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

(0) Obligation:

Clauses:

reverse([], X, X).
reverse(.(X, Y), Z, U) :- reverse(Y, Z, .(X, U)).

Query: reverse(g,a,g)

(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: reverse(T1, T2, T3)\n\nT1 is ground\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
3 [label="3: reverse(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | reverse(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: true | reverse([], T2, T6), SCOPE: 1, CLAUSE: 1\n\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: reverse(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT3 is ground\nX3 is free\nreverse(T1, T2, T3) !=? reverse([], X3, X3)", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: reverse([], T2, T6), SCOPE: 1, CLAUSE: 1\n\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: reverse(T16, T19, .(T15, T18))\n\nT15 is ground\nT16 is ground\nT18 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
2 -> 18 [arrowhead = none ];
18 -> 3;

19 [label="EVAL with clause\nreverse([], X3, X3).\nand substitutionT1 -> [],\nT2 -> T6,\nX3 -> T6,\nT3 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 19 [arrowhead = none ];
19 -> 5;

20 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 20 [arrowhead = none ];
20 -> 9;

21 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 21 [arrowhead = none ];
21 -> 10;

22 [label="EVAL with clause\nreverse(.(X16, X17), X18, X19) :- reverse(X17, X18, .(X16, X19)).\nand substitutionX16 -> T15,\nX17 -> T16,\nT1 -> .(T15, T16),\nT2 -> T19,\nX18 -> T19,\nT3 -> T18,\nX19 -> T18,\nT17 -> T19", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 22 [arrowhead = none ];
22 -> 15;

23 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 23 [arrowhead = none ];
23 -> 17;

24 [label="BACKTRACK\nfor clause: reverse(.(X, Y), Z, U) :- reverse(Y, Z, .(X, U))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 24 [arrowhead = none ];
24 -> 11;

25 [label="INSTANCE with matching:\nT1 -> T16\nT2 -> T19\nT3 -> .(T15, T18)", fontsize=14, style = filled, fillcolor = lightgrey];
15 -> 25 [arrowhead = none , style=dashed];
25 -> 2[style=dashed];

}

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F2_IN(.(z0, z1), z2) → c1(U1'(f2_in(z1, .(z0, z2)), .(z0, z1), z2), F2_IN(z1, .(z0, z2)))

S tuples:

F2_IN(.(z0, z1), z2) → c1(U1'(f2_in(z1, .(z0, z2)), .(z0, z1), z2), F2_IN(z1, .(z0, z2)))

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

Tuples:

F2_IN(.(z0, z1), z2) → c1(F2_IN(z1, .(z0, z2)))

S tuples:

F2_IN(.(z0, z1), z2) → c1(F2_IN(z1, .(z0, z2)))

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), z2) → c1(F2_IN(z1, .(z0, z2)))

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

F2_IN(.(z0, z1), z2) → c1(F2_IN(z1, .(z0, z2)))

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

POL(.(x₁, x₂)) = [3] + x₂   
POL(F2_IN(x₁, x₂)) = [3]x₁ + x₂   
POL(c1(x₁)) = x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F2_IN(.(z0, z1), z2) → c1(F2_IN(z1, .(z0, z2)))

S tuples:none
K tuples:

F2_IN(.(z0, z1), z2) → c1(F2_IN(z1, .(z0, z2)))

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: reverse(T1, T2, T3)\n\nT1 is ground\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
4 [label="4: reverse(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | reverse(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: true | reverse([], T2, T5), SCOPE: 1, CLAUSE: 1\n\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: reverse(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT3 is ground\nX2 is free\nreverse(T1, T2, T3) !=? reverse([], X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: reverse([], T2, T5), SCOPE: 1, CLAUSE: 1\n\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: reverse(T11, T14, .(T10, T13))\n\nT10 is ground\nT11 is ground\nT13 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: reverse(T11, T14, .(T10, T13)), SCOPE: 2, CLAUSE: 0 | reverse(T11, T14, .(T10, T13)), SCOPE: 2, CLAUSE: 1\n\nT10 is ground\nT11 is ground\nT13 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: true | reverse([], T14, .(T22, T23)), SCOPE: 2, CLAUSE: 1\n\nT22 is ground\nT23 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: reverse(T11, T14, .(T10, T13)), SCOPE: 2, CLAUSE: 1\n\nT10 is ground\nT11 is ground\nT13 is ground\nX17 is free\nreverse(T11, T14, .(T10, T13)) !=? reverse([], X17, X17)", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: reverse([], T14, .(T22, T23)), SCOPE: 2, CLAUSE: 1\n\nT22 is ground\nT23 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: reverse(T35, T39, .(T34, .(T37, T38)))\n\nT34 is ground\nT35 is ground\nT37 is ground\nT38 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];
1 -> 24 [arrowhead = none ];
24 -> 4;

25 [label="EVAL with clause\nreverse([], X2, X2).\nand substitutionT1 -> [],\nT2 -> T5,\nX2 -> T5,\nT3 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 25 [arrowhead = none ];
25 -> 6;

26 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 26 [arrowhead = none ];
26 -> 7;

27 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 27 [arrowhead = none ];
27 -> 8;

