Runtime Complexity proof of /tmp/tmp0GAxNe/negationasfailure.pl

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

0 Prolog

↳1 PrologToCdtProblemTransformerProof (UPPER BOUND (ID), 53 ms)

↳2 CdtProblem

    ↳3 CdtGraphRemoveTrailingTuplepartsProof (BOTH BOUNDS(ID, ID), 0 ms)

    ↳4 CdtProblem

    ↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 119 ms)

    ↳6 CdtProblem

    ↳7 SIsEmptyProof (⇔, 0 ms)

    ↳8 BOUNDS(O(1), O(1))

↳9 PrologToCdtProblemTransformerProof (UPPER BOUND (ID), 82 ms)

↳10 CdtProblem

    ↳11 CdtGraphRemoveTrailingTuplepartsProof (BOTH BOUNDS(ID, ID), 0 ms)

    ↳12 CdtProblem

    ↳13 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 93 ms)

    ↳14 CdtProblem

(0) Obligation:

Clauses:

q(X) :- ','(not_zero(X), ','(p(X, Y), q(Y))).
p(0, 0).
p(s(X), X).
zero(0).
not_zero(X) :- ','(zero(X), ','(!, failure(a))).
not_zero(X1).
failure(b).

Query: q(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: q(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
4 [label="4: q(T1), SCOPE: 1, CLAUSE: 0\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ','(not_zero(T4), ','(p(T4, X9), q(X9)))\n\nT4 is ground\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(not_zero(T4), ','(p(T4, X9), q(X9))), SCOPE: 2, CLAUSE: 4 | ','(not_zero(T4), ','(p(T4, X9), q(X9))), SCOPE: 2, CLAUSE: 5\n\nT4 is ground\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(','(zero(T7), ','(!_2, failure(a))), ','(p(T7, X9), q(X9))) | ','(not_zero(T7), ','(p(T7, X9), q(X9))), SCOPE: 2, CLAUSE: 5\n\nT7 is ground\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(','(zero(T7), ','(!_2, failure(a))), ','(p(T7, X9), q(X9))), SCOPE: 3, CLAUSE: 3 | ?_3 | ','(not_zero(T7), ','(p(T7, X9), q(X9))), SCOPE: 2, CLAUSE: 5\n\nT7 is ground\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(','(!_2, failure(a)), ','(p(0, X9), q(X9))) | ?_3 | ','(not_zero(0), ','(p(0, X9), q(X9))), SCOPE: 2, CLAUSE: 5\n\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: ?_3 | ','(not_zero(T7), ','(p(T7, X9), q(X9))), SCOPE: 2, CLAUSE: 5\n\nT7 is ground\nX9 is free\nzero(T7) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: ','(failure(a), ','(p(0, X9), q(X9)))\n\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: ','(failure(a), ','(p(0, X9), q(X9))), SCOPE: 4, CLAUSE: 6\n\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: ','(not_zero(T7), ','(p(T7, X9), q(X9))), SCOPE: 2, CLAUSE: 5\n\nT7 is ground\nX9 is free\nzero(T7) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: ','(p(T10, X9), q(X9))\n\nT10 is ground\nX9 is free\nzero(T10) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: ','(p(T10, X9), q(X9)), SCOPE: 5, CLAUSE: 1 | ','(p(T10, X9), q(X9)), SCOPE: 5, CLAUSE: 2\n\nT10 is ground\nX9 is free\nzero(T10) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: ','(p(T10, X9), q(X9)), SCOPE: 5, CLAUSE: 2\n\nT10 is ground\nX9 is free\nzero(T10) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: q(T13)\n\nT13 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
2 -> 35 [arrowhead = none ];
35 -> 4;

36 [label="ONLY EVAL with clause\nq(X8) :- ','(not_zero(X8), ','(p(X8, X9), q(X9))).\nand substitutionT1 -> T4,\nX8 -> T4", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 36 [arrowhead = none ];
36 -> 6;

37 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
6 -> 37 [arrowhead = none ];
37 -> 8;

38 [label="ONLY EVAL with clause\nnot_zero(X15) :- ','(zero(X15), ','(!_2, failure(a))).\nand substitutionT4 -> T7,\nX15 -> T7", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 38 [arrowhead = none ];
38 -> 11;

39 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
11 -> 39 [arrowhead = none ];
39 -> 15;

40 [label="EVAL with clause\nzero(0).\nand substitutionT7 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
15 -> 40 [arrowhead = none ];
40 -> 18;

41 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 41 [arrowhead = none ];
41 -> 20;

