Left Termination proof of /tmp/tmp3BckbJ/len.pl

Left Termination of the query pattern len_in_2(g, a) w.r.t. the given Prolog program could not be shown:

0 Prolog

↳1 PrologToPrologProblemTransformerProof (⇐)

↳2 Prolog

    ↳3 PrologToPiTRSProof (⇐)

        ↳4 PiTRS

            ↳5 DependencyPairsProof (⇔)

                ↳6 PiDP

                    ↳7 DependencyGraphProof (⇔)

                        ↳8 AND

                            ↳9 PiDP

                                ↳10 UsableRulesProof (⇔)

                                    ↳11 PiDP

                                        ↳12 PiDPToQDPProof (⇐)

                                            ↳13 QDP

                                                ↳14 NonTerminationProof (⇔)

                                                    ↳15 FALSE

                            ↳16 PiDP

                                ↳17 UsableRulesProof (⇔)

                                    ↳18 PiDP

                                        ↳19 PiDPToQDPProof (⇐)

                                            ↳20 QDP

                                                ↳21 QDPSizeChangeProof (⇔)

                                                    ↳22 TRUE

    ↳23 PrologToPiTRSProof (⇐)

        ↳24 PiTRS

            ↳25 DependencyPairsProof (⇔)

                ↳26 PiDP

                    ↳27 DependencyGraphProof (⇔)

                        ↳28 AND

                            ↳29 PiDP

                                ↳30 UsableRulesProof (⇔)

                                    ↳31 PiDP

                                        ↳32 PiDPToQDPProof (⇐)

                                            ↳33 QDP

                                                ↳34 NonTerminationProof (⇔)

                                                    ↳35 FALSE

                            ↳36 PiDP

                                ↳37 UsableRulesProof (⇔)

                                    ↳38 PiDP

                                        ↳39 PiDPToQDPProof (⇐)

                                            ↳40 QDP

                                                ↳41 QDPSizeChangeProof (⇔)

                                                    ↳42 TRUE

(0) Obligation:

Clauses:

len([], 0) :- !.
len(Xs, s(N)) :- ','(tail(Xs, Ys), len(Ys, N)).
tail([], []).
tail(.(X, Xs), Xs).

Queries:

len(g,a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: len(T1, T2)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: len(T1, T2), SCOPE: 1, CLAUSE: 0 | len(T1, T2), SCOPE: 1, CLAUSE: 1 | ?_1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: !_1 | len([], T2), SCOPE: 1, CLAUSE: 1 | ?_1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: len(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nlen(T1, T2) !=? len([], 0)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(tail(T3, X5), len(X5, T5))\n\nT3 is ground\nX5 is free; \nlen(T3, T2) !=? len([], 0)", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\nX3 is free\nX4 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(tail(T3, X5), len(X5, T5)), SCOPE: 2, CLAUSE: 2 | ','(tail(T3, X5), len(X5, T5)), SCOPE: 2, CLAUSE: 3\n\nT3 is ground\nX5 is free; \nlen(T3, T2) !=? len([], 0)", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: len([], T5) | ','(tail([], X5), len(X5, T5)), SCOPE: 2, CLAUSE: 3\n\nX5 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(tail(T3, X5), len(X5, T5)), SCOPE: 2, CLAUSE: 3\n\nT3 is ground\nX5 is free; \nlen(T3, T2) !=? len([], 0)\ntail(T3, X5) !=? tail([], [])", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: len([], T5)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(tail([], X5), len(X5, T5)), SCOPE: 2, CLAUSE: 3\n\nX5 is free", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: len([], T5), SCOPE: 3, CLAUSE: 0 | len([], T5), SCOPE: 3, CLAUSE: 1 | ?_3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: !_3 | len([], T5), SCOPE: 3, CLAUSE: 1 | ?_3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: len([], T5), SCOPE: 3, CLAUSE: 1\n\nlen([], T5) !=? len([], 0)", 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: ','(tail([], X10), len(X10, T7))\n\nX10 is free", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: \n\nX8 is free\nX9 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: ','(tail([], X10), len(X10, T7)), SCOPE: 4, CLAUSE: 2 | ','(tail([], X10), len(X10, T7)), SCOPE: 4, CLAUSE: 3\n\nX10 is free", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: len([], T7) | ','(tail([], X10), len(X10, T7)), SCOPE: 4, CLAUSE: 3\n\nX10 is free", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: len(T9, T5)\n\nT9 is ground\nX5 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: \n\nX5 is free\nX15 is free; \nX16 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 26 [arrowhead = none ];
26 -> 2;

