YES(O(1), O(n^2)) 5.95/1.93 YES(O(1), O(n^2)) 5.95/1.96 5.95/1.96 5.95/1.96 5.95/1.96 5.95/1.97 5.95/1.97 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 5.95/1.97 5.95/1.97 5.95/1.97
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The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

0 CpxTRS
5.95/1.97 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
5.95/1.97 
↳2 CdtProblem
5.95/1.97 
↳3 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
5.95/1.97 
↳4 CdtProblem
5.95/1.97 
↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
5.95/1.97 
↳6 CdtProblem
5.95/1.97 
↳7 CdtKnowledgeProof (BOTH BOUNDS(ID, ID))
5.95/1.97 
↳8 CdtProblem
5.95/1.97 
↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
5.95/1.97 
↳10 CdtProblem
5.95/1.97 
↳11 SIsEmptyProof (BOTH BOUNDS(ID, ID))
5.95/1.97 
↳12 BOUNDS(O(1), O(1))
5.95/1.97
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5.95/1.97

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

sort(nil) → nil 5.95/1.97
sort(cons(x, y)) → insert(x, sort(y)) 5.95/1.97
insert(x, nil) → cons(x, nil) 5.95/1.97
insert(x, cons(v, w)) → choose(x, cons(v, w), x, v) 5.95/1.97
choose(x, cons(v, w), y, 0) → cons(x, cons(v, w)) 5.95/1.97
choose(x, cons(v, w), 0, s(z)) → cons(v, insert(x, w)) 5.95/1.97
choose(x, cons(v, w), s(y), s(z)) → choose(x, cons(v, w), y, z)

Rewrite Strategy: INNERMOST
5.95/1.97
5.95/1.97

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT
5.95/1.97
5.95/1.97

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

sort(nil) → nil 5.95/1.97
sort(cons(z0, z1)) → insert(z0, sort(z1)) 5.95/1.97
insert(z0, nil) → cons(z0, nil) 5.95/1.97
insert(z0, cons(z1, z2)) → choose(z0, cons(z1, z2), z0, z1) 5.95/1.97
choose(z0, cons(z1, z2), z3, 0) → cons(z0, cons(z1, z2)) 5.95/1.97
choose(z0, cons(z1, z2), 0, s(z3)) → cons(z1, insert(z0, z2)) 5.95/1.97
choose(z0, cons(z1, z2), s(z3), s(z4)) → choose(z0, cons(z1, z2), z3, z4)
Tuples:

SORT(cons(z0, z1)) → c1(INSERT(z0, sort(z1)), SORT(z1)) 5.95/1.97
INSERT(z0, cons(z1, z2)) → c3(CHOOSE(z0, cons(z1, z2), z0, z1)) 5.95/1.97
CHOOSE(z0, cons(z1, z2), 0, s(z3)) → c5(INSERT(z0, z2)) 5.95/1.97
CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) → c6(CHOOSE(z0, cons(z1, z2), z3, z4))
S tuples:

SORT(cons(z0, z1)) → c1(INSERT(z0, sort(z1)), SORT(z1)) 5.95/1.97
INSERT(z0, cons(z1, z2)) → c3(CHOOSE(z0, cons(z1, z2), z0, z1)) 5.95/1.97
CHOOSE(z0, cons(z1, z2), 0, s(z3)) → c5(INSERT(z0, z2)) 5.95/1.97
CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) → c6(CHOOSE(z0, cons(z1, z2), z3, z4))
K tuples:none
Defined Rule Symbols:

sort, insert, choose

Defined Pair Symbols:

SORT, INSERT, CHOOSE

Compound Symbols:

c1, c3, c5, c6

5.95/1.97
5.95/1.97

(3) 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.

SORT(cons(z0, z1)) → c1(INSERT(z0, sort(z1)), SORT(z1))
We considered the (Usable) Rules:

sort(nil) → nil 5.95/1.97
sort(cons(z0, z1)) → insert(z0, sort(z1)) 5.95/1.97
insert(z0, nil) → cons(z0, nil) 5.95/1.97
insert(z0, cons(z1, z2)) → choose(z0, cons(z1, z2), z0, z1) 5.95/1.97
choose(z0, cons(z1, z2), z3, 0) → cons(z0, cons(z1, z2)) 5.95/1.97
choose(z0, cons(z1, z2), 0, s(z3)) → cons(z1, insert(z0, z2)) 5.95/1.97
choose(z0, cons(z1, z2), s(z3), s(z4)) → choose(z0, cons(z1, z2), z3, z4)
And the Tuples:

