YES(O(1), O(n^2)) 10.13/3.06 YES(O(1), O(n^2)) 10.66/3.10 10.66/3.10 10.66/3.10 10.66/3.10 10.66/3.10 10.66/3.10 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 10.66/3.10 10.66/3.10 10.66/3.10
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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 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
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↳4 CdtProblem
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↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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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 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
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↳10 CdtProblem
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↳11 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳12 BOUNDS(O(1), O(1))
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(0) Obligation:

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

pred(s(x)) → x 10.66/3.10
minus(x, 0) → x 10.66/3.10
minus(x, s(y)) → pred(minus(x, y)) 10.66/3.10
quot(0, s(y)) → 0 10.66/3.10
quot(s(x), s(y)) → s(quot(minus(x, y), s(y))) 10.66/3.10
log(s(0)) → 0 10.66/3.10
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))

Rewrite Strategy: INNERMOST
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(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT
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(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

pred(s(z0)) → z0 10.66/3.10
minus(z0, 0) → z0 10.66/3.10
minus(z0, s(z1)) → pred(minus(z0, z1)) 10.66/3.10
quot(0, s(z0)) → 0 10.66/3.10
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 10.66/3.10
log(s(0)) → 0 10.66/3.10
log(s(s(z0))) → s(log(s(quot(z0, s(s(0))))))
Tuples:

MINUS(z0, s(z1)) → c2(PRED(minus(z0, z1)), MINUS(z0, z1)) 10.66/3.10
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 10.66/3.10
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
S tuples:

MINUS(z0, s(z1)) → c2(PRED(minus(z0, z1)), MINUS(z0, z1)) 10.66/3.10
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 10.66/3.10
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
K tuples:none
Defined Rule Symbols:

pred, minus, quot, log

Defined Pair Symbols:

MINUS, QUOT, LOG

Compound Symbols:

c2, c4, c6

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

Removed 1 trailing tuple parts
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

pred(s(z0)) → z0 10.66/3.10
minus(z0, 0) → z0 10.66/3.10
minus(z0, s(z1)) → pred(minus(z0, z1)) 10.66/3.10
quot(0, s(z0)) → 0 10.66/3.10
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 10.66/3.10
log(s(0)) → 0 10.66/3.10
log(s(s(z0))) → s(log(s(quot(z0, s(s(0))))))
Tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 10.66/3.10
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0)))) 10.66/3.10
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
S tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 10.66/3.10
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0)))) 10.66/3.10
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

pred, minus, quot, log

Defined Pair Symbols:

QUOT, LOG, MINUS

Compound Symbols:

c4, c6, c2

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(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.

LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
We considered the (Usable) Rules:

quot(0, s(z0)) → 0 10.66/3.10
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 10.66/3.10
minus(z0, 0) → z0 10.66/3.10
minus(z0, s(z1)) → pred(minus(z0, z1)) 10.66/3.10
pred(s(z0)) → z0
And the Tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 10.66/3.10
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0)))) 10.66/3.10
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 10.66/3.10

POL(0) = 0    10.66/3.10
POL(LOG(x1)) = [4]x1    10.66/3.10
POL(MINUS(x1, x2)) = 0    10.66/3.10
POL(QUOT(x1, x2)) = 0    10.66/3.10
POL(c2(x1)) = x1    10.66/3.10
POL(c4(x1, x2)) = x1 + x2    10.66/3.10
POL(c6(x1, x2)) = x1 + x2    10.66/3.10
POL(minus(x1, x2)) = x1    10.66/3.10
POL(pred(x1)) = x1    10.66/3.10
POL(quot(x1, x2)) = x1    10.66/3.10
POL(s(x1)) = [1] + x1   
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

pred(s(z0)) → z0 10.66/3.10
minus(z0, 0) → z0 10.66/3.10
minus(z0, s(z1)) → pred(minus(z0, z1)) 10.66/3.10
quot(0, s(z0)) → 0 10.66/3.10
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 10.66/3.10
log(s(0)) → 0 10.66/3.10
log(s(s(z0))) → s(log(s(quot(z0, s(s(0))))))
Tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 10.66/3.10
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0)))) 10.66/3.10
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
S tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 10.66/3.10
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
K tuples:

LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0))))
Defined Rule Symbols:

pred, minus, quot, log

Defined Pair Symbols:

QUOT, LOG, MINUS

Compound Symbols:

c4, c6, c2

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(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.

