### (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 Cpx (relative) TRS 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))
P(z0, 0, 0) → c2
S tuples:

P(z0, z1, s(z2)) → c(P(z0, z2, z1))
P(z0, s(z1), 0) → c1(P(0, z1, z0))
P(z0, 0, 0) → c2
K tuples:none
Defined Rule Symbols:

p

Defined Pair Symbols:

P

Compound Symbols:

c, c1, c2

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

Removed 1 trailing nodes:

P(z0, 0, 0) → c2

### (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:

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

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
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:none

Defined Pair Symbols:

P

Compound Symbols:

c, c1

### (7) 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.

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

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
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:none

Defined Pair Symbols:

P

Compound Symbols:

c, c1

### (9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty