Left Termination of the query pattern p_in_3(a, a, a) w.r.t. the given Prolog program could not be shown:

(0) Obligation:

Clauses:

p(X, Y, Z) :- ','(append(X, Y, Z), !).
append([], Y, Y).
append(.(H, Xs), Ys, .(H, Zs)) :- append(Xs, Ys, Zs).

Queries:

p(a,a,a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: p(T1, T2, T3)\n\n", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: p(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | ?_1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(append(T7, T8, T9), !_1) | ?_1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: ','(append(T7, T8, T9), !_1), SCOPE: 2, CLAUSE: 1 | ','(append(T7, T8, T9), !_1), SCOPE: 2, CLAUSE: 2 | ?_1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: !_1 | ','(append(T7, T8, T9), !_1), SCOPE: 2, CLAUSE: 2 | ?_1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ','(append(T7, T8, T9), !_1), SCOPE: 2, CLAUSE: 2 | ?_1\n\nX8 is free\nappend(T7, T8, T9) !=? append([], X8, X8)", 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: ','(append(T15, T16, T17), !_1) | ?_1\n\nX8 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: \n\nX8 is free\nX13 is free; \nX14 is free; \nX15 is free; \nX16 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\np(X4, X5, X6) :- ','(append(X4, X5, X6), !).\nand substitution\nT1 -> T7\nX4 -> T7\nT2 -> T8\nX5 -> T8\nT3 -> T9\nX6 -> T9\nT4 -> T7\nT5 -> T8\nT6 -> T9", 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\nappend([], X8, X8).\nand substitution\nT7 -> []\nT8 -> T10\nX8 -> T10\nT9 -> T10", 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\nappend(.(X13, X14), X15, .(X13, X16)) :- append(X14, X15, X16).\nand substitution\nX13 -> T11\nX14 -> T15\nT7 -> .(T11, T15)\nT8 -> T16\nX15 -> T16\nX16 -> T17\nT9 -> .(T11, T17)\nT12 -> T15\nT13 -> T16\nT14 -> T17", 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:\nT7 -> T15\nT8 -> T16\nT9 -> T17", fontsize=14, style = filled, fillcolor = lightgrey];
9 -> 20 [arrowhead = none , style=dashed];
20 -> 3[style=dashed];

}

(2) Obligation:

Clauses:

p3([], T10, T10).
p3(.(T11, T15), T16, .(T11, T17)) :- p3(T15, T16, T17).
p1([], T10, T10).
p1(.(T11, T15), T16, .(T11, T17)) :- p3(T15, T16, T17).

Queries:

p1(a,a,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,f,f)
p3_in: (f,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p1_in_aaa([], T10, T10) → p1_out_aaa([], T10, T10)
p1_in_aaa(.(T11, T15), T16, .(T11, T17)) → U2_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
p3_in_aaa([], T10, T10) → p3_out_aaa([], T10, T10)
p3_in_aaa(.(T11, T15), T16, .(T11, T17)) → U1_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
U1_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p3_out_aaa(.(T11, T15), T16, .(T11, T17))
U2_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p1_out_aaa(.(T11, T15), T16, .(T11, T17))

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
p3_in_aaa(x1, x2, x3)  =  p3_in_aaa
p3_out_aaa(x1, x2, x3)  =  p3_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(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_aaa([], T10, T10) → p1_out_aaa([], T10, T10)
p1_in_aaa(.(T11, T15), T16, .(T11, T17)) → U2_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
p3_in_aaa([], T10, T10) → p3_out_aaa([], T10, T10)
p3_in_aaa(.(T11, T15), T16, .(T11, T17)) → U1_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
U1_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p3_out_aaa(.(T11, T15), T16, .(T11, T17))
U2_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p1_out_aaa(.(T11, T15), T16, .(T11, T17))

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
p3_in_aaa(x1, x2, x3)  =  p3_in_aaa
p3_out_aaa(x1, x2, x3)  =  p3_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(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_AAA(.(T11, T15), T16, .(T11, T17)) → U2_AAA(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
P1_IN_AAA(.(T11, T15), T16, .(T11, T17)) → P3_IN_AAA(T15, T16, T17)
P3_IN_AAA(.(T11, T15), T16, .(T11, T17)) → U1_AAA(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
P3_IN_AAA(.(T11, T15), T16, .(T11, T17)) → P3_IN_AAA(T15, T16, T17)

The TRS R consists of the following rules:

p1_in_aaa([], T10, T10) → p1_out_aaa([], T10, T10)
p1_in_aaa(.(T11, T15), T16, .(T11, T17)) → U2_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
p3_in_aaa([], T10, T10) → p3_out_aaa([], T10, T10)
p3_in_aaa(.(T11, T15), T16, .(T11, T17)) → U1_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
U1_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p3_out_aaa(.(T11, T15), T16, .(T11, T17))
U2_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p1_out_aaa(.(T11, T15), T16, .(T11, T17))