42 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
18 -> 42 [arrowhead = none ];
42 -> 22;

43 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
20 -> 43 [arrowhead = none ];
43 -> 27;

44 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
22 -> 44 [arrowhead = none ];
44 -> 25;

45 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 45 [arrowhead = none ];
45 -> 26;

46 [label="ONLY EVAL with clause\nnot_zero(X24).\nand substitutionT7 -> T10,\nX24 -> T10", fontsize=14, style = filled, fillcolor = lightblue];
27 -> 46 [arrowhead = none ];
46 -> 30;

47 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
30 -> 47 [arrowhead = none ];
47 -> 31;

48 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (zero(T10), zero(0))", fontsize=14, style = filled, fillcolor = lightgreen];
31 -> 48 [arrowhead = none ];
48 -> 32;

49 [label="EVAL with clause\np(s(X30), X30).\nand substitutionX30 -> T13,\nT10 -> s(T13),\nX9 -> T13", fontsize=14, style = filled, fillcolor = lightblue];
32 -> 49 [arrowhead = none ];
49 -> 33;

50 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
32 -> 50 [arrowhead = none ];
50 -> 34;

51 [label="INSTANCE with matching:\nT1 -> T13", fontsize=14, style = filled, fillcolor = lightgrey];
33 -> 51 [arrowhead = none , style=dashed];
51 -> 2[style=dashed];

}

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f2_in(s(z0)) → U1(f2_in(z0), s(z0))
U1(f2_out1, s(z0)) → f2_out1

Tuples:

F2_IN(s(z0)) → c(U1'(f2_in(z0), s(z0)), F2_IN(z0))

S tuples:

F2_IN(s(z0)) → c(U1'(f2_in(z0), s(z0)), F2_IN(z0))

K tuples:none
Defined Rule Symbols:

f2_in, U1

Defined Pair Symbols:

F2_IN

Compound Symbols:

c

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

Removed 1 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f2_in(s(z0)) → U1(f2_in(z0), s(z0))
U1(f2_out1, s(z0)) → f2_out1

Tuples:

F2_IN(s(z0)) → c(F2_IN(z0))

S tuples:

F2_IN(s(z0)) → c(F2_IN(z0))

K tuples:none
Defined Rule Symbols:

f2_in, U1

Defined Pair Symbols:

F2_IN

Compound Symbols:

c

(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(s(z0)) → c(F2_IN(z0))

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

F2_IN(s(z0)) → c(F2_IN(z0))

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

POL(F2_IN(x₁)) = x₁
POL(c(x₁)) = x₁
POL(s(x₁)) = [1] + x₁

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f2_in(s(z0)) → U1(f2_in(z0), s(z0))
U1(f2_out1, s(z0)) → f2_out1

Tuples:

F2_IN(s(z0)) → c(F2_IN(z0))

S tuples:none
K tuples:

F2_IN(s(z0)) → c(F2_IN(z0))

Defined Rule Symbols:

f2_in, U1

Defined Pair Symbols:

F2_IN

Compound Symbols:

c

(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: q(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
3 [label="3: q(T1), SCOPE: 1, CLAUSE: 0\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ','(not_zero(T3), ','(p(T3, X4), q(X4)))\n\nT3 is ground\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(not_zero(T3), ','(p(T3, X4), q(X4))), SCOPE: 2, CLAUSE: 4 | ','(not_zero(T3), ','(p(T3, X4), q(X4))), SCOPE: 2, CLAUSE: 5\n\nT3 is ground\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(','(zero(T6), ','(!_2, failure(a))), ','(p(T6, X4), q(X4))) | ','(not_zero(T6), ','(p(T6, X4), q(X4))), SCOPE: 2, CLAUSE: 5\n\nT6 is ground\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(','(zero(T6), ','(!_2, failure(a))), ','(p(T6, X4), q(X4))), SCOPE: 3, CLAUSE: 3 | ?_3 | ','(not_zero(T6), ','(p(T6, X4), q(X4))), SCOPE: 2, CLAUSE: 5\n\nT6 is ground\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(','(!_2, failure(a)), ','(p(0, X4), q(X4))) | ?_3 | ','(not_zero(0), ','(p(0, X4), q(X4))), SCOPE: 2, CLAUSE: 5\n\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ?_3 | ','(not_zero(T6), ','(p(T6, X4), q(X4))), SCOPE: 2, CLAUSE: 5\n\nT6 is ground\nX4 is free\nzero(T6) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(failure(a), ','(p(0, X4), q(X4)))\n\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(failure(a), ','(p(0, X4), q(X4))), SCOPE: 4, CLAUSE: 6\n\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: ','(not_zero(T6), ','(p(T6, X4), q(X4))), SCOPE: 2, CLAUSE: 5\n\nT6 is ground\nX4 is free\nzero(T6) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: ','(p(T9, X4), q(X4))\n\nT9 is ground\nX4 is free\nzero(T9) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: ','(p(T9, X4), q(X4)), SCOPE: 5, CLAUSE: 1 | ','(p(T9, X4), q(X4)), SCOPE: 5, CLAUSE: 2\n\nT9 is ground\nX4 is free\nzero(T9) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: ','(p(T9, X4), q(X4)), SCOPE: 5, CLAUSE: 2\n\nT9 is ground\nX4 is free\nzero(T9) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: q(T12)\n\nT12 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 30 [arrowhead = none ];
30 -> 3;

