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

(0) Obligation:

Clauses:

append(.(H, X), Y, .(X, Z)) :- append(X, Y, Z).
append([], Y, Y).

Query: append(g,g,a)

(1) LPReorderTransformerProof (EQUIVALENT transformation)

Reordered facts before rules in definite LP [PROLOG].

(2) Obligation:

Clauses:

append([], Y, Y).
append(.(H, X), Y, .(X, Z)) :- append(X, Y, Z).

Query: append(g,g,a)

(3) 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: append(T1, T2, T3)\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
3 [label="3: append(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | append(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: true | append([], T6, T3), SCOPE: 1, CLAUSE: 1\n\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: append(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT2 is ground\nX3 is free\nappend(T1, T2, T3) !=? append([], X3, X3)", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: append([], T6, T3), SCOPE: 1, CLAUSE: 1\n\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: append(T16, T17, T19)\n\nT16 is ground\nT17 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
2 -> 22 [arrowhead = none ];
22 -> 3;

23 [label="EVAL with clause\nappend([], X3, X3).\nand substitutionT1 -> [],\nT2 -> T6,\nX3 -> T6,\nT3 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 23 [arrowhead = none ];
23 -> 10;

24 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 24 [arrowhead = none ];
24 -> 13;

25 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 25 [arrowhead = none ];
25 -> 14;

26 [label="EVAL with clause\nappend(.(X16, X17), X18, .(X17, X19)) :- append(X17, X18, X19).\nand substitutionX16 -> T15,\nX17 -> T16,\nT1 -> .(T15, T16),\nT2 -> T17,\nX18 -> T17,\nX19 -> T19,\nT3 -> .(T16, T19),\nT18 -> T19", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 26 [arrowhead = none ];
26 -> 20;

27 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 27 [arrowhead = none ];
27 -> 21;

28 [label="BACKTRACK\nfor clause: append(.(H, X), Y, .(X, Z)) :- append(X, Y, Z)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 28 [arrowhead = none ];
28 -> 15;

29 [label="INSTANCE with matching:\nT1 -> T16\nT2 -> T17\nT3 -> T19", fontsize=14, style = filled, fillcolor = lightgrey];
20 -> 29 [arrowhead = none , style=dashed];
29 -> 2[style=dashed];

}

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

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

S tuples:

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

K tuples:none
Defined Rule Symbols:

f2_in, U1

Defined Pair Symbols:

F2_IN

Compound Symbols:

c1

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

Removed 1 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

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

S tuples:

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

K tuples:none
Defined Rule Symbols:

f2_in, U1

Defined Pair Symbols:

F2_IN

Compound Symbols:

c1

(7) 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, z2))

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

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

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

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

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

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

S tuples:none
K tuples:

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

Defined Rule Symbols:

f2_in, U1

Defined Pair Symbols:

F2_IN

Compound Symbols:

c1

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(10) BOUNDS(O(1), O(1))

(11) 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: append(T1, T2, T3)\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
4 [label="4: append(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | append(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: true | append([], T5, T3), SCOPE: 1, CLAUSE: 1\n\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: append(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT2 is ground\nX2 is free\nappend(T1, T2, T3) !=? append([], X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: append([], T5, T3), SCOPE: 1, CLAUSE: 1\n\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: append(T11, T12, T14)\n\nT11 is ground\nT12 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: append(T11, T12, T14), SCOPE: 2, CLAUSE: 0 | append(T11, T12, T14), SCOPE: 2, CLAUSE: 1\n\nT11 is ground\nT12 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: true | append([], T17, T14), SCOPE: 2, CLAUSE: 1\n\nT17 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: append(T11, T12, T14), SCOPE: 2, CLAUSE: 1\n\nT11 is ground\nT12 is ground\nX17 is free\nappend(T11, T12, T14) !=? append([], X17, X17)", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: append([], T17, T14), SCOPE: 2, CLAUSE: 1\n\nT17 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: append(T27, T28, T30)\n\nT27 is ground\nT28 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\nappend([], X2, X2).\nand substitutionT1 -> [],\nT2 -> T5,\nX2 -> T5,\nT3 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 25 [arrowhead = none ];
25 -> 5;

26 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 26 [arrowhead = none ];
26 -> 6;

27 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 27 [arrowhead = none ];
27 -> 7;

28 [label="EVAL with clause\nappend(.(X11, X12), X13, .(X12, X14)) :- append(X12, X13, X14).\nand substitutionX11 -> T10,\nX12 -> T11,\nT1 -> .(T10, T11),\nT2 -> T12,\nX13 -> T12,\nX14 -> T14,\nT3 -> .(T11, T14),\nT13 -> T14", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 28 [arrowhead = none ];
28 -> 9;

29 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 29 [arrowhead = none ];
29 -> 11;

30 [label="BACKTRACK\nfor clause: append(.(H, X), Y, .(X, Z)) :- append(X, Y, Z)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 30 [arrowhead = none ];
30 -> 8;

31 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
9 -> 31 [arrowhead = none ];
31 -> 12;

32 [label="EVAL with clause\nappend([], X17, X17).\nand substitutionT11 -> [],\nT12 -> T17,\nX17 -> T17,\nT14 -> T17", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 32 [arrowhead = none ];
32 -> 16;

33 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 33 [arrowhead = none ];
33 -> 17;

34 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 34 [arrowhead = none ];
34 -> 18;

35 [label="EVAL with clause\nappend(.(X30, X31), X32, .(X31, X33)) :- append(X31, X32, X33).\nand substitutionX30 -> T26,\nX31 -> T27,\nT11 -> .(T26, T27),\nT12 -> T28,\nX32 -> T28,\nX33 -> T30,\nT14 -> .(T27, T30),\nT29 -> T30", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 35 [arrowhead = none ];
35 -> 22;

36 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 36 [arrowhead = none ];
36 -> 23;

37 [label="BACKTRACK\nfor clause: append(.(H, X), Y, .(X, Z)) :- append(X, Y, Z)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
18 -> 37 [arrowhead = none ];
37 -> 19;

38 [label="INSTANCE with matching:\nT1 -> T27\nT2 -> T28\nT3 -> T30", fontsize=14, style = filled, fillcolor = lightgrey];
22 -> 38 [arrowhead = none , style=dashed];
38 -> 1[style=dashed];

}

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

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

S tuples:

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

K tuples:none
Defined Rule Symbols:

f1_in, U1

Defined Pair Symbols:

F1_IN

Compound Symbols:

c2

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

Removed 1 trailing tuple parts

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

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

S tuples:

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

K tuples:none
Defined Rule Symbols:

f1_in, U1

Defined Pair Symbols:

F1_IN

Compound Symbols:

c2

(15) 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, z3))

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

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

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

POL(.(x₁, x₂)) = [1] + x₂   
POL(F1_IN(x₁, x₂)) = [2]x₁   
POL(c2(x₁)) = x₁   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

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

S tuples:none
K tuples:

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

Defined Rule Symbols:

f1_in, U1

Defined Pair Symbols:

F1_IN

Compound Symbols:

c2