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:

rev(a) → a
rev(b) → b
rev(++(x, y)) → ++(rev(y), rev(x))
rev(++(x, x)) → rev(x)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(a) → a
rev(b) → b
rev(++(z0, z1)) → ++(rev(z1), rev(z0))
rev(++(z0, z0)) → rev(z0)

Tuples:

REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))

S tuples:

REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))

K tuples:none
Defined Rule Symbols:

rev

Defined Pair Symbols:

REV

Compound Symbols:

c2, c3

(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.

REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))

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

REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))

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

POL(++(x₁, x₂)) = [3] + x₁ + x₂   
POL(REV(x₁)) = [2] + [2]x₁   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(a) → a
rev(b) → b
rev(++(z0, z1)) → ++(rev(z1), rev(z0))
rev(++(z0, z0)) → rev(z0)

Tuples:

REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))

S tuples:none
K tuples:

REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))

Defined Rule Symbols:

rev

Defined Pair Symbols:

REV

Compound Symbols:

c2, c3

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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