Left Termination of the query pattern
len(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).

Query: len(g,a)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: 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: ','(!_1, eq(T5, 0)) | len([], T2), SCOPE: 1, CLAUSE: 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(T5, 0)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: eq(T5, 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\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(tail(T11, X10), len(X10, T13))\n\nT11 is ground\nX2 is free\nX10 is free\nlen(T11, T2) !=? len([], X2)", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(tail(T11, X10), len(X10, T13)), SCOPE: 3, CLAUSE: 2 | ','(tail(T11, X10), len(X10, T13)), SCOPE: 3, CLAUSE: 3\n\nT11 is ground\nX2 is free\nX10 is free\nlen(T11, T2) !=? len([], X2)", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(tail(T11, X10), len(X10, T13)), SCOPE: 3, CLAUSE: 3\n\nT11 is ground\nX2 is free\nX10 is free\nlen(T11, T2) !=? len([], X2)", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: len(T19, T13)\n\nT19 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", 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) :- ','(!_1, eq(X2, 0)).\nand substitutionT1 -> [],\nT2 -> T5,\nX2 -> T5,\nT4 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 17 [arrowhead = none ];
17 -> 3;

18 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
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(X8, s(X9)) :- ','(tail(X8, X10), len(X10, X9)).\nand substitutionT1 -> T11,\nX8 -> T11,\nX9 -> T13,\nT2 -> s(T13),\nT12 -> T13", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 20 [arrowhead = none ];
20 -> 10;

21 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
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(X5, X5).\nand substitutionT5 -> 0,\nX5 -> 0,\nT8 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 23 [arrowhead = none ];
23 -> 7;

24 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
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(T11, T2), len([], X2))", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 27 [arrowhead = none ];
27 -> 13;

28 [label="EVAL with clause\ntail(.(X15, X16), X16).\nand substitutionX15 -> T18,\nX16 -> T19,\nT11 -> .(T18, T19),\nX10 -> T19", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 28 [arrowhead = none ];
28 -> 14;

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

30 [label="INSTANCE with matching:\nT1 -> T19\nT2 -> T13", fontsize=14, style = filled, fillcolor = lightgrey];
14 -> 30 [arrowhead = none , style=dashed];
30 -> 1[style=dashed];

}

(2) Obligation:

Triples:

lenA(.(X1, X2), s(X3)) :- lenA(X2, X3).

Clauses:

lencA([], 0).
lencA(.(X1, X2), s(X3)) :- lencA(X2, X3).

Afs:

lenA(x1, x2)  =  lenA(x1)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes:
lenA_in: (b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

LENA_IN_GA(.(X1, X2), s(X3)) → U1_GA(X1, X2, X3, lenA_in_ga(X2, X3))
LENA_IN_GA(.(X1, X2), s(X3)) → LENA_IN_GA(X2, X3)

R is empty.
The argument filtering Pi contains the following mapping:
lenA_in_ga(x1, x2)  =  lenA_in_ga(x1)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
LENA_IN_GA(x1, x2)  =  LENA_IN_GA(x1)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x1, x2, x4)

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

Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES

(4) Obligation:

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

LENA_IN_GA(.(X1, X2), s(X3)) → U1_GA(X1, X2, X3, lenA_in_ga(X2, X3))
LENA_IN_GA(.(X1, X2), s(X3)) → LENA_IN_GA(X2, X3)

R is empty.
The argument filtering Pi contains the following mapping:
lenA_in_ga(x1, x2)  =  lenA_in_ga(x1)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
LENA_IN_GA(x1, x2)  =  LENA_IN_GA(x1)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x1, x2, x4)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.

(6) Obligation:

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

LENA_IN_GA(.(X1, X2), s(X3)) → LENA_IN_GA(X2, X3)

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

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

(7) PiDPToQDPProof (SOUND transformation)

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

(8) Obligation:

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

LENA_IN_GA(.(X1, X2)) → LENA_IN_GA(X2)

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

(9) 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:

• LENA_IN_GA(.(X1, X2)) → LENA_IN_GA(X2)
  The graph contains the following edges 1 > 1

(10) YES