2_A(x2)
P3_IN_A(x1)  =  P3_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:

Q1_IN_A(s(T5)) → U2_A(T5, p3_in_a(T5))
Q1_IN_A(s(T5)) → P3_IN_A(T5)
P3_IN_A(s(T5)) → U1_A(T5, p3_in_a(T5))
P3_IN_A(s(T5)) → P3_IN_A(T5)

The TRS R consists of the following rules:

q1_in_a(0) → q1_out_a(0)
q1_in_a(s(T5)) → U2_a(T5, p3_in_a(T5))
p3_in_a(0) → p3_out_a(0)
p3_in_a(s(T5)) → U1_a(T5, p3_in_a(T5))
U1_a(T5, p3_out_a(T5)) → p3_out_a(s(T5))
U2_a(T5, p3_out_a(T5)) → q1_out_a(s(T5))

The argument filtering Pi contains the following mapping:
q1_in_a(x1)  =  q1_in_a
q1_out_a(x1)  =  q1_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
p3_in_a(x1)  =  p3_in_a
p3_out_a(x1)  =  p3_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
s(x1)  =  s(x1)
Q1_IN_A(x1)  =  Q1_IN_A
U2_A(x1, x2)  =  U2_A(x2)
P3_IN_A(x1)  =  P3_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 3 less nodes.

(8) Obligation:

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

P3_IN_A(s(T5)) → P3_IN_A(T5)

The TRS R consists of the following rules:

q1_in_a(0) → q1_out_a(0)
q1_in_a(s(T5)) → U2_a(T5, p3_in_a(T5))
p3_in_a(0) → p3_out_a(0)
p3_in_a(s(T5)) → U1_a(T5, p3_in_a(T5))
U1_a(T5, p3_out_a(T5)) → p3_out_a(s(T5))
U2_a(T5, p3_out_a(T5)) → q1_out_a(s(T5))

The argument filtering Pi contains the following mapping:
q1_in_a(x1)  =  q1_in_a
q1_out_a(x1)  =  q1_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
p3_in_a(x1)  =  p3_in_a
p3_out_a(x1)  =  p3_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
s(x1)  =  s(x1)
P3_IN_A(x1)  =  P3_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:

P3_IN_A(s(T5)) → P3_IN_A(T5)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
P3_IN_A(x1)  =  P3_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:

P3_IN_A → P3_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 = P3_IN_A evaluates to t =P3_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 P3_IN_A to P3_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:
q1_in: (f)
p3_in: (f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

q1_in_a(0) → q1_out_a(0)
q1_in_a(s(T5)) → U2_a(T5, p3_in_a(T5))
p3_in_a(0) → p3_out_a(0)
p3_in_a(s(T5)) → U1_a(T5, p3_in_a(T5))
U1_a(T5, p3_out_a(T5)) → p3_out_a(s(T5))
U2_a(T5, p3_out_a(T5)) → q1_out_a(s(T5))

The argument filtering Pi contains the following mapping:
q1_in_a(x1)  =  q1_in_a
q1_out_a(x1)  =  q1_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
p3_in_a(x1)  =  p3_in_a
p3_out_a(x1)  =  p3_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
s(x1)  =  s(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(16) Obligation:

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

q1_in_a(0) → q1_out_a(0)
q1_in_a(s(T5)) → U2_a(T5, p3_in_a(T5))
p3_in_a(0) → p3_out_a(0)
p3_in_a(s(T5)) → U1_a(T5, p3_in_a(T5))
U1_a(T5, p3_out_a(T5)) → p3_out_a(s(T5))
U2_a(T5, p3_out_a(T5)) → q1_out_a(s(T5))

The argument filtering Pi contains the following mapping:
q1_in_a(x1)  =  q1_in_a
q1_out_a(x1)  =  q1_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
p3_in_a(x1)  =  p3_in_a
p3_out_a(x1)  =  p3_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
s(x1)  =  s(x1)

(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:

Q1_IN_A(s(T5)) → U2_A(T5, p3_in_a(T5))
Q1_IN_A(s(T5)) → P3_IN_A(T5)
P3_IN_A(s(T5)) → U1_A(T5, p3_in_a(T5))
P3_IN_A(s(T5)) → P3_IN_A(T5)

The TRS R consists of the following rules:

q1_in_a(0) → q1_out_a(0)
q1_in_a(s(T5)) → U2_a(T5, p3_in_a(T5))
p3_in_a(0) → p3_out_a(0)
p3_in_a(s(T5)) → U1_a(T5, p3_in_a(T5))
U1_a(T5, p3_out_a(T5)) → p3_out_a(s(T5))
U2_a(T5, p3_out_a(T5)) → q1_out_a(s(T5))

The argument filtering Pi contains the following mapping:
q1_in_a(x1)  =  q1_in_a
q1_out_a(x1)  =  q1_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
p3_in_a(x1)  =  p3_in_a
p3_out_a(x1)  =  p3_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
s(x1)  =  s(x1)
Q1_IN_A(x1)  =  Q1_IN_A
U2_A(x1, x2)  =  U2_A(x2)
P3_IN_A(x1)  =  P3_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:

Q1_IN_A(s(T5)) → U2_A(T5, p3_in_a(T5))
Q1_IN_A(s(T5)) → P3_IN_A(T5)
P3_IN_A(s(T5)) → U1_A(T5, p3_in_a(T5))
P3_IN_A(s(T5)) → P3_IN_A(T5)

The TRS R consists of the following rules:

q1_in_a(0) → q1_out_a(0)
q1_in_a(s(T5)) → U2_a(T5, p3_in_a(T5))
p3_in_a(0) → p3_out_a(0)
p3_in_a(s(T5)) → U1_a(T5, p3_in_a(T5))
U1_a(T5, p3_out_a(T5)) → p3_out_a(s(T5))
U2_a(T5, p3_out_a(T5)) → q1_out_a(s(T5))

The argument filtering Pi contains the following mapping:
q1_in_a(x1)  =  q1_in_a
q1_out_a(x1)  =  q1_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
p3_in_a(x1)  =  p3_in_a
p3_out_a(x1)  =  p3_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
s(x1)  =  s(x1)
Q1_IN_A(x1)  =  Q1_IN_A
U2_A(x1, x2)  =  U2_A(x2)
P3_IN_A(x1)  =  P3_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 3 less nodes.

(20) Obligation:

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

P3_IN_A(s(T5)) → P3_IN_A(T5)

The TRS R consists of the following rules:

q1_in_a(0) → q1_out_a(0)
q1_in_a(s(T5)) → U2_a(T5, p3_in_a(T5))
p3_in_a(0) → p3_out_a(0)
p3_in_a(s(T5)) → U1_a(T5, p3_in_a(T5))
U1_a(T5, p3_out_a(T5)) → p3_out_a(s(T5))
U2_a(T5, p3_out_a(T5)) → q1_out_a(s(T5))

The argument filtering Pi contains the following mapping:
q1_in_a(x1)  =  q1_in_a
q1_out_a(x1)  =  q1_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
p3_in_a(x1)  =  p3_in_a
p3_out_a(x1)  =  p3_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
s(x1)  =  s(x1)
P3_IN_A(x1)  =  P3_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:

P3_IN_A(s(T5)) → P3_IN_A(T5)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
P3_IN_A(x1)  =  P3_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:

P3_IN_A → P3_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 = P3_IN_A evaluates to t =P3_IN_A

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 P3_IN_A to P3_IN_A.

(26) FALSE