YES(O(1), O(n^2)) 8.16/2.54 YES(O(1), O(n^2)) 8.16/2.58 8.16/2.58 8.16/2.58 8.16/2.58 8.16/2.58 8.16/2.58 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 8.16/2.58 8.16/2.58 8.16/2.58
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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
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↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
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↳2 CdtProblem
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↳3 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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↳4 CdtProblem
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↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
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↳6 CdtProblem
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↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
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↳8 CdtProblem
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↳9 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳10 BOUNDS(O(1), O(1))
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8.16/2.58

(0) Obligation:

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

min(X, 0) → X 8.16/2.58
min(s(X), s(Y)) → min(X, Y) 8.16/2.58
quot(0, s(Y)) → 0 8.16/2.58
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y))) 8.16/2.58
log(s(0)) → 0 8.16/2.58
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

Rewrite Strategy: INNERMOST
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8.16/2.58

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

Converted CpxTRS to CDT
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8.16/2.58

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

min(z0, 0) → z0 8.16/2.58
min(s(z0), s(z1)) → min(z0, z1) 8.16/2.58
quot(0, s(z0)) → 0 8.16/2.58
quot(s(z0), s(z1)) → s(quot(min(z0, z1), s(z1))) 8.16/2.58
log(s(0)) → 0 8.16/2.58
log(s(s(z0))) → s(log(s(quot(z0, s(s(0))))))
Tuples:

MIN(s(z0), s(z1)) → c1(MIN(z0, z1)) 8.16/2.58
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1)) 8.16/2.58
LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:

MIN(s(z0), s(z1)) → c1(MIN(z0, z1)) 8.16/2.58
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1)) 8.16/2.58
LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
K tuples:none
Defined Rule Symbols:

min, quot, log

Defined Pair Symbols:

MIN, QUOT, LOG

Compound Symbols:

c1, c3, c5

8.16/2.58
8.16/2.58

(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))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
We considered the (Usable) Rules:

quot(0, s(z0)) → 0 8.16/2.58
quot(s(z0), s(z1)) → s(quot(min(z0, z1), s(z1))) 8.16/2.58
min(z0, 0) → z0 8.16/2.58
min(s(z0), s(z1)) → min(z0, z1)
And the Tuples:

MIN(s(z0), s(z1)) → c1(MIN(z0, z1)) 8.16/2.58
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1)) 8.16/2.58
LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
The order we found is given by the following interpretation:
Polynomial interpretation : 8.16/2.58

POL(0) = 0    8.16/2.58
POL(LOG(x1)) = x1    8.16/2.58
POL(MIN(x1, x2)) = 0    8.16/2.58
POL(QUOT(x1, x2)) = 0    8.16/2.58
POL(c1(x1)) = x1    8.16/2.58
POL(c3(x1, x2)) = x1 + x2    8.16/2.58
POL(c5(x1, x2)) = x1 + x2    8.16/2.58
POL(min(x1, x2)) = x1    8.16/2.58
POL(quot(x1, x2)) = x1    8.16/2.58
POL(s(x1)) = [4] + x1   
8.16/2.58
8.16/2.58

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

min(z0, 0) → z0 8.16/2.58
min(s(z0), s(z1)) → min(z0, z1) 8.16/2.58
quot(0, s(z0)) → 0 8.16/2.58
quot(s(z0), s(z1)) → s(quot(min(z0, z1), s(z1))) 8.16/2.58
log(s(0)) → 0 8.16/2.58
log(s(s(z0))) → s(log(s(quot(z0, s(s(0))))))
Tuples:

MIN(s(z0), s(z1)) → c1(MIN(z0, z1)) 8.16/2.58
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1)) 8.16/2.58
LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:

MIN(s(z0), s(z1)) → c1(MIN(z0, z1)) 8.16/2.58
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1))
K tuples:

LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
Defined Rule Symbols:

min, quot, log

Defined Pair Symbols:

MIN, QUOT, LOG

Compound Symbols:

c1, c3, c5

8.16/2.58
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(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.

QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1))
We considered the (Usable) Rules:

quot(0, s(z0)) → 0 8.16/2.58
quot(s(z0), s(z1)) → s(quot(min(z0, z1), s(z1))) 8.16/2.58
min(z0, 0) → z0 8.16/2.58
min(s(z0), s(z1)) → min(z0, z1)
And the Tuples:

MIN(s(z0), s(z1)) → c1(MIN(z0, z1)) 8.16/2.58
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1)) 8.16/2.58
LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
The order we found is given by the following interpretation:
Polynomial interpretation : 8.16/2.58

POL(0) = 0    8.16/2.58
POL(LOG(x1)) = [2]x1 + [2]x12    8.16/2.58
POL(MIN(x1, x2)) = 0    8.16/2.58
POL(QUOT(x1, x2)) = [2]x1·x2    8.16/2.58
POL(c1(x1)) = x1    8.16/2.58
POL(c3(x1, x2)) = x1 + x2    8.16/2.58
POL(c5(x1, x2)) = x1 + x2    8.16/2.58
POL(min(x1, x2)) = x1    8.16/2.58
POL(quot(x1, x2)) = x1    8.16/2.58
POL(s(x1)) = [2] + x1   
8.16/2.58
8.16/2.58

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

min(z0, 0) → z0 8.16/2.58
min(s(z0), s(z1)) → min(z0, z1) 8.16/2.58
quot(0, s(z0)) → 0 8.16/2.58
quot(s(z0), s(z1)) → s(quot(min(z0, z1), s(z1))) 8.16/2.58
log(s(0)) → 0 8.16/2.58
log(s(s(z0))) → s(log(s(quot(z0, s(s(0))))))
Tuples:

MIN(s(z0), s(z1)) → c1(MIN(z0, z1)) 8.16/2.58
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1)) 8.16/2.58
LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:

MIN(s(z0), s(z1)) → c1(MIN(z0, z1))
K tuples:

LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0)))) 8.16/2.58
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1))
Defined Rule Symbols:

min, quot, log

Defined Pair Symbols:

MIN, QUOT, LOG

Compound Symbols:

c1, c3, c5

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8.16/2.58

(7) 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.

MIN(s(z0), s(z1)) → c1(MIN(z0, z1))
We considered the (Usable) Rules:

quot(0, s(z0)) → 0 8.16/2.58
quot(s(z0), s(z1)) → s(quot(min(z0, z1), s(z1))) 8.16/2.58
min(z0, 0) → z0 8.16/2.58
min(s(z0), s(z1)) → min(z0, z1)
And the Tuples:

MIN(s(z0), s(z1)) → c1(MIN(z0, z1)) 8.16/2.58
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1)) 8.16/2.58
LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
The order we found is given by the following interpretation:
Polynomial interpretation : 8.16/2.58

POL(0) = 0    8.16/2.58
POL(LOG(x1)) = [2]x1 + [2]x12    8.16/2.58
POL(MIN(x1, x2)) = x2    8.16/2.58
POL(QUOT(x1, x2)) = [2] + [2]x1 + x1·x2    8.16/2.58
POL(c1(x1)) = x1    8.16/2.58
POL(c3(x1, x2)) = x1 + x2    8.16/2.58
POL(c5(x1, x2)) = x1 + x2    8.16/2.58
POL(min(x1, x2)) = x1    8.16/2.58
POL(quot(x1, x2)) = x1    8.16/2.58
POL(s(x1)) = [2] + x1   
8.16/2.58
8.16/2.58

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

min(z0, 0) → z0 8.16/2.58
min(s(z0), s(z1)) → min(z0, z1) 8.16/2.58
quot(0, s(z0)) → 0 8.16/2.58
quot(s(z0), s(z1)) → s(quot(min(z0, z1), s(z1))) 8.16/2.58
log(s(0)) → 0 8.16/2.58
log(s(s(z0))) → s(log(s(quot(z0, s(s(0))))))
Tuples:

MIN(s(z0), s(z1)) → c1(MIN(z0, z1)) 8.16/2.58
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1)) 8.16/2.58
LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:none
K tuples:

LOG(s(s(z0))) → c5(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0)))) 8.16/2.58
QUOT(s(z0), s(z1)) → c3(QUOT(min(z0, z1), s(z1)), MIN(z0, z1)) 8.16/2.58
MIN(s(z0), s(z1)) → c1(MIN(z0, z1))
Defined Rule Symbols:

min, quot, log

Defined Pair Symbols:

MIN, QUOT, LOG

Compound Symbols:

c1, c3, c5

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8.16/2.58

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

The set S is empty
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(10) BOUNDS(O(1), O(1))

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8.74/2.69 EOF