27 [label="EVAL with clause\nlen([], 0) :- !.\nand substitution\nT1 -> []\nT2 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 27 [arrowhead = none ];
27 -> 3;

28 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 28 [arrowhead = none ];
28 -> 4;

29 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 29 [arrowhead = none ];
29 -> 5;

30 [label="EVAL with clause\nlen(X3, s(X4)) :- ','(tail(X3, X5), len(X5, X4)).\nand substitution\nT1 -> T3\nX3 -> T3\nX4 -> T5\nT2 -> s(T5)\nT4 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 30 [arrowhead = none ];
30 -> 7;

31 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 31 [arrowhead = none ];
31 -> 8;

32 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 32 [arrowhead = none ];
32 -> 6;

33 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 33 [arrowhead = none ];
33 -> 9;

34 [label="EVAL with clause\ntail([], []).\nand substitution\nT3 -> []\nX5 -> []", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 34 [arrowhead = none ];
34 -> 10;

35 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 35 [arrowhead = none ];
35 -> 11;

36 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
10 -> 36 [arrowhead = none ];
36 -> 12;

37 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
10 -> 37 [arrowhead = none ];
37 -> 13;

38 [label="EVAL with clause\ntail(.(X15, X16), X16).\nand substitution\nX15 -> T8\nX16 -> T9\nT3 -> .(T8, T9)\nX5 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 38 [arrowhead = none ];
38 -> 24;

39 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 39 [arrowhead = none ];
39 -> 25;

40 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 40 [arrowhead = none ];
40 -> 14;

41 [label="BACKTRACK\nfor clause: tail(.(X, Xs), Xs)because of non-unification", fontsize=14, style = filled, fillcolor = white];
13 -> 41 [arrowhead = none ];
41 -> 23;

42 [label="EVAL with clause\nlen([], 0) :- !.\nand substitution\nT5 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 42 [arrowhead = none ];
42 -> 15;

43 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 43 [arrowhead = none ];
43 -> 16;

44 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 44 [arrowhead = none ];
44 -> 17;

45 [label="EVAL with clause\nlen(X8, s(X9)) :- ','(tail(X8, X10), len(X10, X9)).\nand substitution\nX8 -> []\nX9 -> T7\nT5 -> s(T7)\nT6 -> T7", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 45 [arrowhead = none ];
45 -> 19;

46 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 46 [arrowhead = none ];
46 -> 20;

47 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 47 [arrowhead = none ];
47 -> 18;

48 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
19 -> 48 [arrowhead = none ];
48 -> 21;

49 [label="EVAL with clause\ntail([], []).\nand substitution\nX10 -> []", fontsize=14, style = filled, fillcolor = lightblue];
21 -> 49 [arrowhead = none ];
49 -> 22;

50 [label="INSTANCE with matching:\nT5 -> T7\nX5 -> X10", fontsize=14, style = filled, fillcolor = lightgrey];
22 -> 50 [arrowhead = none , style=dashed];
50 -> 10[style=dashed];

51 [label="INSTANCE with matching:\nT1 -> T9\nT2 -> T5", fontsize=14, style = filled, fillcolor = lightgrey];
24 -> 51 [arrowhead = none , style=dashed];
51 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

len10(0, X5).
len10(s(T7), X5) :- len10(T7, X10).
len1([], 0).
len1([], s(0)).
len1([], s(s(T7))) :- len10(T7, X10).
len1(.(T8, T9), s(T5)) :- len1(T9, T5).

