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:

w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(x))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

w(r(z0)) → r(w(z0))
b(r(z0)) → r(b(z0))
b(w(z0)) → w(b(z0))

Tuples:

W(r(z0)) → c(W(z0))
B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))

S tuples:

W(r(z0)) → c(W(z0))
B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))

K tuples:none
Defined Rule Symbols:

w, b

Defined Pair Symbols:

W, B

Compound Symbols:

c, 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.

B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))

We considered the (Usable) Rules:

b(w(z0)) → w(b(z0))
w(r(z0)) → r(w(z0))

And the Tuples:

W(r(z0)) → c(W(z0))
B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))

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

POL(B(x₁)) = [2]x₁   
POL(W(x₁)) = [3]   
POL(b(x₁)) = [1] + [5]x₁   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(r(x₁)) = [3] + x₁   
POL(w(x₁)) = [4] + [4]x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

w(r(z0)) → r(w(z0))
b(r(z0)) → r(b(z0))
b(w(z0)) → w(b(z0))

Tuples:

W(r(z0)) → c(W(z0))
B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))

S tuples:

W(r(z0)) → c(W(z0))

K tuples:

B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))

Defined Rule Symbols:

w, b

Defined Pair Symbols:

W, B

Compound Symbols:

c, c1, 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.

W(r(z0)) → c(W(z0))

We considered the (Usable) Rules:

b(w(z0)) → w(b(z0))
w(r(z0)) → r(w(z0))

And the Tuples:

W(r(z0)) → c(W(z0))
B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))

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

POL(B(x₁)) = [4]x₁   
POL(W(x₁)) = [5] + [2]x₁   
POL(b(x₁)) = [1] + [4]x₁   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(r(x₁)) = [1] + x₁   
POL(w(x₁)) = [4] + [4]x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

w(r(z0)) → r(w(z0))
b(r(z0)) → r(b(z0))
b(w(z0)) → w(b(z0))

Tuples:

W(r(z0)) → c(W(z0))
B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))

S tuples:none
K tuples:

B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))
W(r(z0)) → c(W(z0))

Defined Rule Symbols:

w, b

Defined Pair Symbols:

W, B

Compound Symbols:

c, c1, c2

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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