YES(O(1), O(n^2)) 0.00/0.83 YES(O(1), O(n^2)) 0.00/0.84 0.00/0.84 0.00/0.84 0.00/0.84 0.00/0.84 0.00/0.84 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.84 0.00/0.84 0.00/0.84
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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))
0.00/0.84 
↳2 CdtProblem
0.00/0.84 
↳3 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
0.00/0.84 
↳4 CdtProblem
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↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
0.00/0.84 
↳6 CdtProblem
0.00/0.84 
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
0.00/0.84 
↳8 CdtProblem
0.00/0.84 
↳9 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳10 BOUNDS(O(1), O(1))
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0.00/0.84 0.00/0.84
0.00/0.84
0.00/0.84

(0) Obligation:

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

f(0) → true 0.00/0.84
f(1) → false 0.00/0.84
f(s(x)) → f(x) 0.00/0.84
if(true, s(x), s(y)) → s(x) 0.00/0.84
if(false, s(x), s(y)) → s(y) 0.00/0.84
g(x, c(y)) → c(g(x, y)) 0.00/0.84
g(x, c(y)) → g(x, if(f(x), c(g(s(x), y)), c(y)))

Rewrite Strategy: INNERMOST
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0.00/0.84

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

Converted CpxTRS to CDT
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0.00/0.84

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → true 0.00/0.84
f(1) → false 0.00/0.84
f(s(z0)) → f(z0) 0.00/0.84
if(true, s(z0), s(z1)) → s(z0) 0.00/0.84
if(false, s(z0), s(z1)) → s(z1) 0.00/0.84
g(z0, c(z1)) → c(g(z0, z1)) 0.00/0.84
g(z0, c(z1)) → g(z0, if(f(z0), c(g(s(z0), z1)), c(z1)))
Tuples:

F(s(z0)) → c3(F(z0)) 0.00/0.84
G(z0, c(z1)) → c6(G(z0, z1)) 0.00/0.84
G(z0, c(z1)) → c7(G(z0, if(f(z0), c(g(s(z0), z1)), c(z1))), IF(f(z0), c(g(s(z0), z1)), c(z1)), F(z0), G(s(z0), z1))
S tuples:

F(s(z0)) → c3(F(z0)) 0.00/0.84
G(z0, c(z1)) → c6(G(z0, z1)) 0.00/0.84
G(z0, c(z1)) → c7(G(z0, if(f(z0), c(g(s(z0), z1)), c(z1))), IF(f(z0), c(g(s(z0), z1)), c(z1)), F(z0), G(s(z0), z1))
K tuples:none
Defined Rule Symbols:

f, if, g

Defined Pair Symbols:

F, G

Compound Symbols:

c3, c6, c7

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0.00/0.84

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

Removed 2 trailing tuple parts
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0.00/0.84

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → true 0.00/0.84
f(1) → false 0.00/0.84
f(s(z0)) → f(z0) 0.00/0.84
if(true, s(z0), s(z1)) → s(z0) 0.00/0.84
if(false, s(z0), s(z1)) → s(z1) 0.00/0.84
g(z0, c(z1)) → c(g(z0, z1)) 0.00/0.84
g(z0, c(z1)) → g(z0, if(f(z0), c(g(s(z0), z1)), c(z1)))
Tuples:

F(s(z0)) → c3(F(z0)) 0.00/0.84
G(z0, c(z1)) → c6(G(z0, z1)) 0.00/0.84
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))
S tuples:

F(s(z0)) → c3(F(z0)) 0.00/0.84
G(z0, c(z1)) → c6(G(z0, z1)) 0.00/0.84
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))
K tuples:none
Defined Rule Symbols:

f, if, g

Defined Pair Symbols:

F, G

Compound Symbols:

c3, c6, c7

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0.00/0.84

(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.

G(z0, c(z1)) → c6(G(z0, z1)) 0.00/0.84
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))
We considered the (Usable) Rules:none
And the Tuples:

F(s(z0)) → c3(F(z0)) 0.00/0.84
G(z0, c(z1)) → c6(G(z0, z1)) 0.00/0.84
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.84

POL(F(x1)) = [5]    0.00/0.84
POL(G(x1, x2)) = [4]x2    0.00/0.84
POL(c(x1)) = [4] + x1    0.00/0.84
POL(c3(x1)) = x1    0.00/0.84
POL(c6(x1)) = x1    0.00/0.84
POL(c7(x1, x2)) = x1 + x2    0.00/0.84
POL(s(x1)) = 0   
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0.00/0.84

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → true 0.00/0.84
f(1) → false 0.00/0.84
f(s(z0)) → f(z0) 0.00/0.84
if(true, s(z0), s(z1)) → s(z0) 0.00/0.84
if(false, s(z0), s(z1)) → s(z1) 0.00/0.84
g(z0, c(z1)) → c(g(z0, z1)) 0.00/0.84
g(z0, c(z1)) → g(z0, if(f(z0), c(g(s(z0), z1)), c(z1)))
Tuples:

F(s(z0)) → c3(F(z0)) 0.00/0.84
G(z0, c(z1)) → c6(G(z0, z1)) 0.00/0.84
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))
S tuples:

F(s(z0)) → c3(F(z0))
K tuples:

G(z0, c(z1)) → c6(G(z0, z1)) 0.00/0.84
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))
Defined Rule Symbols:

f, if, g

Defined Pair Symbols:

F, G

Compound Symbols:

c3, c6, c7

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0.00/0.84

(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.

F(s(z0)) → c3(F(z0))
We considered the (Usable) Rules:none
And the Tuples:

F(s(z0)) → c3(F(z0)) 0.00/0.84
G(z0, c(z1)) → c6(G(z0, z1)) 0.00/0.84
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.84

POL(F(x1)) = x1    0.00/0.84
POL(G(x1, x2)) = [2]x22 + x1·x2    0.00/0.84
POL(c(x1)) = [1] + x1    0.00/0.84
POL(c3(x1)) = x1    0.00/0.84
POL(c6(x1)) = x1    0.00/0.84
POL(c7(x1, x2)) = x1 + x2    0.00/0.84
POL(s(x1)) = [1] + x1   
0.00/0.84
0.00/0.84

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → true 0.00/0.84
f(1) → false 0.00/0.84
f(s(z0)) → f(z0) 0.00/0.84
if(true, s(z0), s(z1)) → s(z0) 0.00/0.84
if(false, s(z0), s(z1)) → s(z1) 0.00/0.84
g(z0, c(z1)) → c(g(z0, z1)) 0.00/0.84
g(z0, c(z1)) → g(z0, if(f(z0), c(g(s(z0), z1)), c(z1)))
Tuples:

F(s(z0)) → c3(F(z0)) 0.00/0.84
G(z0, c(z1)) → c6(G(z0, z1)) 0.00/0.84
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))
S tuples:none
K tuples:

G(z0, c(z1)) → c6(G(z0, z1)) 0.00/0.84
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1)) 0.00/0.84
F(s(z0)) → c3(F(z0))
Defined Rule Symbols:

f, if, g

Defined Pair Symbols:

F, G

Compound Symbols:

c3, c6, c7

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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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0.00/0.88 EOF