YES(O(1), O(n^2)) 4.11/1.53 YES(O(1), O(n^2)) 4.11/1.57 4.11/1.57 4.11/1.57 4.11/1.57 4.11/1.57 4.11/1.57 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 4.11/1.57 4.11/1.57 4.11/1.57
4.11/1.57
4.11/1.57
4.11/1.57
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

0 CpxTRS
4.11/1.57 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
4.11/1.57 
↳2 CdtProblem
4.11/1.57 
↳3 CdtNarrowingProof (BOTH BOUNDS(ID, ID))
4.11/1.57 
↳4 CdtProblem
4.11/1.57 
↳5 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
4.11/1.57 
↳6 CdtProblem
4.11/1.57 
↳7 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
4.11/1.57 
↳8 CdtProblem
4.11/1.57 
↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
4.11/1.57 
↳10 CdtProblem
4.11/1.57 
↳11 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
4.11/1.57 
↳12 CdtProblem
4.11/1.57 
↳13 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
4.11/1.57 
↳14 CdtProblem
4.11/1.57 
↳15 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
4.11/1.57 
↳16 CdtProblem
4.11/1.57 
↳17 SIsEmptyProof (BOTH BOUNDS(ID, ID))
4.11/1.57 
↳18 BOUNDS(O(1), O(1))
4.11/1.57
4.11/1.57 4.11/1.57
4.11/1.57
4.11/1.57

(0) Obligation:

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

half(0) → 0 4.11/1.57
half(s(0)) → 0 4.11/1.57
half(s(s(x))) → s(half(x)) 4.11/1.57
lastbit(0) → 0 4.11/1.57
lastbit(s(0)) → s(0) 4.11/1.57
lastbit(s(s(x))) → lastbit(x) 4.11/1.57
conv(0) → cons(nil, 0) 4.11/1.57
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

Rewrite Strategy: INNERMOST
4.11/1.57
4.11/1.57

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

Converted CpxTRS to CDT
4.11/1.57
4.11/1.57

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

half(0) → 0 4.11/1.57
half(s(0)) → 0 4.11/1.57
half(s(s(z0))) → s(half(z0)) 4.11/1.57
lastbit(0) → 0 4.11/1.57
lastbit(s(0)) → s(0) 4.11/1.57
lastbit(s(s(z0))) → lastbit(z0) 4.11/1.57
conv(0) → cons(nil, 0) 4.11/1.57
conv(s(z0)) → cons(conv(half(s(z0))), lastbit(s(z0)))
Tuples:

HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.57
LASTBIT(s(s(z0))) → c5(LASTBIT(z0)) 4.11/1.57
CONV(s(z0)) → c7(CONV(half(s(z0))), HALF(s(z0)), LASTBIT(s(z0)))
S tuples:

HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.57
LASTBIT(s(s(z0))) → c5(LASTBIT(z0)) 4.11/1.57
CONV(s(z0)) → c7(CONV(half(s(z0))), HALF(s(z0)), LASTBIT(s(z0)))
K tuples:none
Defined Rule Symbols:

half, lastbit, conv

Defined Pair Symbols:

HALF, LASTBIT, CONV

Compound Symbols:

c2, c5, c7

4.11/1.57
4.11/1.57

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

Use narrowing to replace CONV(s(z0)) → c7(CONV(half(s(z0))), HALF(s(z0)), LASTBIT(s(z0))) by

CONV(s(0)) → c7(CONV(0), HALF(s(0)), LASTBIT(s(0))) 4.11/1.57
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0))))
4.11/1.57
4.11/1.57

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

half(0) → 0 4.11/1.57
half(s(0)) → 0 4.11/1.57
half(s(s(z0))) → s(half(z0)) 4.11/1.57
lastbit(0) → 0 4.11/1.57
lastbit(s(0)) → s(0) 4.11/1.57
lastbit(s(s(z0))) → lastbit(z0) 4.11/1.57
conv(0) → cons(nil, 0) 4.11/1.57
conv(s(z0)) → cons(conv(half(s(z0))), lastbit(s(z0)))
Tuples:

HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.57
LASTBIT(s(s(z0))) → c5(LASTBIT(z0)) 4.11/1.57
CONV(s(0)) → c7(CONV(0), HALF(s(0)), LASTBIT(s(0))) 4.11/1.57
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0))))
S tuples:

HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.57
LASTBIT(s(s(z0))) → c5(LASTBIT(z0)) 4.11/1.57
CONV(s(0)) → c7(CONV(0), HALF(s(0)), LASTBIT(s(0))) 4.11/1.57
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0))))
K tuples:none
Defined Rule Symbols:

half, lastbit, conv

Defined Pair Symbols:

HALF, LASTBIT, CONV

Compound Symbols:

c2, c5, c7

4.11/1.57
4.11/1.57

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

Removed 3 trailing tuple parts
4.11/1.57
4.11/1.57

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

half(0) → 0 4.11/1.57
half(s(0)) → 0 4.11/1.57
half(s(s(z0))) → s(half(z0)) 4.11/1.57
lastbit(0) → 0 4.11/1.57
lastbit(s(0)) → s(0) 4.11/1.57
lastbit(s(s(z0))) → lastbit(z0) 4.11/1.57
conv(0) → cons(nil, 0) 4.11/1.57
conv(s(z0)) → cons(conv(half(s(z0))), lastbit(s(z0)))
Tuples:

HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.57
LASTBIT(s(s(z0))) → c5(LASTBIT(z0)) 4.11/1.57
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0)))) 4.11/1.57
CONV(s(0)) → c7
S tuples:

HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.57
LASTBIT(s(s(z0))) → c5(LASTBIT(z0)) 4.11/1.57
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0)))) 4.11/1.57
CONV(s(0)) → c7
K tuples:none
Defined Rule Symbols:

half, lastbit, conv

Defined Pair Symbols:

HALF, LASTBIT, CONV

Compound Symbols:

c2, c5, c7, c7

4.11/1.57
4.11/1.57

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

Removed 1 trailing nodes:

CONV(s(0)) → c7
4.11/1.57
4.11/1.57

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

half(0) → 0 4.11/1.57
half(s(0)) → 0 4.11/1.57
half(s(s(z0))) → s(half(z0)) 4.11/1.57
lastbit(0) → 0 4.11/1.57
lastbit(s(0)) → s(0) 4.11/1.57
lastbit(s(s(z0))) → lastbit(z0) 4.11/1.57
conv(0) → cons(nil, 0) 4.11/1.57
conv(s(z0)) → cons(conv(half(s(z0))), lastbit(s(z0)))
Tuples:

HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.57
LASTBIT(s(s(z0))) → c5(LASTBIT(z0)) 4.11/1.57
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0)))) 4.11/1.57
CONV(s(0)) → c7
S tuples:

HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.58
LASTBIT(s(s(z0))) → c5(LASTBIT(z0)) 4.11/1.58
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0)))) 4.11/1.58
CONV(s(0)) → c7
K tuples:none
Defined Rule Symbols:

half, lastbit, conv

Defined Pair Symbols:

HALF, LASTBIT, CONV

Compound Symbols:

c2, c5, c7, c7

4.11/1.58
4.11/1.58

(9) 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.

CONV(s(0)) → c7
We considered the (Usable) Rules:

half(0) → 0 4.11/1.58
half(s(0)) → 0 4.11/1.58
half(s(s(z0))) → s(half(z0))
And the Tuples:

HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.58
LASTBIT(s(s(z0))) → c5(LASTBIT(z0)) 4.11/1.58
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0)))) 4.11/1.58
CONV(s(0)) → c7
The order we found is given by the following interpretation:
Polynomial interpretation : 4.11/1.58

POL(0) = [2]    4.11/1.58
POL(CONV(x1)) = [1]    4.11/1.58
POL(HALF(x1)) = 0    4.11/1.58
POL(LASTBIT(x1)) = 0    4.11/1.58
POL(c2(x1)) = x1    4.11/1.58
POL(c5(x1)) = x1    4.11/1.58
POL(c7) = 0    4.11/1.58
POL(c7(x1, x2, x3)) = x1 + x2 + x3    4.11/1.58
POL(half(x1)) = [2] + [2]x1    4.11/1.58
POL(s(x1)) = [4]   
4.11/1.58
4.11/1.58

