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:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

not(true) → false
not(false) → true
odd(0) → false
odd(s(z0)) → not(odd(z0))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
+(s(z0), z1) → s(+(z0, z1))

Tuples:

ODD(s(z0)) → c3(NOT(odd(z0)), ODD(z0))
+'(z0, s(z1)) → c5(+'(z0, z1))
+'(s(z0), z1) → c6(+'(z0, z1))

S tuples:

ODD(s(z0)) → c3(NOT(odd(z0)), ODD(z0))
+'(z0, s(z1)) → c5(+'(z0, z1))
+'(s(z0), z1) → c6(+'(z0, z1))

K tuples:none
Defined Rule Symbols:

not, odd, +

Defined Pair Symbols:

ODD, +'

Compound Symbols:

c3, c5, c6

(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.

+'(z0, s(z1)) → c5(+'(z0, z1))
+'(s(z0), z1) → c6(+'(z0, z1))

We considered the (Usable) Rules:

odd(0) → false
odd(s(z0)) → not(odd(z0))
not(true) → false
not(false) → true

And the Tuples:

ODD(s(z0)) → c3(NOT(odd(z0)), ODD(z0))
+'(z0, s(z1)) → c5(+'(z0, z1))
+'(s(z0), z1) → c6(+'(z0, z1))

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

POL(+'(x₁, x₂)) = x₁ + x₂   
POL(0) = [3]   
POL(NOT(x₁)) = [3]   
POL(ODD(x₁)) = x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(false) = 0   
POL(not(x₁)) = [2]   
POL(odd(x₁)) = [3] + [3]x₁   
POL(s(x₁)) = [3] + x₁   
POL(true) = [1]   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

not(true) → false
not(false) → true
odd(0) → false
odd(s(z0)) → not(odd(z0))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
+(s(z0), z1) → s(+(z0, z1))

Tuples:

ODD(s(z0)) → c3(NOT(odd(z0)), ODD(z0))
+'(z0, s(z1)) → c5(+'(z0, z1))
+'(s(z0), z1) → c6(+'(z0, z1))

S tuples:

ODD(s(z0)) → c3(NOT(odd(z0)), ODD(z0))

K tuples:

+'(z0, s(z1)) → c5(+'(z0, z1))
+'(s(z0), z1) → c6(+'(z0, z1))

Defined Rule Symbols:

not, odd, +

Defined Pair Symbols:

ODD, +'

Compound Symbols:

c3, c5, c6

(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.

ODD(s(z0)) → c3(NOT(odd(z0)), ODD(z0))

We considered the (Usable) Rules:

odd(0) → false
odd(s(z0)) → not(odd(z0))
not(true) → false
not(false) → true

And the Tuples:

ODD(s(z0)) → c3(NOT(odd(z0)), ODD(z0))
+'(z0, s(z1)) → c5(+'(z0, z1))
+'(s(z0), z1) → c6(+'(z0, z1))

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

POL(+'(x₁, x₂)) = [3]x₁ + [4]x₂   
POL(0) = [2]   
POL(NOT(x₁)) = [2]   
POL(ODD(x₁)) = [2]x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(false) = [3]   
POL(not(x₁)) = [2] + [3]x₁   
POL(odd(x₁)) = [2]   
POL(s(x₁)) = [4] + x₁   
POL(true) = [2]   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

not(true) → false
not(false) → true
odd(0) → false
odd(s(z0)) → not(odd(z0))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
+(s(z0), z1) → s(+(z0, z1))

Tuples:

ODD(s(z0)) → c3(NOT(odd(z0)), ODD(z0))
+'(z0, s(z1)) → c5(+'(z0, z1))
+'(s(z0), z1) → c6(+'(z0, z1))

S tuples:none
K tuples:

+'(z0, s(z1)) → c5(+'(z0, z1))
+'(s(z0), z1) → c6(+'(z0, z1))
ODD(s(z0)) → c3(NOT(odd(z0)), ODD(z0))

Defined Rule Symbols:

not, odd, +

Defined Pair Symbols:

ODD, +'

Compound Symbols:

c3, c5, c6

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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