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:

D(t) → 1
D(constant) → 0
D(+(x, y)) → +(D(x), D(y))
D(*(x, y)) → +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) → -(D(x), D(y))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

D(t) → 1
D(constant) → 0
D(+(z0, z1)) → +(D(z0), D(z1))
D(*(z0, z1)) → +(*(z1, D(z0)), *(z0, D(z1)))
D(-(z0, z1)) → -(D(z0), D(z1))

Tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))

S tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))

K tuples:none
Defined Rule Symbols:

D

Defined Pair Symbols:

D'

Compound Symbols:

c2, c3, c4

(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.

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))

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

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))

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

POL(*(x₁, x₂)) = [5] + x₁ + x₂   
POL(+(x₁, x₂)) = [5] + x₁ + x₂   
POL(-(x₁, x₂)) = [4] + x₁ + x₂   
POL(D'(x₁)) = [5] + [4]x₁   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂)) = x₁ + x₂   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

D(t) → 1
D(constant) → 0
D(+(z0, z1)) → +(D(z0), D(z1))
D(*(z0, z1)) → +(*(z1, D(z0)), *(z0, D(z1)))
D(-(z0, z1)) → -(D(z0), D(z1))

Tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))

S tuples:

D'(-(z0, z1)) → c4(D'(z0), D'(z1))

K tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))

Defined Rule Symbols:

D

Defined Pair Symbols:

D'

Compound Symbols:

c2, c3, c4

(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.

D'(-(z0, z1)) → c4(D'(z0), D'(z1))

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

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))

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

POL(*(x₁, x₂)) = [2] + x₁ + x₂   
POL(+(x₁, x₂)) = [1] + x₁ + x₂   
POL(-(x₁, x₂)) = [3] + x₁ + x₂   
POL(D'(x₁)) = [1] + x₁²   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂)) = x₁ + x₂   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

D(t) → 1
D(constant) → 0
D(+(z0, z1)) → +(D(z0), D(z1))
D(*(z0, z1)) → +(*(z1, D(z0)), *(z0, D(z1)))
D(-(z0, z1)) → -(D(z0), D(z1))

Tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))

S tuples:none
K tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))

Defined Rule Symbols:

D

Defined Pair Symbols:

D'

Compound Symbols:

c2, c3, c4

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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