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
p3_in_aaa(x1, x2, x3)  =  p3_in_aaa
p3_out_aaa(x1, x2, x3)  =  p3_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(x2)
P1_IN_AAA(x1, x2, x3)  =  P1_IN_AAA
U2_AAA(x1, x2, x3, x4, x5)  =  U2_AAA(x5)
P3_IN_AAA(x1, x2, x3)  =  P3_IN_AAA
U1_AAA(x1, x2, x3, x4, x5)  =  U1_AAA(x5)

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

(6) Obligation:

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

P1_IN_AAA(.(T11, T15), T16, .(T11, T17)) → U2_AAA(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
P1_IN_AAA(.(T11, T15), T16, .(T11, T17)) → P3_IN_AAA(T15, T16, T17)
P3_IN_AAA(.(T11, T15), T16, .(T11, T17)) → U1_AAA(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
P3_IN_AAA(.(T11, T15), T16, .(T11, T17)) → P3_IN_AAA(T15, T16, T17)

The TRS R consists of the following rules:

p1_in_aaa([], T10, T10) → p1_out_aaa([], T10, T10)
p1_in_aaa(.(T11, T15), T16, .(T11, T17)) → U2_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
p3_in_aaa([], T10, T10) → p3_out_aaa([], T10, T10)
p3_in_aaa(.(T11, T15), T16, .(T11, T17)) → U1_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
U1_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p3_out_aaa(.(T11, T15), T16, .(T11, T17))
U2_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p1_out_aaa(.(T11, T15), T16, .(T11, T17))

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
p3_in_aaa(x1, x2, x3)  =  p3_in_aaa
p3_out_aaa(x1, x2, x3)  =  p3_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(x2)
P1_IN_AAA(x1, x2, x3)  =  P1_IN_AAA
U2_AAA(x1, x2, x3, x4, x5)  =  U2_AAA(x5)
P3_IN_AAA(x1, x2, x3)  =  P3_IN_AAA
U1_AAA(x1, x2, x3, x4, x5)  =  U1_AAA(x5)

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_AAA(.(T11, T15), T16, .(T11, T17)) → P3_IN_AAA(T15, T16, T17)

The TRS R consists of the following rules:

p1_in_aaa([], T10, T10) → p1_out_aaa([], T10, T10)
p1_in_aaa(.(T11, T15), T16, .(T11, T17)) → U2_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
p3_in_aaa([], T10, T10) → p3_out_aaa([], T10, T10)
p3_in_aaa(.(T11, T15), T16, .(T11, T17)) → U1_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
U1_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p3_out_aaa(.(T11, T15), T16, .(T11, T17))
U2_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p1_out_aaa(.(T11, T15), T16, .(T11, T17))

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
p3_in_aaa(x1, x2, x3)  =  p3_in_aaa
p3_out_aaa(x1, x2, x3)  =  p3_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(x2)
P3_IN_AAA(x1, x2, x3)  =  P3_IN_AAA

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_AAA(.(T11, T15), T16, .(T11, T17)) → P3_IN_AAA(T15, T16, T17)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x2)
P3_IN_AAA(x1, x2, x3)  =  P3_IN_AAA

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_AAA → P3_IN_AAA

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_AAA evaluates to t =P3_IN_AAA

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_AAA to P3_IN_AAA.

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

p1_in_aaa([], T10, T10) → p1_out_aaa([], T10, T10)
p1_in_aaa(.(T11, T15), T16, .(T11, T17)) → U2_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
p3_in_aaa([], T10, T10) → p3_out_aaa([], T10, T10)
p3_in_aaa(.(T11, T15), T16, .(T11, T17)) → U1_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
U1_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p3_out_aaa(.(T11, T15), T16, .(T11, T17))
U2_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p1_out_aaa(.(T11, T15), T16, .(T11, T17))

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
p3_in_aaa(x1, x2, x3)  =  p3_in_aaa
p3_out_aaa(x1, x2, x3)  =  p3_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(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_aaa([], T10, T10) → p1_out_aaa([], T10, T10)
p1_in_aaa(.(T11, T15), T16, .(T11, T17)) → U2_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
p3_in_aaa([], T10, T10) → p3_out_aaa([], T10, T10)
p3_in_aaa(.(T11, T15), T16, .(T11, T17)) → U1_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
U1_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p3_out_aaa(.(T11, T15), T16, .(T11, T17))
U2_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p1_out_aaa(.(T11, T15), T16, .(T11, T17))