Queries:

len1(g,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:
len1_in: (b,f)
len10_in: (f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga([], s(0)) → len1_out_ga([], s(0))
len1_in_ga([], s(s(T7))) → U2_ga(T7, len10_in_aa(T7, X10))
len10_in_aa(0, X5) → len10_out_aa(0, X5)
len10_in_aa(s(T7), X5) → U1_aa(T7, X5, len10_in_aa(T7, X10))
U1_aa(T7, X5, len10_out_aa(T7, X10)) → len10_out_aa(s(T7), X5)
U2_ga(T7, len10_out_aa(T7, X10)) → len1_out_ga([], s(s(T7)))
len1_in_ga(.(T8, T9), s(T5)) → U3_ga(T8, T9, T5, len1_in_ga(T9, T5))
U3_ga(T8, T9, T5, len1_out_ga(T9, T5)) → len1_out_ga(.(T8, T9), s(T5))

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga([], s(0)) → len1_out_ga([], s(0))
len1_in_ga([], s(s(T7))) → U2_ga(T7, len10_in_aa(T7, X10))
len10_in_aa(0, X5) → len10_out_aa(0, X5)
len10_in_aa(s(T7), X5) → U1_aa(T7, X5, len10_in_aa(T7, X10))
U1_aa(T7, X5, len10_out_aa(T7, X10)) → len10_out_aa(s(T7), X5)
U2_ga(T7, len10_out_aa(T7, X10)) → len1_out_ga([], s(s(T7)))
len1_in_ga(.(T8, T9), s(T5)) → U3_ga(T8, T9, T5, len1_in_ga(T9, T5))
U3_ga(T8, T9, T5, len1_out_ga(T9, T5)) → len1_out_ga(.(T8, T9), s(T5))

(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:

LEN1_IN_GA([], s(s(T7))) → U2_GA(T7, len10_in_aa(T7, X10))
LEN1_IN_GA([], s(s(T7))) → LEN10_IN_AA(T7, X10)
LEN10_IN_AA(s(T7), X5) → U1_AA(T7, X5, len10_in_aa(T7, X10))
LEN10_IN_AA(s(T7), X5) → LEN10_IN_AA(T7, X10)
LEN1_IN_GA(.(T8, T9), s(T5)) → U3_GA(T8, T9, T5, len1_in_ga(T9, T5))
LEN1_IN_GA(.(T8, T9), s(T5)) → LEN1_IN_GA(T9, T5)

The TRS R consists of the following rules:

len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga([], s(0)) → len1_out_ga([], s(0))
len1_in_ga([], s(s(T7))) → U2_ga(T7, len10_in_aa(T7, X10))
len10_in_aa(0, X5) → len10_out_aa(0, X5)
len10_in_aa(s(T7), X5) → U1_aa(T7, X5, len10_in_aa(T7, X10))
U1_aa(T7, X5, len10_out_aa(T7, X10)) → len10_out_aa(s(T7), X5)
U2_ga(T7, len10_out_aa(T7, X10)) → len1_out_ga([], s(s(T7)))
len1_in_ga(.(T8, T9), s(T5)) → U3_ga(T8, T9, T5, len1_in_ga(T9, T5))
U3_ga(T8, T9, T5, len1_out_ga(T9, T5)) → len1_out_ga(.(T8, T9), s(T5))

The argument filtering Pi contains the following mapping:
len1_in_ga(x1, x2) = len1_in_ga(x1)
[] = []
len1_out_ga(x1, x2) = len1_out_ga(x2)
U2_ga(x1, x2) = U2_ga(x2)
len10_in_aa(x1, x2) = len10_in_aa
len10_out_aa(x1, x2) = len10_out_aa(x1)
U1_aa(x1, x2, x3) = U1_aa(x3)
.(x1, x2) = .(x1, x2)
U3_ga(x1, x2, x3, x4) = U3_ga(x4)
s(x1) = s(x1)
LEN1_IN_GA(x1, x2) = LEN1_IN_GA(x1)
U2_GA(x1, x2) = U2_GA(x2)
LEN10_IN_AA(x1, x2) = LEN10_IN_AA
U1_AA(x1, x2, x3) = U1_AA(x3)
U3_GA(x1, x2, x3, x4) = U3_GA(x4)

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

(6) Obligation:

