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

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, eq(T4, 0)) | 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\nX2 is free; \nlen(T1, T2) !=? len([], X2)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: eq(T4, 0)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: eq(T4, 0), SCOPE: 2, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(tail(T6, X9), len(X9, T8))\n\nT6 is ground\nX2 is free; \nX9 is free; \nlen(T6, T2) !=? len([], X2)", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\nX2 is free\nX7 is free; \nX8 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(tail(T6, X9), len(X9, T8)), SCOPE: 3, CLAUSE: 2 | ','(tail(T6, X9), len(X9, T8)), SCOPE: 3, CLAUSE: 3\n\nT6 is ground\nX2 is free; \nX9 is free; \nlen(T6, T2) !=? len([], X2)", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(tail(T6, X9), len(X9, T8)), SCOPE: 3, CLAUSE: 3\n\nT6 is ground\nX2 is free; \nX9 is free; \nlen(T6, T2) !=? len([], X2)\ntail(T6, X9) !=? tail([], [])", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: len(T10, T8)\n\nT10 is ground\nX2 is free; \nX9 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\nX2 is free\nX9 is free; \nX12 is free; \nX13 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 16 [arrowhead = none ];
16 -> 2;

17 [label="EVAL with clause\nlen([], X2) :- ','(!, eq(X2, 0)).\nand substitution\nT1 -> []\nT2 -> T4\nX2 -> T4\nT3 -> T4", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 17 [arrowhead = none ];
17 -> 3;

18 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 18 [arrowhead = none ];
18 -> 4;

19 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 19 [arrowhead = none ];
19 -> 5;

20 [label="EVAL with clause\nlen(X7, s(X8)) :- ','(tail(X7, X9), len(X9, X8)).\nand substitution\nT1 -> T6\nX7 -> T6\nX8 -> T8\nT2 -> s(T8)\nT7 -> T8", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 20 [arrowhead = none ];
20 -> 10;

21 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 21 [arrowhead = none ];
21 -> 11;

22 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
5 -> 22 [arrowhead = none ];
22 -> 6;

23 [label="EVAL with clause\neq(X4, X4).\nand substitution\nT4 -> 0\nX4 -> 0\nT5 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 23 [arrowhead = none ];
23 -> 7;

24 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 24 [arrowhead = none ];
24 -> 8;

25 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 25 [arrowhead = none ];
25 -> 9;

26 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
10 -> 26 [arrowhead = none ];
26 -> 12;

27 [label="BACKTRACK\nfor clause: tail([], [])\nwith clash: (len(T6, T2), len([], X2))", fontsize=14, style = filled, fillcolor = white];
12 -> 27 [arrowhead = none ];
27 -> 13;

28 [label="EVAL with clause\ntail(.(X12, X13), X13).\nand substitution\nX12 -> T9\nX13 -> T10\nT6 -> .(T9, T10)\nX9 -> T10", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 28 [arrowhead = none ];
28 -> 14;

29 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 29 [arrowhead = none ];
29 -> 15;

30 [label="INSTANCE with matching:\nT1 -> T10\nT2 -> T8", fontsize=14, style = filled, fillcolor = lightgrey];
14 -> 30 [arrowhead = none , style=dashed];
30 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

len1([], 0).
len1(.(T9, T10), s(T8)) :- len1(T10, T8).

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(.(T9, T10), s(T8)) → U1_ga(T9, T10, T8, len1_in_ga(T10, T8))
U1_ga(T9, T10, T8, len1_out_ga(T10, T8)) → len1_out_ga(.(T9, T10), s(T8))

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)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x1, x2, 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(.(T9, T10), s(T8)) → U1_ga(T9, T10, T8, len1_in_ga(T10, T8))
U1_ga(T9, T10, T8, len1_out_ga(T10, T8)) → len1_out_ga(.(T9, T10), s(T8))

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)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x1, x2, 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(.(T9, T10), s(T8)) → U1_GA(T9, T10, T8, len1_in_ga(T10, T8))
LEN1_IN_GA(.(T9, T10), s(T8)) → LEN1_IN_GA(T10, T8)

The TRS R consists of the following rules:

len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga(.(T9, T10), s(T8)) → U1_ga(T9, T10, T8, len1_in_ga(T10, T8))
U1_ga(T9, T10, T8, len1_out_ga(T10, T8)) → len1_out_ga(.(T9, T10), s(T8))

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)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x1, x2, x4)
s(x1)  =  s(x1)
LEN1_IN_GA(x1, x2)  =  LEN1_IN_GA(x1)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x1, x2, 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(.(T9, T10), s(T8)) → U1_GA(T9, T10, T8, len1_in_ga(T10, T8))
LEN1_IN_GA(.(T9, T10), s(T8)) → LEN1_IN_GA(T10, T8)

The TRS R consists of the following rules:

len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga(.(T9, T10), s(T8)) → U1_ga(T9, T10, T8, len1_in_ga(T10, T8))
U1_ga(T9, T10, T8, len1_out_ga(T10, T8)) → len1_out_ga(.(T9, T10), s(T8))

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)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x1, x2, x4)
s(x1)  =  s(x1)
LEN1_IN_GA(x1, x2)  =  LEN1_IN_GA(x1)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x1, x2, 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(.(T9, T10), s(T8)) → LEN1_IN_GA(T10, T8)

The TRS R consists of the following rules:

len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga(.(T9, T10), s(T8)) → U1_ga(T9, T10, T8, len1_in_ga(T10, T8))
U1_ga(T9, T10, T8, len1_out_ga(T10, T8)) → len1_out_ga(.(T9, T10), s(T8))

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)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x1, x2, 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(.(T9, T10), s(T8)) → LEN1_IN_GA(T10, T8)

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(.(T9, T10)) → LEN1_IN_GA(T10)

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(.(T9, T10)) → LEN1_IN_GA(T10)
  The graph contains the following edges 1 > 1

(14) TRUE