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(S(x)) → even(x)
even(S(x)) → odd(x)
odd(0) → 0
even(0) → S(0)

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

odd(S(z0)) → even(z0)
odd(0) → 0
even(S(z0)) → odd(z0)
even(0) → S(0)

Tuples:

ODD(S(z0)) → c(EVEN(z0))
ODD(0) → c1
EVEN(S(z0)) → c2(ODD(z0))
EVEN(0) → c3

S tuples:

ODD(S(z0)) → c(EVEN(z0))
ODD(0) → c1
EVEN(S(z0)) → c2(ODD(z0))
EVEN(0) → c3

K tuples:none
Defined Rule Symbols:

odd, even

Defined Pair Symbols:

ODD, EVEN

Compound Symbols:

c, c1, c2, c3

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

Removed 2 trailing nodes:

EVEN(0) → c3
ODD(0) → c1

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

odd(S(z0)) → even(z0)
odd(0) → 0
even(S(z0)) → odd(z0)
even(0) → S(0)

Tuples:

ODD(S(z0)) → c(EVEN(z0))
EVEN(S(z0)) → c2(ODD(z0))

S tuples:

ODD(S(z0)) → c(EVEN(z0))
EVEN(S(z0)) → c2(ODD(z0))

K tuples:none
Defined Rule Symbols:

odd, even

Defined Pair Symbols:

ODD, EVEN

Compound Symbols:

c, c2

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

odd(S(z0)) → even(z0)
odd(0) → 0
even(S(z0)) → odd(z0)
even(0) → S(0)

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ODD(S(z0)) → c(EVEN(z0))
EVEN(S(z0)) → c2(ODD(z0))

S tuples:

ODD(S(z0)) → c(EVEN(z0))
EVEN(S(z0)) → c2(ODD(z0))

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(S(z0)) → c(EVEN(z0))
EVEN(S(z0)) → c2(ODD(z0))

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

ODD(S(z0)) → c(EVEN(z0))
EVEN(S(z0)) → c2(ODD(z0))

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

POL(EVEN(x₁)) = [4] + [4]x₁   
POL(ODD(x₁)) = [2] + [4]x₁   
POL(S(x₁)) = [5] + x₁   
POL(c(x₁)) = x₁   
POL(c2(x₁)) = x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ODD(S(z0)) → c(EVEN(z0))
EVEN(S(z0)) → c2(ODD(z0))

S tuples:none
K tuples:

ODD(S(z0)) → c(EVEN(z0))
EVEN(S(z0)) → c2(ODD(z0))

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))