The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^3)).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

p(s(x)) → x
s(p(x)) → x
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0
s(p(z0)) → z0
+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
+(p(z0), z1) → p(+(z0, z1))
minus(0) → 0
minus(s(z0)) → p(minus(z0))
minus(p(z0)) → s(minus(z0))
*(0, z0) → 0
*(s(z0), z1) → +(*(z0, z1), z1)
*(p(z0), z1) → +(*(z0, z1), minus(z1))

Tuples:

+'(s(z0), z1) → c3(S(+(z0, z1)), +'(z0, z1))
+'(p(z0), z1) → c4(P(+(z0, z1)), +'(z0, z1))
MINUS(s(z0)) → c6(P(minus(z0)), MINUS(z0))
MINUS(p(z0)) → c7(S(minus(z0)), MINUS(z0))
*'(s(z0), z1) → c9(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c10(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

S tuples:

+'(s(z0), z1) → c3(S(+(z0, z1)), +'(z0, z1))
+'(p(z0), z1) → c4(P(+(z0, z1)), +'(z0, z1))
MINUS(s(z0)) → c6(P(minus(z0)), MINUS(z0))
MINUS(p(z0)) → c7(S(minus(z0)), MINUS(z0))
*'(s(z0), z1) → c9(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c10(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

K tuples:none
Defined Rule Symbols:

p, s, +, minus, *

Defined Pair Symbols:

+', MINUS, *'

Compound Symbols:

c3, c4, c6, c7, c9, c10

(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.

*'(s(z0), z1) → c9(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c10(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

We considered the (Usable) Rules:

*(0, z0) → 0
*(p(z0), z1) → +(*(z0, z1), minus(z1))
*(s(z0), z1) → +(*(z0, z1), z1)
minus(0) → 0
minus(s(z0)) → p(minus(z0))
minus(p(z0)) → s(minus(z0))
s(p(z0)) → z0
p(s(z0)) → z0
+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
+(p(z0), z1) → p(+(z0, z1))

And the Tuples:

+'(s(z0), z1) → c3(S(+(z0, z1)), +'(z0, z1))
+'(p(z0), z1) → c4(P(+(z0, z1)), +'(z0, z1))
MINUS(s(z0)) → c6(P(minus(z0)), MINUS(z0))
MINUS(p(z0)) → c7(S(minus(z0)), MINUS(z0))
*'(s(z0), z1) → c9(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c10(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

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

POL(*(x₁, x₂)) = [4]x₁ + [4]x₂   
POL(*'(x₁, x₂)) = [5]x₁   
POL(+(x₁, x₂)) = [3]x₂   
POL(+'(x₁, x₂)) = 0   
POL(0) = [4]   
POL(MINUS(x₁)) = [2]   
POL(P(x₁)) = 0   
POL(S(x₁)) = 0   
POL(c10(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(c7(x₁, x₂)) = x₁ + x₂   
POL(c9(x₁, x₂)) = x₁ + x₂   
POL(minus(x₁)) = [3] + x₁   
POL(p(x₁)) = [4] + [4]x₁   
POL(s(x₁)) = [5] + [2]x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0
s(p(z0)) → z0
+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
+(p(z0), z1) → p(+(z0, z1))
minus(0) → 0
minus(s(z0)) → p(minus(z0))
minus(p(z0)) → s(minus(z0))
*(0, z0) → 0
*(s(z0), z1) → +(*(z0, z1), z1)
*(p(z0), z1) → +(*(z0, z1), minus(z1))

Tuples:

+'(s(z0), z1) → c3(S(+(z0, z1)), +'(z0, z1))
+'(p(z0), z1) → c4(P(+(z0, z1)), +'(z0, z1))
MINUS(s(z0)) → c6(P(minus(z0)), MINUS(z0))
MINUS(p(z0)) → c7(S(minus(z0)), MINUS(z0))
*'(s(z0), z1) → c9(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c10(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

S tuples:

+'(s(z0), z1) → c3(S(+(z0, z1)), +'(z0, z1))
+'(p(z0), z1) → c4(P(+(z0, z1)), +'(z0, z1))
MINUS(s(z0)) → c6(P(minus(z0)), MINUS(z0))
MINUS(p(z0)) → c7(S(minus(z0)), MINUS(z0))

K tuples:

*'(s(z0), z1) → c9(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c10(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

Defined Rule Symbols:

p, s, +, minus, *

Defined Pair Symbols:

+', MINUS, *'

Compound Symbols:

c3, c4, c6, c7, c9, c10

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

MINUS(s(z0)) → c6(P(minus(z0)), MINUS(z0))
MINUS(p(z0)) → c7(S(minus(z0)), MINUS(z0))

We considered the (Usable) Rules:

*(0, z0) → 0
*(p(z0), z1) → +(*(z0, z1), minus(z1))
*(s(z0), z1) → +(*(z0, z1), z1)
minus(0) → 0
minus(s(z0)) → p(minus(z0))
minus(p(z0)) → s(minus(z0))
s(p(z0)) → z0
p(s(z0)) → z0
+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
+(p(z0), z1) → p(+(z0, z1))

And the Tuples:

+'(s(z0), z1) → c3(S(+(z0, z1)), +'(z0, z1))
+'(p(z0), z1) → c4(P(+(z0, z1)), +'(z0, z1))
MINUS(s(z0)) → c6(P(minus(z0)), MINUS(z0))
MINUS(p(z0)) → c7(S(minus(z0)), MINUS(z0))
*'(s(z0), z1) → c9(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c10(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

