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:

p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0) → p(0, n, m)
p(m, 0, 0) → m

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(z0, z1, s(z2)) → p(z0, z2, z1)
p(z0, s(z1), 0) → p(0, z1, z0)
p(z0, 0, 0) → z0

Tuples:

P(z0, z1, s(z2)) → c(P(z0, z2, z1))
P(z0, s(z1), 0) → c1(P(0, z1, z0))

S tuples:

P(z0, z1, s(z2)) → c(P(z0, z2, z1))
P(z0, s(z1), 0) → c1(P(0, z1, z0))

K tuples:none
Defined Rule Symbols:

p

Defined Pair Symbols:

P

Compound Symbols:

c, c1

(3) 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.

P(z0, z1, s(z2)) → c(P(z0, z2, z1))
P(z0, s(z1), 0) → c1(P(0, z1, z0))

We considered the (Usable) Rules:none
And the Tuples:

P(z0, z1, s(z2)) → c(P(z0, z2, z1))
P(z0, s(z1), 0) → c1(P(0, z1, z0))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [4]   
POL(P(x₁, x₂, x₃)) = [5]x₁ + [4]x₂ + [4]x₃   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(s(x₁)) = [4] + x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(z0, z1, s(z2)) → p(z0, z2, z1)
p(z0, s(z1), 0) → p(0, z1, z0)
p(z0, 0, 0) → z0

Tuples:

P(z0, z1, s(z2)) → c(P(z0, z2, z1))
P(z0, s(z1), 0) → c1(P(0, z1, z0))

S tuples:none
K tuples:

P(z0, z1, s(z2)) → c(P(z0, z2, z1))
P(z0, s(z1), 0) → c1(P(0, z1, z0))

Defined Rule Symbols:

p

Defined Pair Symbols:

P

Compound Symbols:

c, c1

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(6) BOUNDS(O(1), O(1))