Runtime Complexity of the query pattern
len(g,a)
w.r.t. the given Prolog program could be shown to be BOUNDS(O(1), O(n^1)).

(0) Obligation:

Clauses:

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

Query: len(g,a)

(1) PrologToCdtProblemTransformerProof (UPPER BOUND (ID) transformation)

Built complexity over-approximating cdt problems from derivation 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];
4 [label="4: len(T1, T2), SCOPE: 1, CLAUSE: 0 | len(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: true | len([], T2), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: len(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nlen(T1, T2) !=? len([], 0)", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: len([], T2), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: len(T10, T12)\n\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 15 [arrowhead = none ];
15 -> 4;

16 [label="EVAL with clause\nlen([], 0).\nand substitutionT1 -> [],\nT2 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 16 [arrowhead = none ];
16 -> 5;

17 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 17 [arrowhead = none ];
17 -> 7;

18 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 18 [arrowhead = none ];
18 -> 10;

19 [label="EVAL with clause\nlen(.(X11, X12), s(X13)) :- len(X12, X13).\nand substitutionX11 -> T9,\nX12 -> T10,\nT1 -> .(T9, T10),\nX13 -> T12,\nT2 -> s(T12),\nT11 -> T12", fontsize=14, style = filled, fillcolor = lightblue];
7 -> 19 [arrowhead = none ];
19 -> 13;

20 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 20 [arrowhead = none ];
20 -> 14;

21 [label="BACKTRACK\nfor clause: len(.(X1, Ts), s(N)) :- len(Ts, N)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 21 [arrowhead = none ];
21 -> 11;

22 [label="INSTANCE with matching:\nT1 -> T10\nT2 -> T12", fontsize=14, style = filled, fillcolor = lightgrey];
13 -> 22 [arrowhead = none , style=dashed];
22 -> 1[style=dashed];

}

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f1_in([]) → f1_out1(0)
f1_in(.(z0, z1)) → U1(f1_in(z1), .(z0, z1))
U1(f1_out1(z0), .(z1, z2)) → f1_out1(s(z0))

Tuples:

F1_IN(.(z0, z1)) → c1(U1'(f1_in(z1), .(z0, z1)), F1_IN(z1))

S tuples:

F1_IN(.(z0, z1)) → c1(U1'(f1_in(z1), .(z0, z1)), F1_IN(z1))

K tuples:none
Defined Rule Symbols:

f1_in, U1

Defined Pair Symbols:

F1_IN

Compound Symbols:

c1

(3) CdtGraphRemoveTrailingTuplepartsProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f1_in([]) → f1_out1(0)
f1_in(.(z0, z1)) → U1(f1_in(z1), .(z0, z1))
U1(f1_out1(z0), .(z1, z2)) → f1_out1(s(z0))

Tuples:

F1_IN(.(z0, z1)) → c1(F1_IN(z1))

S tuples:

F1_IN(.(z0, z1)) → c1(F1_IN(z1))

K tuples:none
Defined Rule Symbols:

f1_in, U1

Defined Pair Symbols:

F1_IN

Compound Symbols:

c1

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

F1_IN(.(z0, z1)) → c1(F1_IN(z1))

We considered the (Usable) Rules:none
And the Tuples:

F1_IN(.(z0, z1)) → c1(F1_IN(z1))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(.(x₁, x₂)) = [1] + x₂   
POL(F1_IN(x₁)) = x₁   
POL(c1(x₁)) = x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f1_in([]) → f1_out1(0)
f1_in(.(z0, z1)) → U1(f1_in(z1), .(z0, z1))
U1(f1_out1(z0), .(z1, z2)) → f1_out1(s(z0))

Tuples:

F1_IN(.(z0, z1)) → c1(F1_IN(z1))

S tuples:none
K tuples:

F1_IN(.(z0, z1)) → c1(F1_IN(z1))

Defined Rule Symbols:

f1_in, U1

Defined Pair Symbols:

F1_IN

Compound Symbols:

c1

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(8) BOUNDS(O(1), O(1))

(9) PrologToCdtProblemTransformerProof (UPPER BOUND (ID) transformation)

Built complexity over-approximating cdt problems from derivation graph.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
2 [label="2: len(T1, T2)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
3 [label="3: len(T1, T2), SCOPE: 1, CLAUSE: 0 | len(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: true | len([], T2), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: len(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nlen(T1, T2) !=? len([], 0)", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: len([], T2), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: len(T7, T9)\n\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: len(T7, T9), SCOPE: 2, CLAUSE: 0 | len(T7, T9), SCOPE: 2, CLAUSE: 1\n\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: true | len([], T9), SCOPE: 2, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: len(T7, T9), SCOPE: 2, CLAUSE: 1\n\nT7 is ground\nlen(T7, T9) !=? len([], 0)", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: len([], T9), SCOPE: 2, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: len(T17, T19)\n\nT17 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
2 -> 24 [arrowhead = none ];
24 -> 3;