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

POL(*(x₁, x₂)) = [2] + [2]x₁ + [3]x₁²   
POL(*'(x₁, x₂)) = [2]x₁·x₂   
POL(+(x₁, x₂)) = [2]x₁·x₂ + [2]x₁²   
POL(+'(x₁, x₂)) = 0   
POL(0) = [2]   
POL(MINUS(x₁)) = [2]x₁   
POL(P(x₁)) = 0   
POL(S(x₁)) = 0   
POL(c10(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(c7(x₁, x₂)) = x₁ + x₂   
POL(c9(x₁, x₂)) = x₁ + x₂   
POL(minus(x₁)) = 0   
POL(p(x₁)) = [3] + [2]x₁ + [3]x₁²   
POL(s(x₁)) = [2] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0
s(p(z0)) → z0
+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
+(p(z0), z1) → p(+(z0, z1))
minus(0) → 0
minus(s(z0)) → p(minus(z0))
minus(p(z0)) → s(minus(z0))
*(0, z0) → 0
*(s(z0), z1) → +(*(z0, z1), z1)
*(p(z0), z1) → +(*(z0, z1), minus(z1))

Tuples:

+'(s(z0), z1) → c3(S(+(z0, z1)), +'(z0, z1))
+'(p(z0), z1) → c4(P(+(z0, z1)), +'(z0, z1))
MINUS(s(z0)) → c6(P(minus(z0)), MINUS(z0))
MINUS(p(z0)) → c7(S(minus(z0)), MINUS(z0))
*'(s(z0), z1) → c9(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c10(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

S tuples:

+'(s(z0), z1) → c3(S(+(z0, z1)), +'(z0, z1))
+'(p(z0), z1) → c4(P(+(z0, z1)), +'(z0, z1))

K tuples:

*'(s(z0), z1) → c9(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c10(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))
MINUS(s(z0)) → c6(P(minus(z0)), MINUS(z0))
MINUS(p(z0)) → c7(S(minus(z0)), MINUS(z0))

Defined Rule Symbols:

p, s, +, minus, *

Defined Pair Symbols:

+', MINUS, *'

Compound Symbols:

c3, c4, c6, c7, c9, c10

(7) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^3))) transformation)

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

+'(s(z0), z1) → c3(S(+(z0, z1)), +'(z0, z1))
+'(p(z0), z1) → c4(P(+(z0, z1)), +'(z0, z1))

We considered the (Usable) Rules:

*(0, z0) → 0
*(p(z0), z1) → +(*(z0, z1), minus(z1))
*(s(z0), z1) → +(*(z0, z1), z1)
minus(0) → 0
minus(s(z0)) → p(minus(z0))
minus(p(z0)) → s(minus(z0))
s(p(z0)) → z0
p(s(z0)) → z0
+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
+(p(z0), z1) → p(+(z0, z1))

And the Tuples:

+'(s(z0), z1) → c3(S(+(z0, z1)), +'(z0, z1))
+'(p(z0), z1) → c4(P(+(z0, z1)), +'(z0, z1))
MINUS(s(z0)) → c6(P(minus(z0)), MINUS(z0))
MINUS(p(z0)) → c7(S(minus(z0)), MINUS(z0))
*'(s(z0), z1) → c9(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c10(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

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

POL(*(x₁, x₂)) = x₁·x₂   
POL(*'(x₁, x₂)) = x₁·x₂ + x₁²·x₂   
POL(+(x₁, x₂)) = x₁ + x₂   
POL(+'(x₁, x₂)) = x₁   
POL(0) = 0   
POL(MINUS(x₁)) = 0   
POL(P(x₁)) = 0   
POL(S(x₁)) = 0   
POL(c10(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(c7(x₁, x₂)) = x₁ + x₂   
POL(c9(x₁, x₂)) = x₁ + x₂   
POL(minus(x₁)) = x₁   
POL(p(x₁)) = [1] + x₁   
POL(s(x₁)) = [1] + x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0
s(p(z0)) → z0
+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
+(p(z0), z1) → p(+(z0, z1))
minus(0) → 0
minus(s(z0)) → p(minus(z0))
minus(p(z0)) → s(minus(z0))
*(0, z0) → 0
*(s(z0), z1) → +(*(z0, z1), z1)
*(p(z0), z1) → +(*(z0, z1), minus(z1))

Tuples:

+'(s(z0), z1) → c3(S(+(z0, z1)), +'(z0, z1))
+'(p(z0), z1) → c4(P(+(z0, z1)), +'(z0, z1))
MINUS(s(z0)) → c6(P(minus(z0)), MINUS(z0))
MINUS(p(z0)) → c7(S(minus(z0)), MINUS(z0))
*'(s(z0), z1) → c9(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c10(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

S tuples:none
K tuples:

*'(s(z0), z1) → c9(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c10(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))
MINUS(s(z0)) → c6(P(minus(z0)), MINUS(z0))
MINUS(p(z0)) → c7(S(minus(z0)), MINUS(z0))
+'(s(z0), z1) → c3(S(+(z0, z1)), +'(z0, z1))
+'(p(z0), z1) → c4(P(+(z0, z1)), +'(z0, z1))

Defined Rule Symbols:

p, s, +, minus, *

Defined Pair Symbols:

+', MINUS, *'

Compound Symbols:

c3, c4, c6, c7, c9, c10

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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