(0) Obligation:

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

p(s(x)) → x
fac(0) → s(0)
fac(s(x)) → times(s(x), fac(p(s(x))))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0
fac(0) → s(0)
fac(s(z0)) → times(s(z0), fac(p(s(z0))))
Tuples:

FAC(s(z0)) → c2(FAC(p(s(z0))), P(s(z0)))
S tuples:

FAC(s(z0)) → c2(FAC(p(s(z0))), P(s(z0)))
K tuples:none
Defined Rule Symbols:

p, fac

Defined Pair Symbols:

FAC

Compound Symbols:

c2

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

Removed 1 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0
fac(0) → s(0)
fac(s(z0)) → times(s(z0), fac(p(s(z0))))
Tuples:

FAC(s(z0)) → c2(FAC(p(s(z0))))
S tuples:

FAC(s(z0)) → c2(FAC(p(s(z0))))
K tuples:none
Defined Rule Symbols:

p, fac

Defined Pair Symbols:

FAC

Compound Symbols:

c2

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

Use narrowing to replace FAC(s(z0)) → c2(FAC(p(s(z0)))) by

FAC(s(z0)) → c2(FAC(z0))
FAC(s(x0)) → c2

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0
fac(0) → s(0)
fac(s(z0)) → times(s(z0), fac(p(s(z0))))
Tuples:

FAC(s(z0)) → c2(FAC(z0))
FAC(s(x0)) → c2
S tuples:

FAC(s(z0)) → c2(FAC(z0))
FAC(s(x0)) → c2
K tuples:none
Defined Rule Symbols:

p, fac

Defined Pair Symbols:

FAC

Compound Symbols:

c2, c2

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

Removed 1 trailing nodes:

FAC(s(x0)) → c2

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0
fac(0) → s(0)
fac(s(z0)) → times(s(z0), fac(p(s(z0))))
Tuples:

FAC(s(z0)) → c2(FAC(z0))
S tuples:

FAC(s(z0)) → c2(FAC(z0))
K tuples:none
Defined Rule Symbols:

p, fac

Defined Pair Symbols:

FAC

Compound Symbols:

c2

(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.

FAC(s(z0)) → c2(FAC(z0))
We considered the (Usable) Rules:none
And the Tuples:

FAC(s(z0)) → c2(FAC(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(FAC(x1)) = [5]x1   
POL(c2(x1)) = x1   
POL(s(x1)) = [1] + x1   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0
fac(0) → s(0)
fac(s(z0)) → times(s(z0), fac(p(s(z0))))
Tuples:

FAC(s(z0)) → c2(FAC(z0))
S tuples:none
K tuples:

FAC(s(z0)) → c2(FAC(z0))
Defined Rule Symbols:

p, fac

Defined Pair Symbols:

FAC

Compound Symbols:

c2

(11) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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