25 [label="EVAL with clause\nlen([], 0).\nand substitutionT1 -> [],\nT2 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 25 [arrowhead = none ];
25 -> 6;

26 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 26 [arrowhead = none ];
26 -> 8;

27 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 27 [arrowhead = none ];
27 -> 9;

28 [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];
8 -> 28 [arrowhead = none ];
28 -> 15;

29 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 29 [arrowhead = none ];
29 -> 16;

30 [label="BACKTRACK\nfor clause: len(.(X1, Ts), s(N)) :- len(Ts, N)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 30 [arrowhead = none ];
30 -> 12;

31 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
15 -> 31 [arrowhead = none ];
31 -> 17;

32 [label="EVAL with clause\nlen([], 0).\nand substitutionT7 -> [],\nT9 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 32 [arrowhead = none ];
32 -> 18;

33 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 33 [arrowhead = none ];
33 -> 19;

34 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
18 -> 34 [arrowhead = none ];
34 -> 20;

35 [label="EVAL with clause\nlen(.(X20, X21), s(X22)) :- len(X21, X22).\nand substitutionX20 -> T16,\nX21 -> T17,\nT7 -> .(T16, T17),\nX22 -> T19,\nT9 -> s(T19),\nT18 -> T19", fontsize=14, style = filled, fillcolor = lightblue];
19 -> 35 [arrowhead = none ];
35 -> 22;

36 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
19 -> 36 [arrowhead = none ];
36 -> 23;

37 [label="BACKTRACK\nfor clause: len(.(X1, Ts), s(N)) :- len(Ts, N)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
20 -> 37 [arrowhead = none ];
37 -> 21;

38 [label="INSTANCE with matching:\nT1 -> T17\nT2 -> T19", fontsize=14, style = filled, fillcolor = lightgrey];
22 -> 38 [arrowhead = none , style=dashed];
38 -> 2[style=dashed];

}

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f2_in([]) → f2_out1(0)
f2_in(.(z0, [])) → f2_out1(s(0))
f2_in(.(z0, .(z1, z2))) → U1(f2_in(z2), .(z0, .(z1, z2)))
U1(f2_out1(z0), .(z1, .(z2, z3))) → f2_out1(s(s(z0)))

Tuples:

F2_IN(.(z0, .(z1, z2))) → c2(U1'(f2_in(z2), .(z0, .(z1, z2))), F2_IN(z2))

S tuples:

F2_IN(.(z0, .(z1, z2))) → c2(U1'(f2_in(z2), .(z0, .(z1, z2))), F2_IN(z2))

K tuples:none
Defined Rule Symbols:

f2_in, U1

Defined Pair Symbols:

F2_IN

Compound Symbols:

c2

(11) CdtGraphRemoveTrailingTuplepartsProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

f2_in([]) → f2_out1(0)
f2_in(.(z0, [])) → f2_out1(s(0))
f2_in(.(z0, .(z1, z2))) → U1(f2_in(z2), .(z0, .(z1, z2)))
U1(f2_out1(z0), .(z1, .(z2, z3))) → f2_out1(s(s(z0)))

Tuples:

F2_IN(.(z0, .(z1, z2))) → c2(F2_IN(z2))

S tuples:

F2_IN(.(z0, .(z1, z2))) → c2(F2_IN(z2))

K tuples:none
Defined Rule Symbols:

f2_in, U1

Defined Pair Symbols:

F2_IN

Compound Symbols:

c2

(13) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

F2_IN(.(z0, .(z1, z2))) → c2(F2_IN(z2))

We considered the (Usable) Rules:none
And the Tuples:

F2_IN(.(z0, .(z1, z2))) → c2(F2_IN(z2))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(.(x₁, x₂)) = [1] + x₂   
POL(F2_IN(x₁)) = [2]x₁   
POL(c2(x₁)) = x₁   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

f2_in([]) → f2_out1(0)
f2_in(.(z0, [])) → f2_out1(s(0))
f2_in(.(z0, .(z1, z2))) → U1(f2_in(z2), .(z0, .(z1, z2)))
U1(f2_out1(z0), .(z1, .(z2, z3))) → f2_out1(s(s(z0)))

Tuples:

F2_IN(.(z0, .(z1, z2))) → c2(F2_IN(z2))

S tuples:none
K tuples:

F2_IN(.(z0, .(z1, z2))) → c2(F2_IN(z2))

Defined Rule Symbols:

f2_in, U1

Defined Pair Symbols:

F2_IN

Compound Symbols:

c2