YES(O(1), O(n^2)) 0.00/0.96 YES(O(1), O(n^2)) 0.00/1.00 0.00/1.00 0.00/1.00 0.00/1.00 0.00/1.00 0.00/1.00 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/1.00 0.00/1.00 0.00/1.00
0.00/1.00
0.00/1.00
0.00/1.00
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

0 CpxTRS
0.00/1.00 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
0.00/1.00 
↳2 CdtProblem
0.00/1.00 
↳3 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
0.00/1.00 
↳4 CdtProblem
0.00/1.00 
↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
0.00/1.00 
↳6 CdtProblem
0.00/1.00 
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
0.00/1.00 
↳8 CdtProblem
0.00/1.00 
↳9 SIsEmptyProof (BOTH BOUNDS(ID, ID))
0.00/1.00 
↳10 BOUNDS(O(1), O(1))
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0.00/1.00 0.00/1.00
0.00/1.00
0.00/1.00

(0) Obligation:

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

pred(s(x)) → x 0.00/1.00
minus(x, 0) → x 0.00/1.00
minus(x, s(y)) → pred(minus(x, y)) 0.00/1.00
quot(0, s(y)) → 0 0.00/1.00
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))

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

Converted CpxTRS to CDT
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0.00/1.00

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

pred(s(z0)) → z0 0.00/1.00
minus(z0, 0) → z0 0.00/1.00
minus(z0, s(z1)) → pred(minus(z0, z1)) 0.00/1.00
quot(0, s(z0)) → 0 0.00/1.00
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

MINUS(z0, s(z1)) → c2(PRED(minus(z0, z1)), MINUS(z0, z1)) 0.00/1.00
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:

MINUS(z0, s(z1)) → c2(PRED(minus(z0, z1)), MINUS(z0, z1)) 0.00/1.00
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

pred, minus, quot

Defined Pair Symbols:

MINUS, QUOT

Compound Symbols:

c2, c4

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

Removed 1 trailing tuple parts
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0.00/1.00

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

pred(s(z0)) → z0 0.00/1.00
minus(z0, 0) → z0 0.00/1.00
minus(z0, s(z1)) → pred(minus(z0, z1)) 0.00/1.00
quot(0, s(z0)) → 0 0.00/1.00
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 0.00/1.00
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)) 0.00/1.00
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

pred, minus, quot

Defined Pair Symbols:

QUOT, MINUS

Compound Symbols:

c4, c2

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0.00/1.00

(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.

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

minus(z0, 0) → z0 0.00/1.00
minus(z0, s(z1)) → pred(minus(z0, z1)) 0.00/1.00
pred(s(z0)) → z0
And the Tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 0.00/1.00
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/1.00

POL(0) = 0    0.00/1.00
POL(MINUS(x1, x2)) = [1]    0.00/1.00
POL(QUOT(x1, x2)) = [2]x1    0.00/1.00
POL(c2(x1)) = x1    0.00/1.00
POL(c4(x1, x2)) = x1 + x2    0.00/1.00
POL(minus(x1, x2)) = x1    0.00/1.00
POL(pred(x1)) = x1    0.00/1.00
POL(s(x1)) = [1] + x1   
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

pred(s(z0)) → z0 0.00/1.00
minus(z0, 0) → z0 0.00/1.00
minus(z0, s(z1)) → pred(minus(z0, z1)) 0.00/1.00
quot(0, s(z0)) → 0 0.00/1.00
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 0.00/1.00
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
S tuples:

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

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:

pred, minus, quot

Defined Pair Symbols:

QUOT, MINUS

Compound Symbols:

c4, 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.

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

minus(z0, 0) → z0 0.00/1.00
minus(z0, s(z1)) → pred(minus(z0, z1)) 0.00/1.00
pred(s(z0)) → z0
And the Tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 0.00/1.00
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/1.00

POL(0) = [3]    0.00/1.00
POL(MINUS(x1, x2)) = [1] + x2    0.00/1.00
POL(QUOT(x1, x2)) = x1·x2    0.00/1.00
POL(c2(x1)) = x1    0.00/1.00
POL(c4(x1, x2)) = x1 + x2    0.00/1.00
POL(minus(x1, x2)) = x1    0.00/1.00
POL(pred(x1)) = x1    0.00/1.00
POL(s(x1)) = [1] + x1   
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0.00/1.00

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

pred(s(z0)) → z0 0.00/1.00
minus(z0, 0) → z0 0.00/1.00
minus(z0, s(z1)) → pred(minus(z0, z1)) 0.00/1.00
quot(0, s(z0)) → 0 0.00/1.00
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 0.00/1.00
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
S tuples:none
K tuples:

QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 0.00/1.00
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
Defined Rule Symbols:

pred, minus, quot

Defined Pair Symbols:

QUOT, MINUS

Compound Symbols:

c4, c2

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(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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2.45/1.06 EOF