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

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

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

Removed 1 trailing tuple parts

(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))))

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

(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))
F(s(x0)) → c

(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(z0))
F(s(x0)) → c

S tuples:

F(s(z0)) → c(F(z0))
F(s(x0)) → c

K tuples:none
Defined Rule Symbols:

f, p

Defined Pair Symbols:

F

Compound Symbols:

c, c

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

Removed 1 trailing nodes:

F(s(x0)) → c

(8) 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(z0))

S tuples:

F(s(z0)) → c(F(z0))

K tuples:none
Defined Rule Symbols:

f, p

Defined Pair Symbols:

F

Compound Symbols:

c

(9) 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 :

POL(F(x₁)) = x₁   
POL(c(x₁)) = x₁   
POL(s(x₁)) = [2] + x₁   

(10) 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(z0))

S tuples:none
K tuples:

F(s(z0)) → c(F(z0))

Defined Rule Symbols:

f, p

Defined Pair Symbols:

F

Compound Symbols:

c

(11) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(12) BOUNDS(O(1), O(1))