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

(0) Obligation:

Clauses:

len([], 0).
len(.(X1, Ts), s(N)) :- len(Ts, N).

Query: len(g,a)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

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];
5 [label="5: 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: true | 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\nlen(T1, T2) !=? len([], 0)", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: len([], T2), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: len(T7, T9)\n\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: len(T7, T9), SCOPE: 2, CLAUSE: 0 | len(T7, T9), SCOPE: 2, CLAUSE: 1\n\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: len(T7, T9), SCOPE: 2, CLAUSE: 0\n\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: len(T7, T9), SCOPE: 2, CLAUSE: 1\n\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: len(T17, T19)\n\nT17 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 32 [arrowhead = none ];
32 -> 5;

33 [label="EVAL with clause\nlen([], 0).\nand substitutionT1 -> [],\nT2 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 33 [arrowhead = none ];
33 -> 9;

34 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 34 [arrowhead = none ];
34 -> 12;

35 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 35 [arrowhead = none ];
35 -> 17;

36 [label="EVAL with clause\nlen(.(X8, X9), s(X10)) :- len(X9, X10).\nand substitutionX8 -> T6,\nX9 -> T7,\nT1 -> .(T6, T7),\nX10 -> T9,\nT2 -> s(T9),\nT8 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 36 [arrowhead = none ];
36 -> 21;

37 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 37 [arrowhead = none ];
37 -> 22;

38 [label="BACKTRACK\nfor clause: len(.(X1, Ts), s(N)) :- len(Ts, N)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 38 [arrowhead = none ];
38 -> 18;

39 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
21 -> 39 [arrowhead = none ];
39 -> 23;

40 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
23 -> 40 [arrowhead = none ];
40 -> 24;

41 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
23 -> 41 [arrowhead = none ];
41 -> 25;

42 [label="EVAL with clause\nlen([], 0).\nand substitutionT7 -> [],\nT9 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
24 -> 42 [arrowhead = none ];
42 -> 26;

43 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
24 -> 43 [arrowhead = none ];
43 -> 27;

44 [label="EVAL with clause\nlen(.(X17, X18), s(X19)) :- len(X18, X19).\nand substitutionX17 -> T16,\nX18 -> T17,\nT7 -> .(T16, T17),\nX19 -> T19,\nT9 -> s(T19),\nT18 -> T19", fontsize=14, style = filled, fillcolor = lightblue];
25 -> 44 [arrowhead = none ];
44 -> 30;

45 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 45 [arrowhead = none ];
45 -> 31;

46 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
26 -> 46 [arrowhead = none ];
46 -> 28;

47 [label="INSTANCE with matching:\nT1 -> T17\nT2 -> T19", fontsize=14, style = filled, fillcolor = lightgrey];
30 -> 47 [arrowhead = none , style=dashed];
47 -> 1[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(.(T6, []), s(0)) → lenA_out_ga(.(T6, []), s(0))
lenA_in_ga(.(T6, .(T16, T17)), s(s(T19))) → U1_ga(T6, T16, T17, T19, lenA_in_ga(T17, T19))
U1_ga(T6, T16, T17, T19, lenA_out_ga(T17, T19)) → lenA_out_ga(.(T6, .(T16, T17)), s(s(T19)))

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, x5)  =  U1_ga(x1, x2, x3, x5)
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(.(T6, .(T16, T17)), s(s(T19))) → U1_GA(T6, T16, T17, T19, lenA_in_ga(T17, T19))
LENA_IN_GA(.(T6, .(T16, T17)), s(s(T19))) → LENA_IN_GA(T17, T19)

The TRS R consists of the following rules:

lenA_in_ga([], 0) → lenA_out_ga([], 0)
lenA_in_ga(.(T6, []), s(0)) → lenA_out_ga(.(T6, []), s(0))
lenA_in_ga(.(T6, .(T16, T17)), s(s(T19))) → U1_ga(T6, T16, T17, T19, lenA_in_ga(T17, T19))
U1_ga(T6, T16, T17, T19, lenA_out_ga(T17, T19)) → lenA_out_ga(.(T6, .(T16, T17)), s(s(T19)))

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, x5)  =  U1_ga(x1, x2, x3, x5)
s(x1)  =  s(x1)
LENA_IN_GA(x1, x2)  =  LENA_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x1, x2, x3, x5)

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(.(T6, .(T16, T17)), s(s(T19))) → U1_GA(T6, T16, T17, T19, lenA_in_ga(T17, T19))
LENA_IN_GA(.(T6, .(T16, T17)), s(s(T19))) → LENA_IN_GA(T17, T19)

The TRS R consists of the following rules:

lenA_in_ga([], 0) → lenA_out_ga([], 0)
lenA_in_ga(.(T6, []), s(0)) → lenA_out_ga(.(T6, []), s(0))
lenA_in_ga(.(T6, .(T16, T17)), s(s(T19))) → U1_ga(T6, T16, T17, T19, lenA_in_ga(T17, T19))
U1_ga(T6, T16, T17, T19, lenA_out_ga(T17, T19)) → lenA_out_ga(.(T6, .(T16, T17)), s(s(T19)))

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, x5)  =  U1_ga(x1, x2, x3, x5)
s(x1)  =  s(x1)
LENA_IN_GA(x1, x2)  =  LENA_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x1, x2, x3, x5)

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(.(T6, .(T16, T17)), s(s(T19))) → LENA_IN_GA(T17, T19)

The TRS R consists of the following rules:

lenA_in_ga([], 0) → lenA_out_ga([], 0)
lenA_in_ga(.(T6, []), s(0)) → lenA_out_ga(.(T6, []), s(0))
lenA_in_ga(.(T6, .(T16, T17)), s(s(T19))) → U1_ga(T6, T16, T17, T19, lenA_in_ga(T17, T19))
U1_ga(T6, T16, T17, T19, lenA_out_ga(T17, T19)) → lenA_out_ga(.(T6, .(T16, T17)), s(s(T19)))

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, x5)  =  U1_ga(x1, x2, x3, x5)
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(.(T6, .(T16, T17)), s(s(T19))) → LENA_IN_GA(T17, T19)

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(.(T6, .(T16, T17))) → LENA_IN_GA(T17)

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(.(T6, .(T16, T17))) → LENA_IN_GA(T17)
  The graph contains the following edges 1 > 1

(12) YES