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:

b(a, f(b(b(z, y), a))) → z
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(z, z))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

b(a, f(b(b(z0, z1), a))) → z0
c(c(z0, z1, a), a, z2) → f(f(c(z2, a, f(c(z0, z2, z1)))))
f(f(c(a, z0, z1))) → b(z0, b(z1, z1))

Tuples:

C(c(z0, z1, a), a, z2) → c2(F(f(c(z2, a, f(c(z0, z2, z1))))), F(c(z2, a, f(c(z0, z2, z1)))), C(z2, a, f(c(z0, z2, z1))), F(c(z0, z2, z1)), C(z0, z2, z1))
F(f(c(a, z0, z1))) → c3(B(z0, b(z1, z1)), B(z1, z1))

S tuples:

C(c(z0, z1, a), a, z2) → c2(F(f(c(z2, a, f(c(z0, z2, z1))))), F(c(z2, a, f(c(z0, z2, z1)))), C(z2, a, f(c(z0, z2, z1))), F(c(z0, z2, z1)), C(z0, z2, z1))
F(f(c(a, z0, z1))) → c3(B(z0, b(z1, z1)), B(z1, z1))

K tuples:none
Defined Rule Symbols:

b, c, f

Defined Pair Symbols:

C, F

Compound Symbols:

c2, c3

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

Removed 1 trailing nodes:

F(f(c(a, z0, z1))) → c3(B(z0, b(z1, z1)), B(z1, z1))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

b(a, f(b(b(z0, z1), a))) → z0
c(c(z0, z1, a), a, z2) → f(f(c(z2, a, f(c(z0, z2, z1)))))
f(f(c(a, z0, z1))) → b(z0, b(z1, z1))

Tuples:

C(c(z0, z1, a), a, z2) → c2(F(f(c(z2, a, f(c(z0, z2, z1))))), F(c(z2, a, f(c(z0, z2, z1)))), C(z2, a, f(c(z0, z2, z1))), F(c(z0, z2, z1)), C(z0, z2, z1))

S tuples:

C(c(z0, z1, a), a, z2) → c2(F(f(c(z2, a, f(c(z0, z2, z1))))), F(c(z2, a, f(c(z0, z2, z1)))), C(z2, a, f(c(z0, z2, z1))), F(c(z0, z2, z1)), C(z0, z2, z1))

K tuples:none
Defined Rule Symbols:

b, c, f

Defined Pair Symbols:

C

Compound Symbols:

c2

(5) 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.

C(c(z0, z1, a), a, z2) → c2(F(f(c(z2, a, f(c(z0, z2, z1))))), F(c(z2, a, f(c(z0, z2, z1)))), C(z2, a, f(c(z0, z2, z1))), F(c(z0, z2, z1)), C(z0, z2, z1))

We considered the (Usable) Rules:

c(c(z0, z1, a), a, z2) → f(f(c(z2, a, f(c(z0, z2, z1)))))
f(f(c(a, z0, z1))) → b(z0, b(z1, z1))

And the Tuples:

C(c(z0, z1, a), a, z2) → c2(F(f(c(z2, a, f(c(z0, z2, z1))))), F(c(z2, a, f(c(z0, z2, z1)))), C(z2, a, f(c(z0, z2, z1))), F(c(z0, z2, z1)), C(z0, z2, z1))

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

POL(C(x₁, x₂, x₃)) = [2]x₁ + x₂ + [4]x₃   
POL(F(x₁)) = 0   
POL(a) = 0   
POL(b(x₁, x₂)) = 0   
POL(c(x₁, x₂, x₃)) = [1] + [2]x₁ + [4]x₂   
POL(c2(x₁, x₂, x₃, x₄, x₅)) = x₁ + x₂ + x₃ + x₄ + x₅   
POL(f(x₁)) = 0   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

b(a, f(b(b(z0, z1), a))) → z0
c(c(z0, z1, a), a, z2) → f(f(c(z2, a, f(c(z0, z2, z1)))))
f(f(c(a, z0, z1))) → b(z0, b(z1, z1))

Tuples:

C(c(z0, z1, a), a, z2) → c2(F(f(c(z2, a, f(c(z0, z2, z1))))), F(c(z2, a, f(c(z0, z2, z1)))), C(z2, a, f(c(z0, z2, z1))), F(c(z0, z2, z1)), C(z0, z2, z1))

S tuples:none
K tuples:

C(c(z0, z1, a), a, z2) → c2(F(f(c(z2, a, f(c(z0, z2, z1))))), F(c(z2, a, f(c(z0, z2, z1)))), C(z2, a, f(c(z0, z2, z1))), F(c(z0, z2, z1)), C(z0, z2, z1))

Defined Rule Symbols:

b, c, f

Defined Pair Symbols:

C

Compound Symbols:

c2

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(8) BOUNDS(O(1), O(1))