SORT(cons(z0, z1)) → c1(INSERT(z0, sort(z1)), SORT(z1)) 5.95/1.98
INSERT(z0, cons(z1, z2)) → c3(CHOOSE(z0, cons(z1, z2), z0, z1)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), 0, s(z3)) → c5(INSERT(z0, z2)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) → c6(CHOOSE(z0, cons(z1, z2), z3, z4))
The order we found is given by the following interpretation:
Polynomial interpretation : 5.95/1.98

POL(0) = 0    5.95/1.98
POL(CHOOSE(x1, x2, x3, x4)) = 0    5.95/1.98
POL(INSERT(x1, x2)) = 0    5.95/1.98
POL(SORT(x1)) = [2]x1    5.95/1.98
POL(c1(x1, x2)) = x1 + x2    5.95/1.98
POL(c3(x1)) = x1    5.95/1.98
POL(c5(x1)) = x1    5.95/1.98
POL(c6(x1)) = x1    5.95/1.98
POL(choose(x1, x2, x3, x4)) = [3] + [3]x1 + [3]x4    5.95/1.98
POL(cons(x1, x2)) = [1] + x1 + x2    5.95/1.98
POL(insert(x1, x2)) = [5] + [5]x1    5.95/1.98
POL(nil) = 0    5.95/1.98
POL(s(x1)) = x1    5.95/1.98
POL(sort(x1)) = 0   
5.95/1.98
5.95/1.98

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

sort(nil) → nil 5.95/1.98
sort(cons(z0, z1)) → insert(z0, sort(z1)) 5.95/1.98
insert(z0, nil) → cons(z0, nil) 5.95/1.98
insert(z0, cons(z1, z2)) → choose(z0, cons(z1, z2), z0, z1) 5.95/1.98
choose(z0, cons(z1, z2), z3, 0) → cons(z0, cons(z1, z2)) 5.95/1.98
choose(z0, cons(z1, z2), 0, s(z3)) → cons(z1, insert(z0, z2)) 5.95/1.98
choose(z0, cons(z1, z2), s(z3), s(z4)) → choose(z0, cons(z1, z2), z3, z4)
Tuples:

SORT(cons(z0, z1)) → c1(INSERT(z0, sort(z1)), SORT(z1)) 5.95/1.98
INSERT(z0, cons(z1, z2)) → c3(CHOOSE(z0, cons(z1, z2), z0, z1)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), 0, s(z3)) → c5(INSERT(z0, z2)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) → c6(CHOOSE(z0, cons(z1, z2), z3, z4))
S tuples:

INSERT(z0, cons(z1, z2)) → c3(CHOOSE(z0, cons(z1, z2), z0, z1)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), 0, s(z3)) → c5(INSERT(z0, z2)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) → c6(CHOOSE(z0, cons(z1, z2), z3, z4))
K tuples:

SORT(cons(z0, z1)) → c1(INSERT(z0, sort(z1)), SORT(z1))
Defined Rule Symbols:

sort, insert, choose

Defined Pair Symbols:

SORT, INSERT, CHOOSE

Compound Symbols:

c1, c3, c5, c6

5.95/1.98
5.95/1.98

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

CHOOSE(z0, cons(z1, z2), 0, s(z3)) → c5(INSERT(z0, z2))
We considered the (Usable) Rules:

sort(nil) → nil 5.95/1.98
sort(cons(z0, z1)) → insert(z0, sort(z1)) 5.95/1.98
insert(z0, nil) → cons(z0, nil) 5.95/1.98
insert(z0, cons(z1, z2)) → choose(z0, cons(z1, z2), z0, z1) 5.95/1.98
choose(z0, cons(z1, z2), z3, 0) → cons(z0, cons(z1, z2)) 5.95/1.98
choose(z0, cons(z1, z2), 0, s(z3)) → cons(z1, insert(z0, z2)) 5.95/1.98
choose(z0, cons(z1, z2), s(z3), s(z4)) → choose(z0, cons(z1, z2), z3, z4)
And the Tuples:

SORT(cons(z0, z1)) → c1(INSERT(z0, sort(z1)), SORT(z1)) 5.95/1.98
INSERT(z0, cons(z1, z2)) → c3(CHOOSE(z0, cons(z1, z2), z0, z1)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), 0, s(z3)) → c5(INSERT(z0, z2)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) → c6(CHOOSE(z0, cons(z1, z2), z3, z4))
The order we found is given by the following interpretation:
Polynomial interpretation : 5.95/1.98