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

LEN1_IN_GA([], s(s(T7))) → U2_GA(T7, len10_in_aa(T7, X10))
LEN1_IN_GA([], s(s(T7))) → LEN10_IN_AA(T7, X10)
LEN10_IN_AA(s(T7), X5) → U1_AA(T7, X5, len10_in_aa(T7, X10))
LEN10_IN_AA(s(T7), X5) → LEN10_IN_AA(T7, X10)
LEN1_IN_GA(.(T8, T9), s(T5)) → U3_GA(T8, T9, T5, len1_in_ga(T9, T5))
LEN1_IN_GA(.(T8, T9), s(T5)) → LEN1_IN_GA(T9, T5)

The TRS R consists of the following rules:

len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga([], s(0)) → len1_out_ga([], s(0))
len1_in_ga([], s(s(T7))) → U2_ga(T7, len10_in_aa(T7, X10))
len10_in_aa(0, X5) → len10_out_aa(0, X5)
len10_in_aa(s(T7), X5) → U1_aa(T7, X5, len10_in_aa(T7, X10))
U1_aa(T7, X5, len10_out_aa(T7, X10)) → len10_out_aa(s(T7), X5)
U2_ga(T7, len10_out_aa(T7, X10)) → len1_out_ga([], s(s(T7)))
len1_in_ga(.(T8, T9), s(T5)) → U3_ga(T8, T9, T5, len1_in_ga(T9, T5))
U3_ga(T8, T9, T5, len1_out_ga(T9, T5)) → len1_out_ga(.(T8, T9), s(T5))

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 4 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

LEN10_IN_AA(s(T7), X5) → LEN10_IN_AA(T7, X10)

The TRS R consists of the following rules:

len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga([], s(0)) → len1_out_ga([], s(0))
len1_in_ga([], s(s(T7))) → U2_ga(T7, len10_in_aa(T7, X10))
len10_in_aa(0, X5) → len10_out_aa(0, X5)
len10_in_aa(s(T7), X5) → U1_aa(T7, X5, len10_in_aa(T7, X10))
U1_aa(T7, X5, len10_out_aa(T7, X10)) → len10_out_aa(s(T7), X5)
U2_ga(T7, len10_out_aa(T7, X10)) → len1_out_ga([], s(s(T7)))
len1_in_ga(.(T8, T9), s(T5)) → U3_ga(T8, T9, T5, len1_in_ga(T9, T5))
U3_ga(T8, T9, T5, len1_out_ga(T9, T5)) → len1_out_ga(.(T8, T9), s(T5))

The argument filtering Pi contains the following mapping:
len1_in_ga(x1, x2) = len1_in_ga(x1)
[] = []
len1_out_ga(x1, x2) = len1_out_ga(x2)
U2_ga(x1, x2) = U2_ga(x2)
len10_in_aa(x1, x2) = len10_in_aa
len10_out_aa(x1, x2) = len10_out_aa(x1)
U1_aa(x1, x2, x3) = U1_aa(x3)
.(x1, x2) = .(x1, x2)
U3_ga(x1, x2, x3, x4) = U3_ga(x4)
s(x1) = s(x1)
LEN10_IN_AA(x1, x2) = LEN10_IN_AA

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

(10) UsableRulesProof (EQUIVALENT transformation)

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

(11) Obligation:

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

LEN10_IN_AA(s(T7), X5) → LEN10_IN_AA(T7, X10)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
LEN10_IN_AA(x1, x2) = LEN10_IN_AA

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

(12) PiDPToQDPProof (SOUND transformation)

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

(13) Obligation:

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

LEN10_IN_AA → LEN10_IN_AA

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

(14) 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 = LEN10_IN_AA evaluates to t =LEN10_IN_AA

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 LEN10_IN_AA to LEN10_IN_AA.

