The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

(0) Obligation:

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

p(a(x0), p(a(b(x1)), x2)) → p(a(b(a(x2))), p(a(a(x1)), x2))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(a(z0), p(a(b(z1)), z2)) → p(a(b(a(z2))), p(a(a(z1)), z2))

Tuples:

P(a(z0), p(a(b(z1)), z2)) → c(P(a(b(a(z2))), p(a(a(z1)), z2)), P(a(a(z1)), z2))

S tuples:

P(a(z0), p(a(b(z1)), z2)) → c(P(a(b(a(z2))), p(a(a(z1)), z2)), P(a(a(z1)), z2))

K tuples:none
Defined Rule Symbols:

p

Defined Pair Symbols:

P

Compound Symbols:

c

(3) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

P(a(z0), p(a(b(z1)), z2)) → c(P(a(b(a(z2))), p(a(a(z1)), z2)), P(a(a(z1)), z2))

We considered the (Usable) Rules:none
And the Tuples:

P(a(z0), p(a(b(z1)), z2)) → c(P(a(b(a(z2))), p(a(a(z1)), z2)), P(a(a(z1)), z2))

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

POL(P(x₁, x₂)) = [3]x₂²   
POL(a(x₁)) = x₁   
POL(b(x₁)) = [1] + x₁   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(p(x₁, x₂)) = [1] + x₂² + [3]x₁·x₂ + x₁²   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(a(z0), p(a(b(z1)), z2)) → p(a(b(a(z2))), p(a(a(z1)), z2))

Tuples:

P(a(z0), p(a(b(z1)), z2)) → c(P(a(b(a(z2))), p(a(a(z1)), z2)), P(a(a(z1)), z2))

S tuples:none
K tuples:

P(a(z0), p(a(b(z1)), z2)) → c(P(a(b(a(z2))), p(a(a(z1)), z2)), P(a(a(z1)), z2))

Defined Rule Symbols:

p

Defined Pair Symbols:

P

Compound Symbols:

c

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(6) BOUNDS(O(1), O(1))