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:

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

exp(z0, 0) → s(0)
exp(z0, s(z1)) → *(z0, exp(z0, z1))
*(0, z0) → 0
*(s(z0), z1) → +(z1, *(z0, z1))
-(0, z0) → 0
-(z0, 0) → z0
-(s(z0), s(z1)) → -(z0, z1)

Tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1))
*'(s(z0), z1) → c3(*'(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))

S tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1))
*'(s(z0), z1) → c3(*'(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))

K tuples:none
Defined Rule Symbols:

exp, *, -

Defined Pair Symbols:

EXP, *', -'

Compound Symbols:

c1, c3, 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.

-'(s(z0), s(z1)) → c6(-'(z0, z1))

We considered the (Usable) Rules:

exp(z0, 0) → s(0)
exp(z0, s(z1)) → *(z0, exp(z0, z1))
*(0, z0) → 0
*(s(z0), z1) → +(z1, *(z0, z1))

And the Tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1))
*'(s(z0), z1) → c3(*'(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))

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

POL(*(x₁, x₂)) = [2] + x₁   
POL(*'(x₁, x₂)) = 0   
POL(+(x₁, x₂)) = [1] + x₂   
POL(-'(x₁, x₂)) = [4]x₂   
POL(0) = [5]   
POL(EXP(x₁, x₂)) = 0   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(exp(x₁, x₂)) = [1] + [3]x₁ + x₂   
POL(s(x₁)) = [1] + x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

exp(z0, 0) → s(0)
exp(z0, s(z1)) → *(z0, exp(z0, z1))
*(0, z0) → 0
*(s(z0), z1) → +(z1, *(z0, z1))
-(0, z0) → 0
-(z0, 0) → z0
-(s(z0), s(z1)) → -(z0, z1)

Tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1))
*'(s(z0), z1) → c3(*'(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))

S tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1))
*'(s(z0), z1) → c3(*'(z0, z1))

K tuples:

-'(s(z0), s(z1)) → c6(-'(z0, z1))

Defined Rule Symbols:

exp, *, -

Defined Pair Symbols:

EXP, *', -'

Compound Symbols:

c1, c3, 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.

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1))

We considered the (Usable) Rules:

exp(z0, 0) → s(0)
exp(z0, s(z1)) → *(z0, exp(z0, z1))
*(0, z0) → 0
*(s(z0), z1) → +(z1, *(z0, z1))

And the Tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1))
*'(s(z0), z1) → c3(*'(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))

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

POL(*(x₁, x₂)) = [5] + [5]x₁ + x₂   
POL(*'(x₁, x₂)) = 0   
POL(+(x₁, x₂)) = [3]   
POL(-'(x₁, x₂)) = 0   
POL(0) = [3]   
POL(EXP(x₁, x₂)) = x₂   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(exp(x₁, x₂)) = [3] + [5]x₁ + [3]x₂   
POL(s(x₁)) = [1] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

exp(z0, 0) → s(0)
exp(z0, s(z1)) → *(z0, exp(z0, z1))
*(0, z0) → 0
*(s(z0), z1) → +(z1, *(z0, z1))
-(0, z0) → 0
-(z0, 0) → z0
-(s(z0), s(z1)) → -(z0, z1)

Tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1))
*'(s(z0), z1) → c3(*'(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))

S tuples:

*'(s(z0), z1) → c3(*'(z0, z1))

K tuples:

-'(s(z0), s(z1)) → c6(-'(z0, z1))
EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1))

Defined Rule Symbols:

exp, *, -

Defined Pair Symbols:

EXP, *', -'

Compound Symbols:

c1, c3, c6

(7) 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.

*'(s(z0), z1) → c3(*'(z0, z1))

We considered the (Usable) Rules:

exp(z0, 0) → s(0)
exp(z0, s(z1)) → *(z0, exp(z0, z1))
*(0, z0) → 0
*(s(z0), z1) → +(z1, *(z0, z1))

And the Tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1))
*'(s(z0), z1) → c3(*'(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))

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

POL(*(x₁, x₂)) = [2]   
POL(*'(x₁, x₂)) = [2]x₁   
POL(+(x₁, x₂)) = 0   
POL(-'(x₁, x₂)) = [2]x₁ + [2]x₂ + x₁·x₂ + [2]x₁²   
POL(0) = [2]   
POL(EXP(x₁, x₂)) = [3]x₁·x₂   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(exp(x₁, x₂)) = [2]x₂² + [3]x₁·x₂ + x₁²   
POL(s(x₁)) = [3] + x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

exp(z0, 0) → s(0)
exp(z0, s(z1)) → *(z0, exp(z0, z1))
*(0, z0) → 0
*(s(z0), z1) → +(z1, *(z0, z1))
-(0, z0) → 0
-(z0, 0) → z0
-(s(z0), s(z1)) → -(z0, z1)

Tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1))
*'(s(z0), z1) → c3(*'(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))

S tuples:none
K tuples:

-'(s(z0), s(z1)) → c6(-'(z0, z1))
EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1))
*'(s(z0), z1) → c3(*'(z0, z1))

Defined Rule Symbols:

exp, *, -

Defined Pair Symbols:

EXP, *', -'

Compound Symbols:

c1, c3, c6

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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