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([], 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) 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];
12 [label="12: ','(!_1, eq(T6, 0)) | len([], T2), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: len(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nX3 is free\nlen(T1, T2) !=? len([], X3)", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: eq(T6, 0)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: eq(T6, 0), SCOPE: 2, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: ','(tail(T14, X14), len(X14, T16))\n\nT14 is ground\nX3 is free\nX14 is free\nlen(T14, T2) !=? len([], X3)", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: ','(tail(T14, X14), len(X14, T16)), SCOPE: 3, CLAUSE: 2 | ','(tail(T14, X14), len(X14, T16)), SCOPE: 3, CLAUSE: 3\n\nT14 is ground\nX3 is free\nX14 is free\nlen(T14, T2) !=? len([], X3)", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: ','(tail(T14, X14), len(X14, T16)), SCOPE: 3, CLAUSE: 3\n\nT14 is ground\nX3 is free\nX14 is free\nlen(T14, T2) !=? len([], X3)", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: len(T23, T16)\n\nT23 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
2 -> 31 [arrowhead = none ];
31 -> 3;

32 [label="EVAL with clause\nlen([], X3) :- ','(!_1, eq(X3, 0)).\nand substitutionT1 -> [],\nT2 -> T6,\nX3 -> T6,\nT5 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 32 [arrowhead = none ];
32 -> 12;

33 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 33 [arrowhead = none ];
33 -> 13;

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

35 [label="EVAL with clause\nlen(X12, s(X13)) :- ','(tail(X12, X14), len(X14, X13)).\nand substitutionT1 -> T14,\nX12 -> T14,\nX13 -> T16,\nT2 -> s(T16),\nT15 -> T16", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 35 [arrowhead = none ];
35 -> 25;

36 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 36 [arrowhead = none ];
36 -> 26;

37 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
14 -> 37 [arrowhead = none ];
37 -> 15;

38 [label="EVAL with clause\neq(X6, X6).\nand substitutionT6 -> 0,\nX6 -> 0,\nT9 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
15 -> 38 [arrowhead = none ];
38 -> 18;

39 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 39 [arrowhead = none ];
39 -> 19;

40 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
18 -> 40 [arrowhead = none ];
40 -> 21;

41 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
25 -> 41 [arrowhead = none ];
41 -> 27;

42 [label="BACKTRACK\nfor clause: tail([], [])\nwith clash: (len(T14, T2), len([], X3))", fontsize=14, style = filled, fillcolor = lightgreen];
27 -> 42 [arrowhead = none ];
42 -> 28;

43 [label="EVAL with clause\ntail(.(X20, X21), X21).\nand substitutionX20 -> T22,\nX21 -> T23,\nT14 -> .(T22, T23),\nX14 -> T23", fontsize=14, style = filled, fillcolor = lightblue];
28 -> 43 [arrowhead = none ];
43 -> 29;

44 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
28 -> 44 [arrowhead = none ];
44 -> 30;

45 [label="INSTANCE with matching:\nT1 -> T23\nT2 -> T16", fontsize=14, style = filled, fillcolor = lightgrey];
29 -> 45 [arrowhead = none , style=dashed];
45 -> 2[style=dashed];

}

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F2_IN(.(z0, z1)) → c1(U1'(f2_in(z1), .(z0, z1)), F2_IN(z1))

S tuples:

F2_IN(.(z0, z1)) → c1(U1'(f2_in(z1), .(z0, z1)), F2_IN(z1))

K tuples:none
Defined Rule Symbols:

f2_in, U1

Defined Pair Symbols:

F2_IN

Compound Symbols:

c1

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

Removed 1 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F2_IN(.(z0, z1)) → c1(F2_IN(z1))

S tuples:

F2_IN(.(z0, z1)) → c1(F2_IN(z1))

K tuples:none
Defined Rule Symbols:

f2_in, U1

Defined Pair Symbols:

F2_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.

F2_IN(.(z0, z1)) → c1(F2_IN(z1))

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

F2_IN(.(z0, z1)) → c1(F2_IN(z1))

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

POL(.(x₁, x₂)) = [1] + x₂   
POL(F2_IN(x₁)) = x₁   
POL(c1(x₁)) = x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F2_IN(.(z0, z1)) → c1(F2_IN(z1))

S tuples:none
K tuples:

F2_IN(.(z0, z1)) → c1(F2_IN(z1))

Defined Rule Symbols:

f2_in, U1

Defined Pair Symbols:

F2_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];
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: ','(!_1, eq(T5, 0)) | len([], T2), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: 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];
7 [label="7: eq(T5, 0)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: eq(T5, 0), SCOPE: 2, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(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];
17 [label="17: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: ','(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];
22 [label="22: ','(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];
23 [label="23: len(T19, T13)\n\nT19 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 25 [arrowhead = none ];
25 -> 4;

26 [label="EVAL with clause\nlen([], X2) :- ','(!_1, eq(X2, 0)).\nand substitutionT1 -> [],\nT2 -> T5,\nX2 -> T5,\nT4 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 26 [arrowhead = none ];
26 -> 5;

27 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 27 [arrowhead = none ];
27 -> 6;

28 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 28 [arrowhead = none ];
28 -> 7;

29 [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];
6 -> 29 [arrowhead = none ];
29 -> 16;

30 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 30 [arrowhead = none ];
30 -> 17;

31 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 31 [arrowhead = none ];
31 -> 8;

32 [label="EVAL with clause\neq(X5, X5).\nand substitutionT5 -> 0,\nX5 -> 0,\nT8 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 32 [arrowhead = none ];
32 -> 9;

33 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 33 [arrowhead = none ];
33 -> 10;

34 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 34 [arrowhead = none ];
34 -> 11;

35 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
16 -> 35 [arrowhead = none ];
35 -> 20;

36 [label="BACKTRACK\nfor clause: tail([], [])\nwith clash: (len(T11, T2), len([], X2))", fontsize=14, style = filled, fillcolor = lightgreen];
20 -> 36 [arrowhead = none ];
36 -> 22;

37 [label="EVAL with clause\ntail(.(X15, X16), X16).\nand substitutionX15 -> T18,\nX16 -> T19,\nT11 -> .(T18, T19),\nX10 -> T19", fontsize=14, style = filled, fillcolor = lightblue];
22 -> 37 [arrowhead = none ];
37 -> 23;

38 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
22 -> 38 [arrowhead = none ];
38 -> 24;

39 [label="INSTANCE with matching:\nT1 -> T19\nT2 -> T13", fontsize=14, style = filled, fillcolor = lightgrey];
23 -> 39 [arrowhead = none , style=dashed];
39 -> 1[style=dashed];

}

(10) 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

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

Removed 1 trailing tuple parts

(12) 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

(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.

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   

(14) 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