31 [label="ONLY EVAL with clause\nq(X3) :- ','(not_zero(X3), ','(p(X3, X4), q(X4))).\nand substitutionT1 -> T3,\nX3 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 31 [arrowhead = none ];
31 -> 5;

32 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
5 -> 32 [arrowhead = none ];
32 -> 7;

33 [label="ONLY EVAL with clause\nnot_zero(X7) :- ','(zero(X7), ','(!_2, failure(a))).\nand substitutionT3 -> T6,\nX7 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
7 -> 33 [arrowhead = none ];
33 -> 9;

34 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
9 -> 34 [arrowhead = none ];
34 -> 10;

35 [label="EVAL with clause\nzero(0).\nand substitutionT6 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 35 [arrowhead = none ];
35 -> 12;

36 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 36 [arrowhead = none ];
36 -> 13;

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

38 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 38 [arrowhead = none ];
38 -> 19;

39 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
14 -> 39 [arrowhead = none ];
39 -> 16;

40 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 40 [arrowhead = none ];
40 -> 17;

41 [label="ONLY EVAL with clause\nnot_zero(X16).\nand substitutionT6 -> T9,\nX16 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
19 -> 41 [arrowhead = none ];
41 -> 21;

42 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
21 -> 42 [arrowhead = none ];
42 -> 23;

43 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (zero(T9), zero(0))", fontsize=14, style = filled, fillcolor = lightgreen];
23 -> 43 [arrowhead = none ];
43 -> 24;

44 [label="EVAL with clause\np(s(X22), X22).\nand substitutionX22 -> T12,\nT9 -> s(T12),\nX4 -> T12", fontsize=14, style = filled, fillcolor = lightblue];
24 -> 44 [arrowhead = none ];
44 -> 28;

45 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
24 -> 45 [arrowhead = none ];
45 -> 29;

46 [label="INSTANCE with matching:\nT1 -> T12", fontsize=14, style = filled, fillcolor = lightgrey];
28 -> 46 [arrowhead = none , style=dashed];
46 -> 1[style=dashed];

}

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f1_in(s(z0)) → U1(f1_in(z0), s(z0))
U1(f1_out1, s(z0)) → f1_out1

Tuples:

F1_IN(s(z0)) → c(U1'(f1_in(z0), s(z0)), F1_IN(z0))

S tuples:

F1_IN(s(z0)) → c(U1'(f1_in(z0), s(z0)), F1_IN(z0))

K tuples:none
Defined Rule Symbols:

f1_in, U1

Defined Pair Symbols:

F1_IN

Compound Symbols:

c

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

Removed 1 trailing tuple parts

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

f1_in(s(z0)) → U1(f1_in(z0), s(z0))
U1(f1_out1, s(z0)) → f1_out1

Tuples:

F1_IN(s(z0)) → c(F1_IN(z0))

S tuples:

F1_IN(s(z0)) → c(F1_IN(z0))

K tuples:none
Defined Rule Symbols:

f1_in, U1

Defined Pair Symbols:

F1_IN

Compound Symbols:

c

(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(s(z0)) → c(F1_IN(z0))

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

F1_IN(s(z0)) → c(F1_IN(z0))

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

POL(F1_IN(x₁)) = x₁
POL(c(x₁)) = x₁
POL(s(x₁)) = [1] + x₁

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

f1_in(s(z0)) → U1(f1_in(z0), s(z0))
U1(f1_out1, s(z0)) → f1_out1

Tuples:

F1_IN(s(z0)) → c(F1_IN(z0))

S tuples:none
K tuples:

F1_IN(s(z0)) → c(F1_IN(z0))

Defined Rule Symbols:

f1_in, U1

Defined Pair Symbols:

F1_IN

Compound Symbols:

c