28 [label="EVAL with clause\nreverse(.(X11, X12), X13, X14) :- reverse(X12, X13, .(X11, X14)).\nand substitutionX11 -> T10,\nX12 -> T11,\nT1 -> .(T10, T11),\nT2 -> T14,\nX13 -> T14,\nT3 -> T13,\nX14 -> T13,\nT12 -> T14", fontsize=14, style = filled, fillcolor = lightblue];
7 -> 28 [arrowhead = none ];
28 -> 13;

29 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 29 [arrowhead = none ];
29 -> 14;

30 [label="BACKTRACK\nfor clause: reverse(.(X, Y), Z, U) :- reverse(Y, Z, .(X, U))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 30 [arrowhead = none ];
30 -> 12;

31 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
13 -> 31 [arrowhead = none ];
31 -> 16;

32 [label="EVAL with clause\nreverse([], X17, X17).\nand substitutionT11 -> [],\nT14 -> .(T22, T23),\nX17 -> .(T22, T23),\nT10 -> T22,\nT13 -> T23,\nT21 -> .(T22, T23)", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 32 [arrowhead = none ];
32 -> 18;

33 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 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\nreverse(.(X30, X31), X32, X33) :- reverse(X31, X32, .(X30, X33)).\nand substitutionX30 -> T34,\nX31 -> T35,\nT11 -> .(T34, T35),\nT14 -> T39,\nX32 -> T39,\nT10 -> T37,\nT13 -> T38,\nX33 -> .(T37, T38),\nT36 -> T39", 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: reverse(.(X, Y), Z, U) :- reverse(Y, Z, .(X, U))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
20 -> 37 [arrowhead = none ];
37 -> 21;

38 [label="INSTANCE with matching:\nT1 -> T35\nT2 -> T39\nT3 -> .(T34, .(T37, T38))", fontsize=14, style = filled, fillcolor = lightgrey];
22 -> 38 [arrowhead = none , style=dashed];
38 -> 1[style=dashed];

}

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f1_in([], z0) → f1_out1(z0)
f1_in(.(z0, []), z1) → f1_out1(.(z0, z1))
f1_in(.(z0, .(z1, z2)), z3) → U1(f1_in(z2, .(z1, .(z0, z3))), .(z0, .(z1, z2)), z3)
U1(f1_out1(z0), .(z1, .(z2, z3)), z4) → f1_out1(z0)

Tuples:

F1_IN(.(z0, .(z1, z2)), z3) → c2(U1'(f1_in(z2, .(z1, .(z0, z3))), .(z0, .(z1, z2)), z3), F1_IN(z2, .(z1, .(z0, z3))))

S tuples:

F1_IN(.(z0, .(z1, z2)), z3) → c2(U1'(f1_in(z2, .(z1, .(z0, z3))), .(z0, .(z1, z2)), z3), F1_IN(z2, .(z1, .(z0, z3))))

K tuples:none
Defined Rule Symbols:

f1_in, U1

Defined Pair Symbols:

F1_IN

Compound Symbols:

c2

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

Removed 1 trailing tuple parts

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

f1_in([], z0) → f1_out1(z0)
f1_in(.(z0, []), z1) → f1_out1(.(z0, z1))
f1_in(.(z0, .(z1, z2)), z3) → U1(f1_in(z2, .(z1, .(z0, z3))), .(z0, .(z1, z2)), z3)
U1(f1_out1(z0), .(z1, .(z2, z3)), z4) → f1_out1(z0)

Tuples:

F1_IN(.(z0, .(z1, z2)), z3) → c2(F1_IN(z2, .(z1, .(z0, z3))))

S tuples:

F1_IN(.(z0, .(z1, z2)), z3) → c2(F1_IN(z2, .(z1, .(z0, z3))))

K tuples:none
Defined Rule Symbols:

f1_in, U1

Defined Pair Symbols:

F1_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.

F1_IN(.(z0, .(z1, z2)), z3) → c2(F1_IN(z2, .(z1, .(z0, z3))))

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

F1_IN(.(z0, .(z1, z2)), z3) → c2(F1_IN(z2, .(z1, .(z0, z3))))

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

POL(.(x₁, x₂)) = [3] + x₂   
POL(F1_IN(x₁, x₂)) = [3]x₁ + [2]x₂   
POL(c2(x₁)) = x₁   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

f1_in([], z0) → f1_out1(z0)
f1_in(.(z0, []), z1) → f1_out1(.(z0, z1))
f1_in(.(z0, .(z1, z2)), z3) → U1(f1_in(z2, .(z1, .(z0, z3))), .(z0, .(z1, z2)), z3)
U1(f1_out1(z0), .(z1, .(z2, z3)), z4) → f1_out1(z0)

Tuples:

F1_IN(.(z0, .(z1, z2)), z3) → c2(F1_IN(z2, .(z1, .(z0, z3))))

S tuples:none
K tuples:

F1_IN(.(z0, .(z1, z2)), z3) → c2(F1_IN(z2, .(z1, .(z0, z3))))

Defined Rule Symbols:

f1_in, U1

Defined Pair Symbols:

F1_IN

Compound Symbols:

c2