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

half(0) → 0 4.11/1.58
half(s(0)) → 0 4.11/1.58
half(s(s(z0))) → s(half(z0)) 4.11/1.58
lastbit(0) → 0 4.11/1.58
lastbit(s(0)) → s(0) 4.11/1.58
lastbit(s(s(z0))) → lastbit(z0) 4.11/1.58
conv(0) → cons(nil, 0) 4.11/1.58
conv(s(z0)) → cons(conv(half(s(z0))), lastbit(s(z0)))
Tuples:

HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.58
LASTBIT(s(s(z0))) → c5(LASTBIT(z0)) 4.11/1.58
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0)))) 4.11/1.58
CONV(s(0)) → c7
S tuples:

HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.58
LASTBIT(s(s(z0))) → c5(LASTBIT(z0)) 4.11/1.58
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0))))
K tuples:

CONV(s(0)) → c7
Defined Rule Symbols:

half, lastbit, conv

Defined Pair Symbols:

HALF, LASTBIT, CONV

Compound Symbols:

c2, c5, c7, c7

4.11/1.58
4.11/1.58

(11) 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.

CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0))))
We considered the (Usable) Rules:

half(0) → 0 4.11/1.58
half(s(0)) → 0 4.11/1.58
half(s(s(z0))) → s(half(z0))
And the Tuples:

HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.58
LASTBIT(s(s(z0))) → c5(LASTBIT(z0)) 4.11/1.58
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0)))) 4.11/1.58
CONV(s(0)) → c7
The order we found is given by the following interpretation:
Polynomial interpretation : 4.11/1.58

POL(0) = 0    4.11/1.58
POL(CONV(x1)) = x1    4.11/1.58
POL(HALF(x1)) = 0    4.11/1.58
POL(LASTBIT(x1)) = 0    4.11/1.58
POL(c2(x1)) = x1    4.11/1.58
POL(c5(x1)) = x1    4.11/1.58
POL(c7) = 0    4.11/1.58
POL(c7(x1, x2, x3)) = x1 + x2 + x3    4.11/1.58
POL(half(x1)) = x1    4.11/1.58
POL(s(x1)) = [4] + x1   
4.11/1.58
4.11/1.58

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

half(0) → 0 4.11/1.58
half(s(0)) → 0 4.11/1.58
half(s(s(z0))) → s(half(z0)) 4.11/1.58
lastbit(0) → 0 4.11/1.58
lastbit(s(0)) → s(0) 4.11/1.58
lastbit(s(s(z0))) → lastbit(z0) 4.11/1.58
conv(0) → cons(nil, 0) 4.11/1.58
conv(s(z0)) → cons(conv(half(s(z0))), lastbit(s(z0)))
Tuples:

HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.58
LASTBIT(s(s(z0))) → c5(LASTBIT(z0)) 4.11/1.58
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0)))) 4.11/1.58
CONV(s(0)) → c7
S tuples:

HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.58
LASTBIT(s(s(z0))) → c5(LASTBIT(z0))
K tuples:

CONV(s(0)) → c7 4.11/1.58
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0))))
Defined Rule Symbols:

half, lastbit, conv

Defined Pair Symbols:

HALF, LASTBIT, CONV

Compound Symbols:

c2, c5, c7, c7

4.11/1.58
4.11/1.58

(13) 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.

HALF(s(s(z0))) → c2(HALF(z0))
We considered the (Usable) Rules:

half(0) → 0 4.11/1.58
half(s(0)) → 0 4.11/1.58
half(s(s(z0))) → s(half(z0))
And the Tuples:

HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.58
LASTBIT(s(s(z0))) → c5(LASTBIT(z0)) 4.11/1.58
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0)))) 4.11/1.58
CONV(s(0)) → c7
The order we found is given by the following interpretation:
Polynomial interpretation : 4.11/1.58