(15) FALSE

(16) Obligation:

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

LEN1_IN_GA(.(T8, T9), s(T5)) → LEN1_IN_GA(T9, T5)

The TRS R consists of the following rules:

len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga([], s(0)) → len1_out_ga([], s(0))
len1_in_ga([], s(s(T7))) → U2_ga(T7, len10_in_aa(T7, X10))
len10_in_aa(0, X5) → len10_out_aa(0, X5)
len10_in_aa(s(T7), X5) → U1_aa(T7, X5, len10_in_aa(T7, X10))
U1_aa(T7, X5, len10_out_aa(T7, X10)) → len10_out_aa(s(T7), X5)
U2_ga(T7, len10_out_aa(T7, X10)) → len1_out_ga([], s(s(T7)))
len1_in_ga(.(T8, T9), s(T5)) → U3_ga(T8, T9, T5, len1_in_ga(T9, T5))
U3_ga(T8, T9, T5, len1_out_ga(T9, T5)) → len1_out_ga(.(T8, T9), s(T5))

(17) UsableRulesProof (EQUIVALENT transformation)

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

(18) Obligation:

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

LEN1_IN_GA(.(T8, T9), s(T5)) → LEN1_IN_GA(T9, T5)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
LEN1_IN_GA(x1, x2) = LEN1_IN_GA(x1)

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

(19) PiDPToQDPProof (SOUND transformation)

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

(20) Obligation:

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

LEN1_IN_GA(.(T8, T9)) → LEN1_IN_GA(T9)

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

(21) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

LEN1_IN_GA(.(T8, T9)) → LEN1_IN_GA(T9)
The graph contains the following edges 1 > 1

(22) TRUE

(23) PrologToPiTRSProof (SOUND transformation)

len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga([], s(0)) → len1_out_ga([], s(0))
len1_in_ga([], s(s(T7))) → U2_ga(T7, len10_in_aa(T7, X10))
len10_in_aa(0, X5) → len10_out_aa(0, X5)
len10_in_aa(s(T7), X5) → U1_aa(T7, X5, len10_in_aa(T7, X10))
U1_aa(T7, X5, len10_out_aa(T7, X10)) → len10_out_aa(s(T7), X5)
U2_ga(T7, len10_out_aa(T7, X10)) → len1_out_ga([], s(s(T7)))
len1_in_ga(.(T8, T9), s(T5)) → U3_ga(T8, T9, T5, len1_in_ga(T9, T5))
U3_ga(T8, T9, T5, len1_out_ga(T9, T5)) → len1_out_ga(.(T8, T9), s(T5))

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(24) Obligation:

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

len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga([], s(0)) → len1_out_ga([], s(0))
len1_in_ga([], s(s(T7))) → U2_ga(T7, len10_in_aa(T7, X10))
len10_in_aa(0, X5) → len10_out_aa(0, X5)
len10_in_aa(s(T7), X5) → U1_aa(T7, X5, len10_in_aa(T7, X10))
U1_aa(T7, X5, len10_out_aa(T7, X10)) → len10_out_aa(s(T7), X5)
U2_ga(T7, len10_out_aa(T7, X10)) → len1_out_ga([], s(s(T7)))
len1_in_ga(.(T8, T9), s(T5)) → U3_ga(T8, T9, T5, len1_in_ga(T9, T5))
U3_ga(T8, T9, T5, len1_out_ga(T9, T5)) → len1_out_ga(.(T8, T9), s(T5))

(25) 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:

LEN1_IN_GA([], s(s(T7))) → U2_GA(T7, len10_in_aa(T7, X10))
LEN1_IN_GA([], s(s(T7))) → LEN10_IN_AA(T7, X10)
LEN10_IN_AA(s(T7), X5) → U1_AA(T7, X5, len10_in_aa(T7, X10))
LEN10_IN_AA(s(T7), X5) → LEN10_IN_AA(T7, X10)
LEN1_IN_GA(.(T8, T9), s(T5)) → U3_GA(T8, T9, T5, len1_in_ga(T9, T5))
LEN1_IN_GA(.(T8, T9), s(T5)) → LEN1_IN_GA(T9, T5)

The TRS R consists of the following rules:

len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga([], s(0)) → len1_out_ga([], s(0))
len1_in_ga([], s(s(T7))) → U2_ga(T7, len10_in_aa(T7, X10))
len10_in_aa(0, X5) → len10_out_aa(0, X5)
len10_in_aa(s(T7), X5) → U1_aa(T7, X5, len10_in_aa(T7, X10))
U1_aa(T7, X5, len10_out_aa(T7, X10)) → len10_out_aa(s(T7), X5)
U2_ga(T7, len10_out_aa(T7, X10)) → len1_out_ga([], s(s(T7)))
len1_in_ga(.(T8, T9), s(T5)) → U3_ga(T8, T9, T5, len1_in_ga(T9, T5))
U3_ga(T8, T9, T5, len1_out_ga(T9, T5)) → len1_out_ga(.(T8, T9), s(T5))