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
We considered the (Usable) Rules:

quot(0, s(z0)) → 0 10.66/3.10
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 10.66/3.10
minus(z0, 0) → z0 10.66/3.10
minus(z0, s(z1)) → pred(minus(z0, z1)) 10.66/3.10
pred(s(z0)) → z0
And the Tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 10.66/3.10
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0)))) 10.66/3.10
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 10.66/3.10

POL(0) = 0    10.66/3.10
POL(LOG(x1)) = [3]x1 + x12    10.66/3.10
POL(MINUS(x1, x2)) = 0    10.66/3.10
POL(QUOT(x1, x2)) = x1    10.66/3.10
POL(c2(x1)) = x1    10.66/3.10
POL(c4(x1, x2)) = x1 + x2    10.66/3.10
POL(c6(x1, x2)) = x1 + x2    10.66/3.10
POL(minus(x1, x2)) = x1    10.66/3.10
POL(pred(x1)) = x1    10.66/3.10
POL(quot(x1, x2)) = x1    10.66/3.10
POL(s(x1)) = [1] + x1   
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(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

pred(s(z0)) → z0 10.66/3.10
minus(z0, 0) → z0 10.66/3.10
minus(z0, s(z1)) → pred(minus(z0, z1)) 10.66/3.10
quot(0, s(z0)) → 0 10.66/3.10
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 10.66/3.10
log(s(0)) → 0 10.66/3.10
log(s(s(z0))) → s(log(s(quot(z0, s(s(0))))))
Tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 10.66/3.10
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0)))) 10.66/3.10
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
S tuples:

MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
K tuples:

LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0)))) 10.66/3.10
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:

pred, minus, quot, log

Defined Pair Symbols:

QUOT, LOG, MINUS

Compound Symbols:

c4, c6, c2

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(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.

MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
We considered the (Usable) Rules:

quot(0, s(z0)) → 0 10.66/3.10
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 10.66/3.10
minus(z0, 0) → z0 10.66/3.10
minus(z0, s(z1)) → pred(minus(z0, z1)) 10.66/3.10
pred(s(z0)) → z0
And the Tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 10.66/3.10
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0)))) 10.66/3.10
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 10.66/3.10

POL(0) = 0    10.66/3.10
POL(LOG(x1)) = [2]x12    10.66/3.10
POL(MINUS(x1, x2)) = x2    10.66/3.10
POL(QUOT(x1, x2)) = x22 + x1·x2    10.66/3.10
POL(c2(x1)) = x1    10.66/3.10
POL(c4(x1, x2)) = x1 + x2    10.66/3.10
POL(c6(x1, x2)) = x1 + x2    10.66/3.10
POL(minus(x1, x2)) = x1    10.66/3.10
POL(pred(x1)) = x1    10.66/3.10
POL(quot(x1, x2)) = x1    10.66/3.10
POL(s(x1)) = [1] + x1   
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(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

pred(s(z0)) → z0 10.66/3.10
minus(z0, 0) → z0 10.66/3.10
minus(z0, s(z1)) → pred(minus(z0, z1)) 10.66/3.10
quot(0, s(z0)) → 0 10.66/3.10
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 10.66/3.10
log(s(0)) → 0 10.66/3.10
log(s(s(z0))) → s(log(s(quot(z0, s(s(0))))))
Tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 10.66/3.10
LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0)))) 10.66/3.10
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
S tuples:none
K tuples:

LOG(s(s(z0))) → c6(LOG(s(quot(z0, s(s(0))))), QUOT(z0, s(s(0)))) 10.66/3.10
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 10.66/3.10
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
Defined Rule Symbols:

pred, minus, quot, log

Defined Pair Symbols:

QUOT, LOG, MINUS

Compound Symbols:

c4, c6, c2

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

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

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10.66/3.20 EOF