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:

odd(Cons(x, xs)) → even(xs)
odd(Nil) → False
even(Cons(x, xs)) → odd(xs)
notEmpty(Cons(x, xs)) → True
notEmpty(Nil) → False
even(Nil) → True
evenodd(x) → even(x)

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

odd(Cons(z0, z1)) → even(z1)
odd(Nil) → False
even(Cons(z0, z1)) → odd(z1)
even(Nil) → True
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
evenodd(z0) → even(z0)

Tuples:

ODD(Cons(z0, z1)) → c(EVEN(z1))
ODD(Nil) → c1
EVEN(Cons(z0, z1)) → c2(ODD(z1))
EVEN(Nil) → c3
NOTEMPTY(Cons(z0, z1)) → c4
NOTEMPTY(Nil) → c5
EVENODD(z0) → c6(EVEN(z0))

S tuples:

ODD(Cons(z0, z1)) → c(EVEN(z1))
ODD(Nil) → c1
EVEN(Cons(z0, z1)) → c2(ODD(z1))
EVEN(Nil) → c3
NOTEMPTY(Cons(z0, z1)) → c4
NOTEMPTY(Nil) → c5
EVENODD(z0) → c6(EVEN(z0))

K tuples:none
Defined Rule Symbols:

odd, even, notEmpty, evenodd

Defined Pair Symbols:

ODD, EVEN, NOTEMPTY, EVENODD

Compound Symbols:

c, c1, c2, c3, c4, c5, c6

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

EVENODD(z0) → c6(EVEN(z0))

Removed 4 trailing nodes:

EVEN(Nil) → c3
NOTEMPTY(Nil) → c5
ODD(Nil) → c1
NOTEMPTY(Cons(z0, z1)) → c4

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

odd(Cons(z0, z1)) → even(z1)
odd(Nil) → False
even(Cons(z0, z1)) → odd(z1)
even(Nil) → True
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
evenodd(z0) → even(z0)

Tuples:

ODD(Cons(z0, z1)) → c(EVEN(z1))
EVEN(Cons(z0, z1)) → c2(ODD(z1))

S tuples:

ODD(Cons(z0, z1)) → c(EVEN(z1))
EVEN(Cons(z0, z1)) → c2(ODD(z1))

K tuples:none
Defined Rule Symbols:

odd, even, notEmpty, evenodd

Defined Pair Symbols:

ODD, EVEN

Compound Symbols:

c, c2

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

odd(Cons(z0, z1)) → even(z1)
odd(Nil) → False
even(Cons(z0, z1)) → odd(z1)
even(Nil) → True
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
evenodd(z0) → even(z0)

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ODD(Cons(z0, z1)) → c(EVEN(z1))
EVEN(Cons(z0, z1)) → c2(ODD(z1))

S tuples:

ODD(Cons(z0, z1)) → c(EVEN(z1))
EVEN(Cons(z0, z1)) → c2(ODD(z1))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

ODD, EVEN

Compound Symbols:

c, c2

(7) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

ODD(Cons(z0, z1)) → c(EVEN(z1))
EVEN(Cons(z0, z1)) → c2(ODD(z1))

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

ODD(Cons(z0, z1)) → c(EVEN(z1))
EVEN(Cons(z0, z1)) → c2(ODD(z1))

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

POL(Cons(x₁, x₂)) = [5] + x₂   
POL(EVEN(x₁)) = [4] + [4]x₁   
POL(ODD(x₁)) = [2] + [4]x₁   
POL(c(x₁)) = x₁   
POL(c2(x₁)) = x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ODD(Cons(z0, z1)) → c(EVEN(z1))
EVEN(Cons(z0, z1)) → c2(ODD(z1))

S tuples:none
K tuples:

ODD(Cons(z0, z1)) → c(EVEN(z1))
EVEN(Cons(z0, z1)) → c2(ODD(z1))

Defined Rule Symbols:none

Defined Pair Symbols:

ODD, EVEN

Compound Symbols:

c, c2

(9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(10) BOUNDS(O(1), O(1))