The argument filtering Pi contains the following mapping:
len1_in_ga(x1, x2) = len1_in_ga(x1)
[] = []
len1_out_ga(x1, x2) = len1_out_ga(x1, x2)
U2_ga(x1, x2) = U2_ga(x2)
len10_in_aa(x1, x2) = len10_in_aa
len10_out_aa(x1, x2) = len10_out_aa(x1)
U1_aa(x1, x2, x3) = U1_aa(x3)
.(x1, x2) = .(x1, x2)
U3_ga(x1, x2, x3, x4) = U3_ga(x1, x2, x4)
s(x1) = s(x1)
LEN1_IN_GA(x1, x2) = LEN1_IN_GA(x1)
U2_GA(x1, x2) = U2_GA(x2)
LEN10_IN_AA(x1, x2) = LEN10_IN_AA
U1_AA(x1, x2, x3) = U1_AA(x3)
U3_GA(x1, x2, x3, x4) = U3_GA(x1, x2, x4)

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

(26) Obligation:

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

LEN1_IN_GA([], s(s(T7))) → U2_GA(T7, len10_in_aa(T7, X10))
LEN1_IN_GA([], s(s(T7))) → LEN10_IN_AA(T7, X10)
LEN10_IN_AA(s(T7), X5) → U1_AA(T7, X5, len10_in_aa(T7, X10))
LEN10_IN_AA(s(T7), X5) → LEN10_IN_AA(T7, X10)
LEN1_IN_GA(.(T8, T9), s(T5)) → U3_GA(T8, T9, T5, len1_in_ga(T9, T5))
LEN1_IN_GA(.(T8, T9), s(T5)) → LEN1_IN_GA(T9, T5)

The TRS R consists of the following rules:

len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga([], s(0)) → len1_out_ga([], s(0))
len1_in_ga([], s(s(T7))) → U2_ga(T7, len10_in_aa(T7, X10))
len10_in_aa(0, X5) → len10_out_aa(0, X5)
len10_in_aa(s(T7), X5) → U1_aa(T7, X5, len10_in_aa(T7, X10))
U1_aa(T7, X5, len10_out_aa(T7, X10)) → len10_out_aa(s(T7), X5)
U2_ga(T7, len10_out_aa(T7, X10)) → len1_out_ga([], s(s(T7)))
len1_in_ga(.(T8, T9), s(T5)) → U3_ga(T8, T9, T5, len1_in_ga(T9, T5))
U3_ga(T8, T9, T5, len1_out_ga(T9, T5)) → len1_out_ga(.(T8, T9), s(T5))

(27) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 4 less nodes.

(28) Complex Obligation (AND)

(29) Obligation:

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

LEN10_IN_AA(s(T7), X5) → LEN10_IN_AA(T7, X10)

The TRS R consists of the following rules:

len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga([], s(0)) → len1_out_ga([], s(0))
len1_in_ga([], s(s(T7))) → U2_ga(T7, len10_in_aa(T7, X10))
len10_in_aa(0, X5) → len10_out_aa(0, X5)
len10_in_aa(s(T7), X5) → U1_aa(T7, X5, len10_in_aa(T7, X10))
U1_aa(T7, X5, len10_out_aa(T7, X10)) → len10_out_aa(s(T7), X5)
U2_ga(T7, len10_out_aa(T7, X10)) → len1_out_ga([], s(s(T7)))
len1_in_ga(.(T8, T9), s(T5)) → U3_ga(T8, T9, T5, len1_in_ga(T9, T5))
U3_ga(T8, T9, T5, len1_out_ga(T9, T5)) → len1_out_ga(.(T8, T9), s(T5))