POL(0) = 0    5.95/1.98
POL(CHOOSE(x1, x2, x3, x4)) = x2    5.95/1.98
POL(INSERT(x1, x2)) = x2    5.95/1.98
POL(SORT(x1)) = x12    5.95/1.98
POL(c1(x1, x2)) = x1 + x2    5.95/1.98
POL(c3(x1)) = x1    5.95/1.98
POL(c5(x1)) = x1    5.95/1.98
POL(c6(x1)) = x1    5.95/1.98
POL(choose(x1, x2, x3, x4)) = [1] + x2    5.95/1.98
POL(cons(x1, x2)) = [1] + x2    5.95/1.98
POL(insert(x1, x2)) = [1] + x2    5.95/1.98
POL(nil) = 0    5.95/1.98
POL(s(x1)) = 0    5.95/1.98
POL(sort(x1)) = [2]x1   
5.95/1.98
5.95/1.98

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

sort(nil) → nil 5.95/1.98
sort(cons(z0, z1)) → insert(z0, sort(z1)) 5.95/1.98
insert(z0, nil) → cons(z0, nil) 5.95/1.98
insert(z0, cons(z1, z2)) → choose(z0, cons(z1, z2), z0, z1) 5.95/1.98
choose(z0, cons(z1, z2), z3, 0) → cons(z0, cons(z1, z2)) 5.95/1.98
choose(z0, cons(z1, z2), 0, s(z3)) → cons(z1, insert(z0, z2)) 5.95/1.98
choose(z0, cons(z1, z2), s(z3), s(z4)) → choose(z0, cons(z1, z2), z3, z4)
Tuples:

SORT(cons(z0, z1)) → c1(INSERT(z0, sort(z1)), SORT(z1)) 5.95/1.98
INSERT(z0, cons(z1, z2)) → c3(CHOOSE(z0, cons(z1, z2), z0, z1)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), 0, s(z3)) → c5(INSERT(z0, z2)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) → c6(CHOOSE(z0, cons(z1, z2), z3, z4))
S tuples:

INSERT(z0, cons(z1, z2)) → c3(CHOOSE(z0, cons(z1, z2), z0, z1)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) → c6(CHOOSE(z0, cons(z1, z2), z3, z4))
K tuples:

SORT(cons(z0, z1)) → c1(INSERT(z0, sort(z1)), SORT(z1)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), 0, s(z3)) → c5(INSERT(z0, z2))
Defined Rule Symbols:

sort, insert, choose

Defined Pair Symbols:

SORT, INSERT, CHOOSE

Compound Symbols:

c1, c3, c5, c6

5.95/1.98
5.95/1.98

(7) CdtKnowledgeProof (BOTH BOUNDS(ID, ID) transformation)

The following tuples could be moved from S to K by knowledge propagation:

INSERT(z0, cons(z1, z2)) → c3(CHOOSE(z0, cons(z1, z2), z0, z1)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), 0, s(z3)) → c5(INSERT(z0, z2))
5.95/1.98
5.95/1.98

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

sort(nil) → nil 5.95/1.98
sort(cons(z0, z1)) → insert(z0, sort(z1)) 5.95/1.98
insert(z0, nil) → cons(z0, nil) 5.95/1.98
insert(z0, cons(z1, z2)) → choose(z0, cons(z1, z2), z0, z1) 5.95/1.98
choose(z0, cons(z1, z2), z3, 0) → cons(z0, cons(z1, z2)) 5.95/1.98
choose(z0, cons(z1, z2), 0, s(z3)) → cons(z1, insert(z0, z2)) 5.95/1.98
choose(z0, cons(z1, z2), s(z3), s(z4)) → choose(z0, cons(z1, z2), z3, z4)
Tuples:

SORT(cons(z0, z1)) → c1(INSERT(z0, sort(z1)), SORT(z1)) 5.95/1.98
INSERT(z0, cons(z1, z2)) → c3(CHOOSE(z0, cons(z1, z2), z0, z1)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), 0, s(z3)) → c5(INSERT(z0, z2)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) → c6(CHOOSE(z0, cons(z1, z2), z3, z4))
S tuples:

CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) → c6(CHOOSE(z0, cons(z1, z2), z3, z4))
K tuples:

SORT(cons(z0, z1)) → c1(INSERT(z0, sort(z1)), SORT(z1)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), 0, s(z3)) → c5(INSERT(z0, z2)) 5.95/1.98
INSERT(z0, cons(z1, z2)) → c3(CHOOSE(z0, cons(z1, z2), z0, z1))
Defined Rule Symbols:

sort, insert, choose

Defined Pair Symbols:

