### (0) Obligation:

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

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

:(:(z0, z1), z2) → :(z0, :(z1, z2))
:(+(z0, z1), z2) → +(:(z0, z2), :(z1, z2))
:(z0, +(z1, f(z2))) → :(g(z0, z2), +(z1, a))
Tuples:

:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
:'(z0, +(z1, f(z2))) → c2(:'(g(z0, z2), +(z1, a)))
S tuples:

:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
:'(z0, +(z1, f(z2))) → c2(:'(g(z0, z2), +(z1, a)))
K tuples:none
Defined Rule Symbols:

:

Defined Pair Symbols:

:'

Compound Symbols:

c, c1, c2

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

Removed 1 trailing nodes:

:'(z0, +(z1, f(z2))) → c2(:'(g(z0, z2), +(z1, a)))

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

:(:(z0, z1), z2) → :(z0, :(z1, z2))
:(+(z0, z1), z2) → +(:(z0, z2), :(z1, z2))
:(z0, +(z1, f(z2))) → :(g(z0, z2), +(z1, a))
Tuples:

:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
S tuples:

:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
K tuples:none
Defined Rule Symbols:

:

Defined Pair Symbols:

:'

Compound Symbols:

c, c1

### (5) 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.

:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
We considered the (Usable) Rules:none
And the Tuples:

:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(+(x1, x2)) = [4] + x1 + x2
POL(:(x1, x2)) = [4] + x1 + [2]x2
POL(:'(x1, x2)) = [2]x1
POL(a) = [3]
POL(c(x1, x2)) = x1 + x2
POL(c1(x1, x2)) = x1 + x2
POL(f(x1)) = [4]
POL(g(x1, x2)) = [5]

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

:(:(z0, z1), z2) → :(z0, :(z1, z2))
:(+(z0, z1), z2) → +(:(z0, z2), :(z1, z2))
:(z0, +(z1, f(z2))) → :(g(z0, z2), +(z1, a))
Tuples:

:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
S tuples:none
K tuples:

:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
Defined Rule Symbols:

:

Defined Pair Symbols:

:'

Compound Symbols:

c, c1

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

The set S is empty