POL(0) = 0    4.11/1.58
POL(CONV(x1)) = x12    4.11/1.58
POL(HALF(x1)) = x1    4.11/1.58
POL(LASTBIT(x1)) = 0    4.11/1.58
POL(c2(x1)) = x1    4.11/1.58
POL(c5(x1)) = x1    4.11/1.58
POL(c7) = 0    4.11/1.58
POL(c7(x1, x2, x3)) = x1 + x2 + x3    4.11/1.58
POL(half(x1)) = x1    4.11/1.58
POL(s(x1)) = [1] + x1   
4.11/1.58
4.11/1.58

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

half(0) → 0 4.11/1.58
half(s(0)) → 0 4.11/1.58
half(s(s(z0))) → s(half(z0)) 4.11/1.58
lastbit(0) → 0 4.11/1.58
lastbit(s(0)) → s(0) 4.11/1.58
lastbit(s(s(z0))) → lastbit(z0) 4.11/1.58
conv(0) → cons(nil, 0) 4.11/1.58
conv(s(z0)) → cons(conv(half(s(z0))), lastbit(s(z0)))
Tuples:

HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.58
LASTBIT(s(s(z0))) → c5(LASTBIT(z0)) 4.11/1.58
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0)))) 4.11/1.58
CONV(s(0)) → c7
S tuples:

LASTBIT(s(s(z0))) → c5(LASTBIT(z0))
K tuples:

CONV(s(0)) → c7 4.11/1.58
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0)))) 4.11/1.58
HALF(s(s(z0))) → c2(HALF(z0))
Defined Rule Symbols:

half, lastbit, conv

Defined Pair Symbols:

HALF, LASTBIT, CONV

Compound Symbols:

c2, c5, c7, c7

4.11/1.58
4.11/1.58

(15) 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.

LASTBIT(s(s(z0))) → c5(LASTBIT(z0))
We considered the (Usable) Rules:

half(0) → 0 4.11/1.58
half(s(0)) → 0 4.11/1.58
half(s(s(z0))) → s(half(z0))
And the Tuples:

HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.58
LASTBIT(s(s(z0))) → c5(LASTBIT(z0)) 4.11/1.58
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0)))) 4.11/1.58
CONV(s(0)) → c7
The order we found is given by the following interpretation:
Polynomial interpretation : 4.11/1.58

POL(0) = 0    4.11/1.58
POL(CONV(x1)) = x12    4.11/1.58
POL(HALF(x1)) = 0    4.11/1.58
POL(LASTBIT(x1)) = [2]x1    4.11/1.58
POL(c2(x1)) = x1    4.11/1.58
POL(c5(x1)) = x1    4.11/1.58
POL(c7) = 0    4.11/1.58
POL(c7(x1, x2, x3)) = x1 + x2 + x3    4.11/1.58
POL(half(x1)) = x1    4.11/1.58
POL(s(x1)) = [2] + x1   
4.11/1.58
4.11/1.58

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

half(0) → 0 4.11/1.58
half(s(0)) → 0 4.11/1.58
half(s(s(z0))) → s(half(z0)) 4.11/1.58
lastbit(0) → 0 4.11/1.58
lastbit(s(0)) → s(0) 4.11/1.58
lastbit(s(s(z0))) → lastbit(z0) 4.11/1.58
conv(0) → cons(nil, 0) 4.11/1.58
conv(s(z0)) → cons(conv(half(s(z0))), lastbit(s(z0)))
Tuples:

HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.58
LASTBIT(s(s(z0))) → c5(LASTBIT(z0)) 4.11/1.58
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0)))) 4.11/1.58
CONV(s(0)) → c7
S tuples:none
K tuples:

CONV(s(0)) → c7 4.11/1.58
CONV(s(s(z0))) → c7(CONV(s(half(z0))), HALF(s(s(z0))), LASTBIT(s(s(z0)))) 4.11/1.58
HALF(s(s(z0))) → c2(HALF(z0)) 4.11/1.58
LASTBIT(s(s(z0))) → c5(LASTBIT(z0))
Defined Rule Symbols:

half, lastbit, conv

Defined Pair Symbols:

HALF, LASTBIT, CONV

Compound Symbols:

c2, c5, c7, c7

4.11/1.58
4.11/1.58

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

The set S is empty
4.11/1.58
4.11/1.58

(18) BOUNDS(O(1), O(1))

4.11/1.58
4.11/1.58
4.44/1.64 EOF