The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

0 CpxTRS
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
↳2 CdtProblem
↳3 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳4 CdtProblem
↳5 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
↳6 CdtProblem
↳7 CdtUsableRulesProof (⇔, 0 ms)
↳8 CdtProblem
↳9 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 0 ms)
↳10 CdtProblem
↳11 CdtUsableRulesProof (⇔, 0 ms)
↳12 CdtProblem
↳13 CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))), 0 ms)
↳14 CdtProblem
↳15 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
↳16 BOUNDS(O(1), O(1))

(0) Obligation:

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

f(s(x)) → s(s(f(p(s(x)))))
f(0) → 0
p(s(x)) → x

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(s(z0)) → s(s(f(p(s(z0)))))
f(0) → 0
p(s(z0)) → z0
Tuples:

F(s(z0)) → c(F(p(s(z0))), P(s(z0)))
F(0) → c1
P(s(z0)) → c2
S tuples:

F(s(z0)) → c(F(p(s(z0))), P(s(z0)))
F(0) → c1
P(s(z0)) → c2
K tuples:none
Defined Rule Symbols:

f, p

Defined Pair Symbols:

F, P

Compound Symbols:

c, c1, c2

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

Removed 2 trailing nodes:

F(0) → c1
P(s(z0)) → c2

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(s(z0)) → s(s(f(p(s(z0)))))
f(0) → 0
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

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

Removed 1 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(s(z0)) → s(s(f(p(s(z0)))))
f(0) → 0
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

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(s(z0)) → s(s(f(p(s(z0)))))
f(0) → 0

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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:

p

Defined Pair Symbols:

F

Compound Symbols:

c

(9) 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))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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:

p

Defined Pair Symbols:

F

Compound Symbols:

c

(11) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

p(s(z0)) → z0

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(s(z0)) → c(F(z0))
S tuples:

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

Defined Pair Symbols:

F

Compound Symbols:

c

(13) CdtRuleRemovalProof (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 :

POL(F(x1)) = [5]x1   
POL(c(x1)) = x1   
POL(s(x1)) = [1] + x1   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

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

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

Defined Pair Symbols:

F

Compound Symbols:

c

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

The set S is empty

(16) BOUNDS(O(1), O(1))