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
p3_in_aaa(x1, x2, x3)  =  p3_in_aaa
p3_out_aaa(x1, x2, x3)  =  p3_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(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_AAA(.(T11, T15), T16, .(T11, T17)) → U2_AAA(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
P1_IN_AAA(.(T11, T15), T16, .(T11, T17)) → P3_IN_AAA(T15, T16, T17)
P3_IN_AAA(.(T11, T15), T16, .(T11, T17)) → U1_AAA(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
P3_IN_AAA(.(T11, T15), T16, .(T11, T17)) → P3_IN_AAA(T15, T16, T17)

The TRS R consists of the following rules:

p1_in_aaa([], T10, T10) → p1_out_aaa([], T10, T10)
p1_in_aaa(.(T11, T15), T16, .(T11, T17)) → U2_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
p3_in_aaa([], T10, T10) → p3_out_aaa([], T10, T10)
p3_in_aaa(.(T11, T15), T16, .(T11, T17)) → U1_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
U1_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p3_out_aaa(.(T11, T15), T16, .(T11, T17))
U2_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p1_out_aaa(.(T11, T15), T16, .(T11, T17))

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
p3_in_aaa(x1, x2, x3)  =  p3_in_aaa
p3_out_aaa(x1, x2, x3)  =  p3_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(x2)
P1_IN_AAA(x1, x2, x3)  =  P1_IN_AAA
U2_AAA(x1, x2, x3, x4, x5)  =  U2_AAA(x5)
P3_IN_AAA(x1, x2, x3)  =  P3_IN_AAA
U1_AAA(x1, x2, x3, x4, x5)  =  U1_AAA(x5)

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

(18) Obligation:

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

P1_IN_AAA(.(T11, T15), T16, .(T11, T17)) → U2_AAA(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
P1_IN_AAA(.(T11, T15), T16, .(T11, T17)) → P3_IN_AAA(T15, T16, T17)
P3_IN_AAA(.(T11, T15), T16, .(T11, T17)) → U1_AAA(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
P3_IN_AAA(.(T11, T15), T16, .(T11, T17)) → P3_IN_AAA(T15, T16, T17)

The TRS R consists of the following rules:

p1_in_aaa([], T10, T10) → p1_out_aaa([], T10, T10)
p1_in_aaa(.(T11, T15), T16, .(T11, T17)) → U2_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
p3_in_aaa([], T10, T10) → p3_out_aaa([], T10, T10)
p3_in_aaa(.(T11, T15), T16, .(T11, T17)) → U1_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
U1_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p3_out_aaa(.(T11, T15), T16, .(T11, T17))
U2_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p1_out_aaa(.(T11, T15), T16, .(T11, T17))

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
p3_in_aaa(x1, x2, x3)  =  p3_in_aaa
p3_out_aaa(x1, x2, x3)  =  p3_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(x2)
P1_IN_AAA(x1, x2, x3)  =  P1_IN_AAA
U2_AAA(x1, x2, x3, x4, x5)  =  U2_AAA(x5)
P3_IN_AAA(x1, x2, x3)  =  P3_IN_AAA
U1_AAA(x1, x2, x3, x4, x5)  =  U1_AAA(x5)

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_AAA(.(T11, T15), T16, .(T11, T17)) → P3_IN_AAA(T15, T16, T17)

The TRS R consists of the following rules:

p1_in_aaa([], T10, T10) → p1_out_aaa([], T10, T10)
p1_in_aaa(.(T11, T15), T16, .(T11, T17)) → U2_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
p3_in_aaa([], T10, T10) → p3_out_aaa([], T10, T10)
p3_in_aaa(.(T11, T15), T16, .(T11, T17)) → U1_aaa(T11, T15, T16, T17, p3_in_aaa(T15, T16, T17))
U1_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p3_out_aaa(.(T11, T15), T16, .(T11, T17))
U2_aaa(T11, T15, T16, T17, p3_out_aaa(T15, T16, T17)) → p1_out_aaa(.(T11, T15), T16, .(T11, T17))

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
p3_in_aaa(x1, x2, x3)  =  p3_in_aaa
p3_out_aaa(x1, x2, x3)  =  p3_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(x2)
P3_IN_AAA(x1, x2, x3)  =  P3_IN_AAA

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_AAA(.(T11, T15), T16, .(T11, T17)) → P3_IN_AAA(T15, T16, T17)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x2)
P3_IN_AAA(x1, x2, x3)  =  P3_IN_AAA

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_AAA → P3_IN_AAA

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_AAA evaluates to t =P3_IN_AAA

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

• Semiunifier: [ ]
• Matcher: [ ]

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Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from P3_IN_AAA to P3_IN_AAA.

(26) FALSE