### (0) Obligation:

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

q(0, s(y), s(z)) → 0
q(s(x), s(y), z) → q(x, y, z)
q(x, 0, s(z)) → s(q(x, s(z), s(z)))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

q(0, s(z0), s(z1)) → 0
q(s(z0), s(z1), z2) → q(z0, z1, z2)
q(z0, 0, s(z1)) → s(q(z0, s(z1), s(z1)))
Tuples:

Q(s(z0), s(z1), z2) → c1(Q(z0, z1, z2))
Q(z0, 0, s(z1)) → c2(Q(z0, s(z1), s(z1)))
S tuples:

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

q

Defined Pair Symbols:

Q

Compound Symbols:

c1, c2

### (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.

Q(s(z0), s(z1), z2) → c1(Q(z0, z1, z2))
We considered the (Usable) Rules:none
And the Tuples:

Q(s(z0), s(z1), z2) → c1(Q(z0, z1, z2))
Q(z0, 0, s(z1)) → c2(Q(z0, s(z1), s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(Q(x1, x2, x3)) = x1
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(s(x1)) =  + x1

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

q(0, s(z0), s(z1)) → 0
q(s(z0), s(z1), z2) → q(z0, z1, z2)
q(z0, 0, s(z1)) → s(q(z0, s(z1), s(z1)))
Tuples:

Q(s(z0), s(z1), z2) → c1(Q(z0, z1, z2))
Q(z0, 0, s(z1)) → c2(Q(z0, s(z1), s(z1)))
S tuples:

Q(z0, 0, s(z1)) → c2(Q(z0, s(z1), s(z1)))
K tuples:

Q(s(z0), s(z1), z2) → c1(Q(z0, z1, z2))
Defined Rule Symbols:

q

Defined Pair Symbols:

Q

Compound Symbols:

c1, c2

### (5) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

Q(z0, 0, s(z1)) → c2(Q(z0, s(z1), s(z1)))
Q(s(z0), s(z1), z2) → c1(Q(z0, z1, z2))
Now S is empty