YES(O(1), O(n^1)) 2.34/1.02 YES(O(1), O(n^1)) 2.34/1.04 2.34/1.04 2.34/1.04 2.34/1.04 2.34/1.04 2.34/1.04 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 2.34/1.04 2.34/1.04 2.34/1.04
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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))
2.34/1.04 
↳2 CdtProblem
2.34/1.04 
↳3 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
2.34/1.04 
↳4 CdtProblem
2.34/1.04 
↳5 CdtNarrowingProof (BOTH BOUNDS(ID, ID))
2.34/1.04 
↳6 CdtProblem
2.34/1.04 
↳7 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
2.34/1.04 
↳8 CdtProblem
2.34/1.04 
↳9 CdtLeafRemovalProof (ComplexityIfPolyImplication)
2.34/1.04 
↳10 CdtProblem
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↳11 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
2.34/1.04 
↳12 CdtProblem
2.34/1.04 
↳13 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳14 BOUNDS(O(1), O(1))
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2.34/1.04

(0) Obligation:

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

f(x, 0) → s(0) 2.34/1.04
f(s(x), s(y)) → s(f(x, y)) 2.34/1.04
g(0, x) → g(f(x, x), x)

Rewrite Strategy: INNERMOST
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2.34/1.04

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

Converted CpxTRS to CDT
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(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, 0) → s(0) 2.34/1.04
f(s(z0), s(z1)) → s(f(z0, z1)) 2.34/1.04
g(0, z0) → g(f(z0, z0), z0)
Tuples:

F(s(z0), s(z1)) → c1(F(z0, z1)) 2.34/1.04
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
S tuples:

F(s(z0), s(z1)) → c1(F(z0, z1)) 2.34/1.04
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

2.34/1.04
2.34/1.04

(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.

G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
We considered the (Usable) Rules:

f(z0, 0) → s(0) 2.34/1.04
f(s(z0), s(z1)) → s(f(z0, z1))
And the Tuples:

F(s(z0), s(z1)) → c1(F(z0, z1)) 2.34/1.05
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 2.34/1.05

POL(0) = [4]    2.34/1.05
POL(F(x1, x2)) = [1]    2.34/1.05
POL(G(x1, x2)) = [4]x1    2.34/1.05
POL(c1(x1)) = x1    2.34/1.05
POL(c2(x1, x2)) = x1 + x2    2.34/1.05
POL(f(x1, x2)) = 0    2.34/1.05
POL(s(x1)) = 0   
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, 0) → s(0) 2.34/1.05
f(s(z0), s(z1)) → s(f(z0, z1)) 2.34/1.05
g(0, z0) → g(f(z0, z0), z0)
Tuples:

F(s(z0), s(z1)) → c1(F(z0, z1)) 2.34/1.05
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
S tuples:

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

G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

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(5) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0)) by

G(0, 0) → c2(G(s(0), 0), F(0, 0)) 2.34/1.05
G(0, s(z0)) → c2(G(s(f(z0, z0)), s(z0)), F(s(z0), s(z0)))
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, 0) → s(0) 2.34/1.05
f(s(z0), s(z1)) → s(f(z0, z1)) 2.34/1.05
g(0, z0) → g(f(z0, z0), z0)
Tuples:

F(s(z0), s(z1)) → c1(F(z0, z1)) 2.34/1.05
G(0, 0) → c2(G(s(0), 0), F(0, 0)) 2.34/1.05
G(0, s(z0)) → c2(G(s(f(z0, z0)), s(z0)), F(s(z0), s(z0)))
S tuples:

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

G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

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

Removed 3 trailing tuple parts
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(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, 0) → s(0) 2.34/1.05
f(s(z0), s(z1)) → s(f(z0, z1)) 2.34/1.05
g(0, z0) → g(f(z0, z0), z0)
Tuples:

F(s(z0), s(z1)) → c1(F(z0, z1)) 2.34/1.05
G(0, 0) → c2 2.34/1.05
G(0, s(z0)) → c2(F(s(z0), s(z0)))
S tuples:

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

G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2, c2

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(9) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

G(0, s(z0)) → c2(F(s(z0), s(z0)))
Removed 1 trailing nodes:

G(0, 0) → c2
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(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, 0) → s(0) 2.34/1.05
f(s(z0), s(z1)) → s(f(z0, z1)) 2.34/1.05
g(0, z0) → g(f(z0, z0), z0)
Tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))
S tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F

Compound Symbols:

c1

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(11) 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.

F(s(z0), s(z1)) → c1(F(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 2.34/1.05

POL(F(x1, x2)) = [2]x2    2.34/1.05
POL(c1(x1)) = x1    2.34/1.05
POL(s(x1)) = [1] + x1   
2.34/1.05
2.34/1.05

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, 0) → s(0) 2.34/1.05
f(s(z0), s(z1)) → s(f(z0, z1)) 2.34/1.05
g(0, z0) → g(f(z0, z0), z0)
Tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))
S tuples:none
K tuples:

F(s(z0), s(z1)) → c1(F(z0, z1))
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F

Compound Symbols:

c1

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

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

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2.34/1.10 EOF