YES(O(1), O(n^1)) 0.00/0.82 YES(O(1), O(n^1)) 0.00/0.83 0.00/0.83 0.00/0.83 0.00/0.83 0.00/0.83 0.00/0.83 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.83 0.00/0.83 0.00/0.83
0.00/0.83 0.00/0.83 0.00/0.83
0.00/0.83
0.00/0.83

(0) Obligation:

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

f(s(x)) → s(s(f(p(s(x))))) 0.00/0.83
f(0) → 0 0.00/0.83
p(s(x)) → x

Rewrite Strategy: INNERMOST
0.00/0.83
0.00/0.83

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

Converted CpxTRS to CDT
0.00/0.83
0.00/0.83

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(s(z0)) → s(s(f(p(s(z0))))) 0.00/0.83
f(0) → 0 0.00/0.83
p(s(z0)) → z0
Tuples:

F(s(z0)) → c(F(p(s(z0))), P(s(z0)))
S tuples:

F(s(z0)) → c(F(p(s(z0))), P(s(z0)))
K tuples:none
Defined Rule Symbols:

f, p

Defined Pair Symbols:

F

Compound Symbols:

c

0.00/0.83
0.00/0.83

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

Removed 1 trailing tuple parts
0.00/0.83
0.00/0.83

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(s(z0)) → s(s(f(p(s(z0))))) 0.00/0.83
f(0) → 0 0.00/0.83
p(s(z0)) → z0
Tuples:

F(s(z0)) → c(F(p(s(z0))))
S tuples:

F(s(z0)) → c(F(p(s(z0))))
K tuples:none
Defined Rule Symbols:

f, p

Defined Pair Symbols:

F

Compound Symbols:

c

0.00/0.83
0.00/0.83

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

Use narrowing to replace F(s(z0)) → c(F(p(s(z0)))) by

F(s(z0)) → c(F(z0))
0.00/0.83
0.00/0.83

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(s(z0)) → s(s(f(p(s(z0))))) 0.00/0.83
f(0) → 0 0.00/0.83
p(s(z0)) → z0
Tuples:

F(s(z0)) → c(F(z0))
S tuples:

F(s(z0)) → c(F(z0))
K tuples:none
Defined Rule Symbols:

f, p

Defined Pair Symbols:

F

Compound Symbols:

c

0.00/0.83
0.00/0.83

(7) 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)) → c(F(z0))
We considered the (Usable) Rules:none
And the Tuples:

F(s(z0)) → c(F(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.83

POL(F(x1)) = [3]x1    0.00/0.83
POL(c(x1)) = x1    0.00/0.83
POL(s(x1)) = [1] + x1   
0.00/0.83
0.00/0.83

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(s(z0)) → s(s(f(p(s(z0))))) 0.00/0.83
f(0) → 0 0.00/0.83
p(s(z0)) → z0
Tuples:

F(s(z0)) → c(F(z0))
S tuples:none
K tuples:

F(s(z0)) → c(F(z0))
Defined Rule Symbols:

f, p

Defined Pair Symbols:

F

Compound Symbols:

c

0.00/0.83
0.00/0.83

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

The set S is empty
0.00/0.83
0.00/0.83

(10) BOUNDS(O(1), O(1))

0.00/0.83
0.00/0.83
0.00/0.84 EOF