The argument filtering Pi contains the following mapping:
len1_in_ga(x1, x2) = len1_in_ga(x1)
[] = []
len1_out_ga(x1, x2) = len1_out_ga(x1, x2)
U2_ga(x1, x2) = U2_ga(x2)
len10_in_aa(x1, x2) = len10_in_aa
len10_out_aa(x1, x2) = len10_out_aa(x1)
U1_aa(x1, x2, x3) = U1_aa(x3)
.(x1, x2) = .(x1, x2)
U3_ga(x1, x2, x3, x4) = U3_ga(x1, x2, x4)
s(x1) = s(x1)
LEN10_IN_AA(x1, x2) = LEN10_IN_AA

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

(30) UsableRulesProof (EQUIVALENT transformation)

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

(31) Obligation:

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

LEN10_IN_AA(s(T7), X5) → LEN10_IN_AA(T7, X10)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
LEN10_IN_AA(x1, x2) = LEN10_IN_AA

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

(32) PiDPToQDPProof (SOUND transformation)

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

(33) Obligation:

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

LEN10_IN_AA → LEN10_IN_AA

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

(34) NonTerminationProof (EQUIVALENT transformation)

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from LEN10_IN_AA to LEN10_IN_AA.

(35) FALSE

(36) Obligation:

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

LEN1_IN_GA(.(T8, T9), s(T5)) → LEN1_IN_GA(T9, T5)

The TRS R consists of the following rules:

len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga([], s(0)) → len1_out_ga([], s(0))
len1_in_ga([], s(s(T7))) → U2_ga(T7, len10_in_aa(T7, X10))
len10_in_aa(0, X5) → len10_out_aa(0, X5)
len10_in_aa(s(T7), X5) → U1_aa(T7, X5, len10_in_aa(T7, X10))
U1_aa(T7, X5, len10_out_aa(T7, X10)) → len10_out_aa(s(T7), X5)
U2_ga(T7, len10_out_aa(T7, X10)) → len1_out_ga([], s(s(T7)))
len1_in_ga(.(T8, T9), s(T5)) → U3_ga(T8, T9, T5, len1_in_ga(T9, T5))
U3_ga(T8, T9, T5, len1_out_ga(T9, T5)) → len1_out_ga(.(T8, T9), s(T5))

(37) UsableRulesProof (EQUIVALENT transformation)

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

(38) Obligation:

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

LEN1_IN_GA(.(T8, T9), s(T5)) → LEN1_IN_GA(T9, T5)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
LEN1_IN_GA(x1, x2) = LEN1_IN_GA(x1)

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

(39) PiDPToQDPProof (SOUND transformation)

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

(40) Obligation:

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

LEN1_IN_GA(.(T8, T9)) → LEN1_IN_GA(T9)

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

(41) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

LEN1_IN_GA(.(T8, T9)) → LEN1_IN_GA(T9)
The graph contains the following edges 1 > 1

(0) Obligation:

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

(2) Obligation:

(3) PrologToPiTRSProof (SOUND transformation)

(4) Obligation:

(5) DependencyPairsProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Complex Obligation (AND)

(9) Obligation:

(10) UsableRulesProof (EQUIVALENT transformation)

(11) Obligation:

(12) PiDPToQDPProof (SOUND transformation)

(13) Obligation:

(14) NonTerminationProof (EQUIVALENT transformation)

(15) FALSE

(16) Obligation:

(17) UsableRulesProof (EQUIVALENT transformation)

(18) Obligation:

(19) PiDPToQDPProof (SOUND transformation)

(20) Obligation:

(21) QDPSizeChangeProof (EQUIVALENT transformation)

(22) TRUE

(23) PrologToPiTRSProof (SOUND transformation)

(24) Obligation:

(25) DependencyPairsProof (EQUIVALENT transformation)

(26) Obligation:

(27) DependencyGraphProof (EQUIVALENT transformation)

(28) Complex Obligation (AND)

(29) Obligation:

(30) UsableRulesProof (EQUIVALENT transformation)

(31) Obligation:

(32) PiDPToQDPProof (SOUND transformation)

(33) Obligation:

(34) NonTerminationProof (EQUIVALENT transformation)

(35) FALSE

(36) Obligation:

(37) UsableRulesProof (EQUIVALENT transformation)

(38) Obligation:

(39) PiDPToQDPProof (SOUND transformation)

(40) Obligation:

(41) QDPSizeChangeProof (EQUIVALENT transformation)

(42) TRUE