Left Termination of the query pattern len_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

len([], 0).
len(Xs, s(N)) :- ','(no(empty(Xs)), ','(tail(Xs, Ys), len(Ys, N))).
tail([], []).
tail(.(X, Xs), Xs).
empty([]).
no(X) :- ','(X, ','(!, failure(a))).
no(X1).
failure(b).

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\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: len(T1, T2), SCOPE: 1, CLAUSE: 0\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: len(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", 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: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(no(empty(T3)), ','(tail(T3, X6), len(X6, T5)))\n\nT3 is ground\nX6 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\nX4 is free\nX5 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(no(empty(T3)), ','(tail(T3, X6), len(X6, T5))), SCOPE: 2, CLAUSE: 5 | ','(no(empty(T3)), ','(tail(T3, X6), len(X6, T5))), SCOPE: 2, CLAUSE: 6 | ?_2\n\nT3 is ground\nX6 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(empty(T6), ','(!_2, ','(failure(a), ','(tail(T6, X6), len(X6, T5))))) | ','(no(empty(T6)), ','(tail(T6, X6), len(X6, T5))), SCOPE: 2, CLAUSE: 6 | ?_2\n\nT6 is ground\nX6 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(empty(T6), ','(!_2, ','(failure(a), ','(tail(T6, X6), len(X6, T5))))), SCOPE: 3, CLAUSE: 4 | ','(no(empty(T6)), ','(tail(T6, X6), len(X6, T5))), SCOPE: 2, CLAUSE: 6 | ?_2\n\nT6 is ground\nX6 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(!_2, ','(failure(a), ','(tail([], X6), len(X6, T5)))) | ','(no(empty([])), ','(tail([], X6), len(X6, T5))), SCOPE: 2, CLAUSE: 6 | ?_2\n\nX6 is free", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(no(empty(T6)), ','(tail(T6, X6), len(X6, T5))), SCOPE: 2, CLAUSE: 6\n\nT6 is ground\nX6 is free; \nempty(T6) !=? empty([])", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(failure(a), ','(tail([], X6), len(X6, T5)))\n\nX6 is free", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(failure(a), ','(tail([], X6), len(X6, T5))), SCOPE: 4, CLAUSE: 7\n\nX6 is free", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(tail(T7, X6), len(X6, T5))\n\nT7 is ground\nX6 is free; \nempty(T7) !=? empty([])", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: ','(tail(T7, X6), len(X6, T5)), SCOPE: 5, CLAUSE: 2 | ','(tail(T7, X6), len(X6, T5)), SCOPE: 5, CLAUSE: 3\n\nT7 is ground\nX6 is free; \nempty(T7) !=? empty([])", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: ','(tail(T7, X6), len(X6, T5)), SCOPE: 5, CLAUSE: 3\n\nT7 is ground\nX6 is free; \nempty(T7) !=? empty([])\ntail(T7, X6) !=? tail([], [])", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: len(T9, T5)\n\nT9 is ground\nX6 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\nX6 is free\nX13 is free; \nX14 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 23 [arrowhead = none ];
23 -> 2;

24 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
2 -> 24 [arrowhead = none ];
24 -> 3;

25 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
2 -> 25 [arrowhead = none ];
25 -> 4;

26 [label="EVAL with clause\nlen([], 0).\nand substitution\nT1 -> []\nT2 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 26 [arrowhead = none ];
26 -> 5;

27 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 27 [arrowhead = none ];
27 -> 6;

28 [label="EVAL with clause\nlen(X4, s(X5)) :- ','(no(empty(X4)), ','(tail(X4, X6), len(X6, X5))).\nand substitution\nT1 -> T3\nX4 -> T3\nX5 -> T5\nT2 -> s(T5)\nT4 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 28 [arrowhead = none ];
28 -> 8;

29 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 29 [arrowhead = none ];
29 -> 9;

30 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 30 [arrowhead = none ];
30 -> 7;

31 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
8 -> 31 [arrowhead = none ];
31 -> 10;

32 [label="EVAL with clause\nno(X8) :- ','(X8, ','(!, failure(a))).\nand substitution\nT3 -> T6\nX8 -> empty(T6)", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 32 [arrowhead = none ];
32 -> 11;

33 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
11 -> 33 [arrowhead = none ];
33 -> 12;

34 [label="EVAL with clause\nempty([]).\nand substitution\nT6 -> []", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 34 [arrowhead = none ];
34 -> 13;

35 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 35 [arrowhead = none ];
35 -> 14;

36 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 36 [arrowhead = none ];
36 -> 15;

37 [label="EVAL with clause\nno(X10).\nand substitution\nT6 -> T7\nX10 -> empty(T7)", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 37 [arrowhead = none ];
37 -> 18;

38 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
15 -> 38 [arrowhead = none ];
38 -> 16;

39 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = white];
16 -> 39 [arrowhead = none ];
39 -> 17;

40 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
18 -> 40 [arrowhead = none ];
40 -> 19;

41 [label="BACKTRACK\nfor clause: tail([], [])\nwith clash: (empty(T7), empty([]))", fontsize=14, style = filled, fillcolor = white];
19 -> 41 [arrowhead = none ];
41 -> 20;

42 [label="EVAL with clause\ntail(.(X13, X14), X14).\nand substitution\nX13 -> T8\nX14 -> T9\nT7 -> .(T8, T9)\nX6 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
20 -> 42 [arrowhead = none ];
42 -> 21;

43 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
20 -> 43 [arrowhead = none ];
43 -> 22;

44 [label="INSTANCE with matching:\nT1 -> T9\nT2 -> T5", fontsize=14, style = filled, fillcolor = lightgrey];
21 -> 44 [arrowhead = none , style=dashed];
44 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

len1([], 0).
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)
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(.(T8, T9), s(T5)) → U1_ga(T8, T9, T5, len1_in_ga(T9, T5))
U1_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)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
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:

len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga(.(T8, T9), s(T5)) → U1_ga(T8, T9, T5, len1_in_ga(T9, T5))
U1_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)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
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:

LEN1_IN_GA(.(T8, T9), s(T5)) → U1_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(.(T8, T9), s(T5)) → U1_ga(T8, T9, T5, len1_in_ga(T9, T5))
U1_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)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
s(x1)  =  s(x1)
LEN1_IN_GA(x1, x2)  =  LEN1_IN_GA(x1)
U1_GA(x1, x2, x3, x4)  =  U1_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(.(T8, T9), s(T5)) → U1_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(.(T8, T9), s(T5)) → U1_ga(T8, T9, T5, len1_in_ga(T9, T5))
U1_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)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
s(x1)  =  s(x1)
LEN1_IN_GA(x1, x2)  =  LEN1_IN_GA(x1)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x4)

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:

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(.(T8, T9), s(T5)) → U1_ga(T8, T9, T5, len1_in_ga(T9, T5))
U1_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)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
s(x1)  =  s(x1)
LEN1_IN_GA(x1, x2)  =  LEN1_IN_GA(x1)

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:

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

(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:

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

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

(13) 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

(14) TRUE