YES(O(1), O(n^1)) 0.00/0.77 YES(O(1), O(n^1)) 0.00/0.78 0.00/0.78 0.00/0.78 0.00/0.78 0.00/0.78 0.00/0.78 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.78 0.00/0.78 0.00/0.78
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The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

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 CdtKnowledgeProof (⇔)
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↳6 BOUNDS(O(1), O(1))
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(0) Obligation:

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

div(0, y) → 0 0.00/0.78
div(x, y) → quot(x, y, y) 0.00/0.78
quot(0, s(y), z) → 0 0.00/0.78
quot(s(x), s(y), z) → quot(x, y, z) 0.00/0.78
quot(x, 0, s(z)) → s(div(x, s(z)))

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:

div(0, z0) → 0 0.00/0.78
div(z0, z1) → quot(z0, z1, z1) 0.00/0.78
quot(0, s(z0), z1) → 0 0.00/0.78
quot(s(z0), s(z1), z2) → quot(z0, z1, z2) 0.00/0.78
quot(z0, 0, s(z1)) → s(div(z0, s(z1)))
Tuples:

DIV(z0, z1) → c1(QUOT(z0, z1, z1)) 0.00/0.78
QUOT(s(z0), s(z1), z2) → c3(QUOT(z0, z1, z2)) 0.00/0.78
QUOT(z0, 0, s(z1)) → c4(DIV(z0, s(z1)))
S tuples:

DIV(z0, z1) → c1(QUOT(z0, z1, z1)) 0.00/0.78
QUOT(s(z0), s(z1), z2) → c3(QUOT(z0, z1, z2)) 0.00/0.78
QUOT(z0, 0, s(z1)) → c4(DIV(z0, s(z1)))
K tuples:none
Defined Rule Symbols:

div, quot

Defined Pair Symbols:

DIV, QUOT

Compound Symbols:

c1, c3, c4

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(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.

QUOT(s(z0), s(z1), z2) → c3(QUOT(z0, z1, z2))
We considered the (Usable) Rules:none
And the Tuples:

DIV(z0, z1) → c1(QUOT(z0, z1, z1)) 0.00/0.78
QUOT(s(z0), s(z1), z2) → c3(QUOT(z0, z1, z2)) 0.00/0.78
QUOT(z0, 0, s(z1)) → c4(DIV(z0, s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.78

POL(0) = [1]    0.00/0.78
POL(DIV(x1, x2)) = [2]x1    0.00/0.78
POL(QUOT(x1, x2, x3)) = [2]x1    0.00/0.78
POL(c1(x1)) = x1    0.00/0.78
POL(c3(x1)) = x1    0.00/0.78
POL(c4(x1)) = x1    0.00/0.78
POL(s(x1)) = [1] + x1   
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

div(0, z0) → 0 0.00/0.78
div(z0, z1) → quot(z0, z1, z1) 0.00/0.78
quot(0, s(z0), z1) → 0 0.00/0.78
quot(s(z0), s(z1), z2) → quot(z0, z1, z2) 0.00/0.78
quot(z0, 0, s(z1)) → s(div(z0, s(z1)))
Tuples:

DIV(z0, z1) → c1(QUOT(z0, z1, z1)) 0.00/0.78
QUOT(s(z0), s(z1), z2) → c3(QUOT(z0, z1, z2)) 0.00/0.78
QUOT(z0, 0, s(z1)) → c4(DIV(z0, s(z1)))
S tuples:

DIV(z0, z1) → c1(QUOT(z0, z1, z1)) 0.00/0.78
QUOT(z0, 0, s(z1)) → c4(DIV(z0, s(z1)))
K tuples:

QUOT(s(z0), s(z1), z2) → c3(QUOT(z0, z1, z2))
Defined Rule Symbols:

div, quot

Defined Pair Symbols:

DIV, QUOT

Compound Symbols:

c1, c3, c4

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(5) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

QUOT(z0, 0, s(z1)) → c4(DIV(z0, s(z1))) 0.00/0.78
DIV(z0, z1) → c1(QUOT(z0, z1, z1)) 0.00/0.78
QUOT(s(z0), s(z1), z2) → c3(QUOT(z0, z1, z2))
Now S is empty
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(6) BOUNDS(O(1), O(1))

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0.00/0.79 EOF