The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^1).

(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(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

(8) BOUNDS(1, 1)