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:

__(__(X, Y), Z) → __(X, __(Y, Z))
__(X, nil) → X
__(nil, X) → X
U11(tt) → U12(tt)
U12(tt) → tt
isNePal(__(I, __(P, I))) → U11(tt)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

__(__(z0, z1), z2) → __(z0, __(z1, z2))
__(z0, nil) → z0
__(nil, z0) → z0
U11(tt) → U12(tt)
U12(tt) → tt
isNePal(__(z0, __(z1, z0))) → U11(tt)

Tuples:

__'(__(z0, z1), z2) → c(__'(z0, __(z1, z2)), __'(z1, z2))
U11'(tt) → c3(U12'(tt))
ISNEPAL(__(z0, __(z1, z0))) → c5(U11'(tt))

S tuples:

__'(__(z0, z1), z2) → c(__'(z0, __(z1, z2)), __'(z1, z2))
U11'(tt) → c3(U12'(tt))
ISNEPAL(__(z0, __(z1, z0))) → c5(U11'(tt))

K tuples:none
Defined Rule Symbols:

__, U11, U12, isNePal

Defined Pair Symbols:

__', U11', ISNEPAL

Compound Symbols:

c, c3, c5

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

Removed 2 trailing nodes:

ISNEPAL(__(z0, __(z1, z0))) → c5(U11'(tt))
U11'(tt) → c3(U12'(tt))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

__(__(z0, z1), z2) → __(z0, __(z1, z2))
__(z0, nil) → z0
__(nil, z0) → z0
U11(tt) → U12(tt)
U12(tt) → tt
isNePal(__(z0, __(z1, z0))) → U11(tt)

Tuples:

__'(__(z0, z1), z2) → c(__'(z0, __(z1, z2)), __'(z1, z2))

S tuples:

__'(__(z0, z1), z2) → c(__'(z0, __(z1, z2)), __'(z1, z2))

K tuples:none
Defined Rule Symbols:

__, U11, U12, isNePal

Defined Pair Symbols:

__'

Compound Symbols:

c

(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.

__'(__(z0, z1), z2) → c(__'(z0, __(z1, z2)), __'(z1, z2))

We considered the (Usable) Rules:

__(__(z0, z1), z2) → __(z0, __(z1, z2))
__(z0, nil) → z0

And the Tuples:

__'(__(z0, z1), z2) → c(__'(z0, __(z1, z2)), __'(z1, z2))

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

POL(__(x₁, x₂)) = [4] + [2]x₁ + [4]x₂   
POL(__'(x₁, x₂)) = [5] + [4]x₁   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(nil) = 0   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

__(__(z0, z1), z2) → __(z0, __(z1, z2))
__(z0, nil) → z0
__(nil, z0) → z0
U11(tt) → U12(tt)
U12(tt) → tt
isNePal(__(z0, __(z1, z0))) → U11(tt)

Tuples:

__'(__(z0, z1), z2) → c(__'(z0, __(z1, z2)), __'(z1, z2))

S tuples:none
K tuples:

__'(__(z0, z1), z2) → c(__'(z0, __(z1, z2)), __'(z1, z2))

Defined Rule Symbols:

__, U11, U12, isNePal

Defined Pair Symbols:

__'

Compound Symbols:

c

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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