YES(O(1), O(n^2)) 0.00/0.88 YES(O(1), O(n^2)) 0.00/0.89 0.00/0.89 0.00/0.89 0.00/0.89 0.00/0.89 0.00/0.89 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.89 0.00/0.89 0.00/0.89
0.00/0.89 0.00/0.89 0.00/0.89
0.00/0.89
0.00/0.89

(0) Obligation:

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

half(0) → 0 0.00/0.89
half(s(s(x))) → s(half(x)) 0.00/0.89
log(s(0)) → 0 0.00/0.89
log(s(s(x))) → s(log(s(half(x))))

Rewrite Strategy: INNERMOST
0.00/0.89
0.00/0.89

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

Converted CpxTRS to CDT
0.00/0.89
0.00/0.89

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

half(0) → 0 0.00/0.89
half(s(s(z0))) → s(half(z0)) 0.00/0.89
log(s(0)) → 0 0.00/0.89
log(s(s(z0))) → s(log(s(half(z0))))
Tuples:

HALF(s(s(z0))) → c1(HALF(z0)) 0.00/0.89
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
S tuples:

HALF(s(s(z0))) → c1(HALF(z0)) 0.00/0.89
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
K tuples:none
Defined Rule Symbols:

half, log

Defined Pair Symbols:

HALF, LOG

Compound Symbols:

c1, c3

0.00/0.89
0.00/0.89

(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.

LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
We considered the (Usable) Rules:

half(0) → 0 0.00/0.89
half(s(s(z0))) → s(half(z0))
And the Tuples:

HALF(s(s(z0))) → c1(HALF(z0)) 0.00/0.89
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.89

POL(0) = 0    0.00/0.89
POL(HALF(x1)) = 0    0.00/0.89
POL(LOG(x1)) = x1    0.00/0.89
POL(c1(x1)) = x1    0.00/0.89
POL(c3(x1, x2)) = x1 + x2    0.00/0.89
POL(half(x1)) = x1    0.00/0.89
POL(s(x1)) = [4] + x1   
0.00/0.89
0.00/0.89

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

half(0) → 0 0.00/0.89
half(s(s(z0))) → s(half(z0)) 0.00/0.89
log(s(0)) → 0 0.00/0.89
log(s(s(z0))) → s(log(s(half(z0))))
Tuples:

HALF(s(s(z0))) → c1(HALF(z0)) 0.00/0.89
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
S tuples:

HALF(s(s(z0))) → c1(HALF(z0))
K tuples:

LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
Defined Rule Symbols:

half, log

Defined Pair Symbols:

HALF, LOG

Compound Symbols:

c1, c3

0.00/0.89
0.00/0.89

(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.

HALF(s(s(z0))) → c1(HALF(z0))
We considered the (Usable) Rules:

half(0) → 0 0.00/0.89
half(s(s(z0))) → s(half(z0))
And the Tuples:

HALF(s(s(z0))) → c1(HALF(z0)) 0.00/0.89
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.89

POL(0) = 0    0.00/0.89
POL(HALF(x1)) = x1    0.00/0.89
POL(LOG(x1)) = x12    0.00/0.89
POL(c1(x1)) = x1    0.00/0.89
POL(c3(x1, x2)) = x1 + x2    0.00/0.89
POL(half(x1)) = x1    0.00/0.89
POL(s(x1)) = [1] + x1   
0.00/0.89
0.00/0.89

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

half(0) → 0 0.00/0.89
half(s(s(z0))) → s(half(z0)) 0.00/0.89
log(s(0)) → 0 0.00/0.89
log(s(s(z0))) → s(log(s(half(z0))))
Tuples:

HALF(s(s(z0))) → c1(HALF(z0)) 0.00/0.89
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
S tuples:none
K tuples:

LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0)) 0.00/0.89
HALF(s(s(z0))) → c1(HALF(z0))
Defined Rule Symbols:

half, log

Defined Pair Symbols:

HALF, LOG

Compound Symbols:

c1, c3

0.00/0.89
0.00/0.89

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

The set S is empty
0.00/0.89
0.00/0.89

(8) BOUNDS(O(1), O(1))

0.00/0.89
0.00/0.89
0.00/0.94 EOF