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) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
3 [label="A: len(T1, T2)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
6 [label="6: len(T1, T2), SCOPE: 1, CLAUSE: 0 | len(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(!_1, eq(T5, 0)) | len([], T2), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: 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];
13 [label="13: eq(T5, 0)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: eq(T5, 0), SCOPE: 2, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: ','(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];
35 [label="35: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="36: ','(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];
39 [label="39: ','(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];
42 [label="42: len(T19, T13)\n\nT19 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
43 [label="43: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
44 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 44 [arrowhead = none ];
44 -> 6;

45 [label="EVAL with clause\nlen([], X2) :- ','(!_1, eq(X2, 0)).\nand substitutionT1 -> [],\nT2 -> T5,\nX2 -> T5,\nT4 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 45 [arrowhead = none ];
45 -> 9;

46 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 46 [arrowhead = none ];
46 -> 12;

47 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 47 [arrowhead = none ];
47 -> 13;

48 [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];
12 -> 48 [arrowhead = none ];
48 -> 34;

49 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 49 [arrowhead = none ];
49 -> 35;

50 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
13 -> 50 [arrowhead = none ];
50 -> 17;

51 [label="EVAL with clause\neq(X5, X5).\nand substitutionT5 -> 0,\nX5 -> 0,\nT8 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 51 [arrowhead = none ];
51 -> 20;

52 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 52 [arrowhead = none ];
52 -> 25;

53 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
20 -> 53 [arrowhead = none ];
53 -> 27;

54 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
34 -> 54 [arrowhead = none ];
54 -> 36;

55 [label="BACKTRACK\nfor clause: tail([], [])\nwith clash: (len(T11, T2), len([], X2))", fontsize=14, style = filled, fillcolor = lightgreen];
36 -> 55 [arrowhead = none ];
55 -> 39;

56 [label="EVAL with clause\ntail(.(X15, X16), X16).\nand substitutionX15 -> T18,\nX16 -> T19,\nT11 -> .(T18, T19),\nX10 -> T19", fontsize=14, style = filled, fillcolor = lightblue];
39 -> 56 [arrowhead = none ];
56 -> 42;

57 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
39 -> 57 [arrowhead = none ];
57 -> 43;

58 [label="INSTANCE with matching:\nT1 -> T19\nT2 -> T13", fontsize=14, style = filled, fillcolor = lightgrey];
42 -> 58 [arrowhead = none , style=dashed];
58 -> 3[style=dashed];

}

(2) Obligation:

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

lenA_in_ga([], 0) → lenA_out_ga([], 0)
lenA_in_ga(.(T18, T19), s(T13)) → U1_ga(T18, T19, T13, lenA_in_ga(T19, T13))
U1_ga(T18, T19, T13, lenA_out_ga(T19, T13)) → lenA_out_ga(.(T18, T19), s(T13))

The argument filtering Pi contains the following mapping:
lenA_in_ga(x1, x2)  =  lenA_in_ga(x1)
[]  =  []
lenA_out_ga(x1, x2)  =  lenA_out_ga(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x1, x2, x4)
s(x1)  =  s(x1)

(3) 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:

LENA_IN_GA(.(T18, T19), s(T13)) → U1_GA(T18, T19, T13, lenA_in_ga(T19, T13))
LENA_IN_GA(.(T18, T19), s(T13)) → LENA_IN_GA(T19, T13)

The TRS R consists of the following rules:

lenA_in_ga([], 0) → lenA_out_ga([], 0)
lenA_in_ga(.(T18, T19), s(T13)) → U1_ga(T18, T19, T13, lenA_in_ga(T19, T13))
U1_ga(T18, T19, T13, lenA_out_ga(T19, T13)) → lenA_out_ga(.(T18, T19), s(T13))

The argument filtering Pi contains the following mapping:
lenA_in_ga(x1, x2)  =  lenA_in_ga(x1)
[]  =  []
lenA_out_ga(x1, x2)  =  lenA_out_ga(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x1, x2, x4)
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

(4) Obligation:

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

LENA_IN_GA(.(T18, T19), s(T13)) → U1_GA(T18, T19, T13, lenA_in_ga(T19, T13))
LENA_IN_GA(.(T18, T19), s(T13)) → LENA_IN_GA(T19, T13)

The TRS R consists of the following rules:

lenA_in_ga([], 0) → lenA_out_ga([], 0)
lenA_in_ga(.(T18, T19), s(T13)) → U1_ga(T18, T19, T13, lenA_in_ga(T19, T13))
U1_ga(T18, T19, T13, lenA_out_ga(T19, T13)) → lenA_out_ga(.(T18, T19), s(T13))

The argument filtering Pi contains the following mapping:
lenA_in_ga(x1, x2)  =  lenA_in_ga(x1)
[]  =  []
lenA_out_ga(x1, x2)  =  lenA_out_ga(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x1, x2, x4)
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(.(T18, T19), s(T13)) → LENA_IN_GA(T19, T13)

The TRS R consists of the following rules:

lenA_in_ga([], 0) → lenA_out_ga([], 0)
lenA_in_ga(.(T18, T19), s(T13)) → U1_ga(T18, T19, T13, lenA_in_ga(T19, T13))
U1_ga(T18, T19, T13, lenA_out_ga(T19, T13)) → lenA_out_ga(.(T18, T19), s(T13))

The argument filtering Pi contains the following mapping:
lenA_in_ga(x1, x2)  =  lenA_in_ga(x1)
[]  =  []
lenA_out_ga(x1, x2)  =  lenA_out_ga(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x1, x2, x4)
s(x1)  =  s(x1)
LENA_IN_GA(x1, x2)  =  LENA_IN_GA(x1)

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

(7) UsableRulesProof (EQUIVALENT transformation)

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

(8) Obligation:

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

LENA_IN_GA(.(T18, T19), s(T13)) → LENA_IN_GA(T19, T13)

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

(9) PiDPToQDPProof (SOUND transformation)

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

(10) Obligation:

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

LENA_IN_GA(.(T18, T19)) → LENA_IN_GA(T19)

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

(11) 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(.(T18, T19)) → LENA_IN_GA(T19)
  The graph contains the following edges 1 > 1

(12) YES