YES(O(1), O(1)) 0.00/0.95 YES(O(1), O(1)) 0.00/0.97 0.00/0.97 0.00/0.97 0.00/0.97 0.00/0.97 0.00/0.97 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.97 0.00/0.97 0.00/0.97
0.00/0.97 0.00/0.97 0.00/0.97
0.00/0.97
0.00/0.97

(0) Obligation:

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

f(0) → cons(0, n__f(s(0))) 0.00/0.97
f(s(0)) → f(p(s(0))) 0.00/0.97
p(s(X)) → X 0.00/0.97
f(X) → n__f(X) 0.00/0.97
activate(n__f(X)) → f(X) 0.00/0.97
activate(X) → X

Rewrite Strategy: INNERMOST
0.00/0.97
0.00/0.97

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

Converted CpxTRS to CDT
0.00/0.97
0.00/0.97

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0, n__f(s(0))) 0.00/0.97
f(s(0)) → f(p(s(0))) 0.00/0.97
f(z0) → n__f(z0) 0.00/0.97
p(s(z0)) → z0 0.00/0.97
activate(n__f(z0)) → f(z0) 0.00/0.97
activate(z0) → z0
Tuples:

F(s(0)) → c1(F(p(s(0))), P(s(0))) 0.00/0.97
ACTIVATE(n__f(z0)) → c4(F(z0))
S tuples:

F(s(0)) → c1(F(p(s(0))), P(s(0))) 0.00/0.97
ACTIVATE(n__f(z0)) → c4(F(z0))
K tuples:none
Defined Rule Symbols:

f, p, activate

Defined Pair Symbols:

F, ACTIVATE

Compound Symbols:

c1, c4

0.00/0.97
0.00/0.97

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

Removed 1 trailing tuple parts
0.00/0.97
0.00/0.97

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0, n__f(s(0))) 0.00/0.97
f(s(0)) → f(p(s(0))) 0.00/0.97
f(z0) → n__f(z0) 0.00/0.97
p(s(z0)) → z0 0.00/0.97
activate(n__f(z0)) → f(z0) 0.00/0.97
activate(z0) → z0
Tuples:

ACTIVATE(n__f(z0)) → c4(F(z0)) 0.00/0.97
F(s(0)) → c1(F(p(s(0))))
S tuples:

ACTIVATE(n__f(z0)) → c4(F(z0)) 0.00/0.97
F(s(0)) → c1(F(p(s(0))))
K tuples:none
Defined Rule Symbols:

f, p, activate

Defined Pair Symbols:

ACTIVATE, F

Compound Symbols:

c4, c1

0.00/0.97
0.00/0.97

(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

ACTIVATE(n__f(z0)) → c4(F(z0))
0.00/0.97
0.00/0.97

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0, n__f(s(0))) 0.00/0.97
f(s(0)) → f(p(s(0))) 0.00/0.97
f(z0) → n__f(z0) 0.00/0.97
p(s(z0)) → z0 0.00/0.97
activate(n__f(z0)) → f(z0) 0.00/0.97
activate(z0) → z0
Tuples:

F(s(0)) → c1(F(p(s(0))))
S tuples:

F(s(0)) → c1(F(p(s(0))))
K tuples:none
Defined Rule Symbols:

f, p, activate

Defined Pair Symbols:

F

Compound Symbols:

c1

0.00/0.97
0.00/0.97

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

Use narrowing to replace F(s(0)) → c1(F(p(s(0)))) by

F(s(0)) → c1(F(0))
0.00/0.97
0.00/0.97

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0, n__f(s(0))) 0.00/0.97
f(s(0)) → f(p(s(0))) 0.00/0.97
f(z0) → n__f(z0) 0.00/0.97
p(s(z0)) → z0 0.00/0.97
activate(n__f(z0)) → f(z0) 0.00/0.97
activate(z0) → z0
Tuples:

F(s(0)) → c1(F(0))
S tuples:

F(s(0)) → c1(F(0))
K tuples:none
Defined Rule Symbols:

f, p, activate

Defined Pair Symbols:

F

Compound Symbols:

c1

0.00/0.97
0.00/0.97

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

Removed 1 trailing tuple parts
0.00/0.97
0.00/0.97

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0, n__f(s(0))) 0.00/0.97
f(s(0)) → f(p(s(0))) 0.00/0.97
f(z0) → n__f(z0) 0.00/0.97
p(s(z0)) → z0 0.00/0.97
activate(n__f(z0)) → f(z0) 0.00/0.97
activate(z0) → z0
Tuples:

F(s(0)) → c1
S tuples:

F(s(0)) → c1
K tuples:none
Defined Rule Symbols:

f, p, activate

Defined Pair Symbols:

F

Compound Symbols:

c1

0.00/0.97
0.00/0.97

(11) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing nodes:

F(s(0)) → c1
0.00/0.97
0.00/0.97

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0, n__f(s(0))) 0.00/0.97
f(s(0)) → f(p(s(0))) 0.00/0.97
f(z0) → n__f(z0) 0.00/0.97
p(s(z0)) → z0 0.00/0.97
activate(n__f(z0)) → f(z0) 0.00/0.97
activate(z0) → z0
Tuples:none
S tuples:none
K tuples:none
Defined Rule Symbols:

f, p, activate

Defined Pair Symbols:none

Compound Symbols:none

0.00/0.97
0.00/0.97

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

The set S is empty
0.00/0.97
0.00/0.97

(14) BOUNDS(O(1), O(1))

0.00/0.97
0.00/0.97
2.32/1.01 EOF