SORT, INSERT, CHOOSE

Compound Symbols:

c1, c3, c5, c6

5.95/1.98
5.95/1.98

(9) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) → c6(CHOOSE(z0, cons(z1, z2), z3, z4))
We considered the (Usable) Rules:

sort(nil) → nil 5.95/1.98
sort(cons(z0, z1)) → insert(z0, sort(z1)) 5.95/1.98
insert(z0, nil) → cons(z0, nil) 5.95/1.98
insert(z0, cons(z1, z2)) → choose(z0, cons(z1, z2), z0, z1) 5.95/1.98
choose(z0, cons(z1, z2), z3, 0) → cons(z0, cons(z1, z2)) 5.95/1.98
choose(z0, cons(z1, z2), 0, s(z3)) → cons(z1, insert(z0, z2)) 5.95/1.98
choose(z0, cons(z1, z2), s(z3), s(z4)) → choose(z0, cons(z1, z2), z3, z4)
And the Tuples:

SORT(cons(z0, z1)) → c1(INSERT(z0, sort(z1)), SORT(z1)) 5.95/1.98
INSERT(z0, cons(z1, z2)) → c3(CHOOSE(z0, cons(z1, z2), z0, z1)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), 0, s(z3)) → c5(INSERT(z0, z2)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) → c6(CHOOSE(z0, cons(z1, z2), z3, z4))
The order we found is given by the following interpretation:
Polynomial interpretation : 5.95/1.98

POL(0) = 0    5.95/1.98
POL(CHOOSE(x1, x2, x3, x4)) = [2]x3 + x1·x2    5.95/1.98
POL(INSERT(x1, x2)) = [2]x1 + x1·x2    5.95/1.98
POL(SORT(x1)) = x12    5.95/1.98
POL(c1(x1, x2)) = x1 + x2    5.95/1.98
POL(c3(x1)) = x1    5.95/1.98
POL(c5(x1)) = x1    5.95/1.98
POL(c6(x1)) = x1    5.95/1.98
POL(choose(x1, x2, x3, x4)) = [2] + x1 + x2    5.95/1.98
POL(cons(x1, x2)) = [2] + x1 + x2    5.95/1.98
POL(insert(x1, x2)) = [2] + x1 + x2    5.95/1.98
POL(nil) = 0    5.95/1.98
POL(s(x1)) = [2] + x1    5.95/1.98
POL(sort(x1)) = x1   
5.95/1.98
5.95/1.98

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

sort(nil) → nil 5.95/1.98
sort(cons(z0, z1)) → insert(z0, sort(z1)) 5.95/1.98
insert(z0, nil) → cons(z0, nil) 5.95/1.98
insert(z0, cons(z1, z2)) → choose(z0, cons(z1, z2), z0, z1) 5.95/1.98
choose(z0, cons(z1, z2), z3, 0) → cons(z0, cons(z1, z2)) 5.95/1.98
choose(z0, cons(z1, z2), 0, s(z3)) → cons(z1, insert(z0, z2)) 5.95/1.98
choose(z0, cons(z1, z2), s(z3), s(z4)) → choose(z0, cons(z1, z2), z3, z4)
Tuples:

SORT(cons(z0, z1)) → c1(INSERT(z0, sort(z1)), SORT(z1)) 5.95/1.98
INSERT(z0, cons(z1, z2)) → c3(CHOOSE(z0, cons(z1, z2), z0, z1)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), 0, s(z3)) → c5(INSERT(z0, z2)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) → c6(CHOOSE(z0, cons(z1, z2), z3, z4))
S tuples:none
K tuples:

SORT(cons(z0, z1)) → c1(INSERT(z0, sort(z1)), SORT(z1)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), 0, s(z3)) → c5(INSERT(z0, z2)) 5.95/1.98
INSERT(z0, cons(z1, z2)) → c3(CHOOSE(z0, cons(z1, z2), z0, z1)) 5.95/1.98
CHOOSE(z0, cons(z1, z2), s(z3), s(z4)) → c6(CHOOSE(z0, cons(z1, z2), z3, z4))
Defined Rule Symbols:

sort, insert, choose

Defined Pair Symbols:

SORT, INSERT, CHOOSE

Compound Symbols:

c1, c3, c5, c6

5.95/1.98
5.95/1.98

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

The set S is empty
5.95/1.98
5.95/1.98

(12) BOUNDS(O(1), O(1))

5.95/1.98
5.95/1.98
6.32/2.03 EOF