The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

(0) Obligation:

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

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, y))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(z0, 0) → z0
+(minus(z0), z0) → 0
minus(0) → 0
minus(minus(z0)) → z0
minus(+(z0, z1)) → +(minus(z1), minus(z0))
*(z0, 1) → z0
*(z0, 0) → 0
*(z0, +(z1, z2)) → +(*(z0, z1), *(z0, z2))
*(z0, minus(z1)) → minus(*(z0, z1))

Tuples:

MINUS(+(z0, z1)) → c4(+'(minus(z1), minus(z0)), MINUS(z1), MINUS(z0))
*'(z0, +(z1, z2)) → c7(+'(*(z0, z1), *(z0, z2)), *'(z0, z1), *'(z0, z2))
*'(z0, minus(z1)) → c8(MINUS(*(z0, z1)), *'(z0, z1))

S tuples:

MINUS(+(z0, z1)) → c4(+'(minus(z1), minus(z0)), MINUS(z1), MINUS(z0))
*'(z0, +(z1, z2)) → c7(+'(*(z0, z1), *(z0, z2)), *'(z0, z1), *'(z0, z2))
*'(z0, minus(z1)) → c8(MINUS(*(z0, z1)), *'(z0, z1))

K tuples:none
Defined Rule Symbols:

+, minus, *

Defined Pair Symbols:

MINUS, *'

Compound Symbols:

c4, c7, c8

(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, +(z1, z2)) → c7(+'(*(z0, z1), *(z0, z2)), *'(z0, z1), *'(z0, z2))
*'(z0, minus(z1)) → c8(MINUS(*(z0, z1)), *'(z0, z1))

We considered the (Usable) Rules:

*(z0, 1) → z0
*(z0, 0) → 0
*(z0, +(z1, z2)) → +(*(z0, z1), *(z0, z2))
*(z0, minus(z1)) → minus(*(z0, z1))
minus(0) → 0
minus(minus(z0)) → z0
minus(+(z0, z1)) → +(minus(z1), minus(z0))
+(z0, 0) → z0
+(minus(z0), z0) → 0

And the Tuples:

MINUS(+(z0, z1)) → c4(+'(minus(z1), minus(z0)), MINUS(z1), MINUS(z0))
*'(z0, +(z1, z2)) → c7(+'(*(z0, z1), *(z0, z2)), *'(z0, z1), *'(z0, z2))
*'(z0, minus(z1)) → c8(MINUS(*(z0, z1)), *'(z0, z1))

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

POL(*(x₁, x₂)) = [4] + [4]x₂   
POL(*'(x₁, x₂)) = [3] + [5]x₂   
POL(+(x₁, x₂)) = [3] + x₁ + [2]x₂   
POL(+'(x₁, x₂)) = 0   
POL(0) = 0   
POL(1) = 0   
POL(MINUS(x₁)) = 0   
POL(c4(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c8(x₁, x₂)) = x₁ + x₂   
POL(minus(x₁)) = [5] + x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(z0, 0) → z0
+(minus(z0), z0) → 0
minus(0) → 0
minus(minus(z0)) → z0
minus(+(z0, z1)) → +(minus(z1), minus(z0))
*(z0, 1) → z0
*(z0, 0) → 0
*(z0, +(z1, z2)) → +(*(z0, z1), *(z0, z2))
*(z0, minus(z1)) → minus(*(z0, z1))

Tuples:

MINUS(+(z0, z1)) → c4(+'(minus(z1), minus(z0)), MINUS(z1), MINUS(z0))
*'(z0, +(z1, z2)) → c7(+'(*(z0, z1), *(z0, z2)), *'(z0, z1), *'(z0, z2))
*'(z0, minus(z1)) → c8(MINUS(*(z0, z1)), *'(z0, z1))

S tuples:

MINUS(+(z0, z1)) → c4(+'(minus(z1), minus(z0)), MINUS(z1), MINUS(z0))

K tuples:

*'(z0, +(z1, z2)) → c7(+'(*(z0, z1), *(z0, z2)), *'(z0, z1), *'(z0, z2))
*'(z0, minus(z1)) → c8(MINUS(*(z0, z1)), *'(z0, z1))

Defined Rule Symbols:

+, minus, *

Defined Pair Symbols:

MINUS, *'

Compound Symbols:

c4, c7, c8

(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(+(z0, z1)) → c4(+'(minus(z1), minus(z0)), MINUS(z1), MINUS(z0))

We considered the (Usable) Rules:

*(z0, 1) → z0
*(z0, 0) → 0
*(z0, +(z1, z2)) → +(*(z0, z1), *(z0, z2))
*(z0, minus(z1)) → minus(*(z0, z1))
minus(0) → 0
minus(minus(z0)) → z0
minus(+(z0, z1)) → +(minus(z1), minus(z0))
+(z0, 0) → z0
+(minus(z0), z0) → 0

And the Tuples:

MINUS(+(z0, z1)) → c4(+'(minus(z1), minus(z0)), MINUS(z1), MINUS(z0))
*'(z0, +(z1, z2)) → c7(+'(*(z0, z1), *(z0, z2)), *'(z0, z1), *'(z0, z2))
*'(z0, minus(z1)) → c8(MINUS(*(z0, z1)), *'(z0, z1))

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

POL(*(x₁, x₂)) = [2]x₁·x₂   
POL(*'(x₁, x₂)) = x₁ + [2]x₁·x₂   
POL(+(x₁, x₂)) = [1] + [2]x₁ + [2]x₂ + x₁²   
POL(+'(x₁, x₂)) = 0   
POL(0) = 0   
POL(1) = [2]   
POL(MINUS(x₁)) = x₁   
POL(c4(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c8(x₁, x₂)) = x₁ + x₂   
POL(minus(x₁)) = [2]x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(z0, 0) → z0
+(minus(z0), z0) → 0
minus(0) → 0
minus(minus(z0)) → z0
minus(+(z0, z1)) → +(minus(z1), minus(z0))
*(z0, 1) → z0
*(z0, 0) → 0
*(z0, +(z1, z2)) → +(*(z0, z1), *(z0, z2))
*(z0, minus(z1)) → minus(*(z0, z1))

Tuples:

MINUS(+(z0, z1)) → c4(+'(minus(z1), minus(z0)), MINUS(z1), MINUS(z0))
*'(z0, +(z1, z2)) → c7(+'(*(z0, z1), *(z0, z2)), *'(z0, z1), *'(z0, z2))
*'(z0, minus(z1)) → c8(MINUS(*(z0, z1)), *'(z0, z1))

S tuples:none
K tuples:

*'(z0, +(z1, z2)) → c7(+'(*(z0, z1), *(z0, z2)), *'(z0, z1), *'(z0, z2))
*'(z0, minus(z1)) → c8(MINUS(*(z0, z1)), *'(z0, z1))
MINUS(+(z0, z1)) → c4(+'(minus(z1), minus(z0)), MINUS(z1), MINUS(z0))

Defined Rule Symbols:

+, minus, *

Defined Pair Symbols:

MINUS, *'

Compound Symbols:

c4, c7, c8

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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