### (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 Cpx (relative) TRS 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, 0) → c
EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1))
*'(0, z0) → c2
*'(s(z0), z1) → c3(*'(z0, z1))
-'(0, z0) → c4
-'(z0, 0) → c5
-'(s(z0), s(z1)) → c6(-'(z0, z1))
S tuples:

EXP(z0, 0) → c
EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1))
*'(0, z0) → c2
*'(s(z0), z1) → c3(*'(z0, z1))
-'(0, z0) → c4
-'(z0, 0) → c5
-'(s(z0), s(z1)) → c6(-'(z0, z1))
K tuples:none
Defined Rule Symbols:

exp, *, -

Defined Pair Symbols:

EXP, *', -'

Compound Symbols:

c, c1, c2, c3, c4, c5, c6

### (3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 4 trailing nodes:

EXP(z0, 0) → c
-'(z0, 0) → c5
*'(0, z0) → c2
-'(0, z0) → c4

### (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))
-'(s(z0), s(z1)) → c6(-'(z0, z1))
K tuples:none
Defined Rule Symbols:

exp, *, -

Defined Pair Symbols:

EXP, *', -'

Compound Symbols:

c1, c3, c6

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

-(0, z0) → 0
-(z0, 0) → z0
-(s(z0), s(z1)) → -(z0, z1)

### (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))
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

### (7) CdtRuleRemovalProof (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:none
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

### (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))
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

### (9) CdtRuleRemovalProof (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:none
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

### (10) 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))
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

### (11) CdtRuleRemovalProof (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:none
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

### (12) 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))
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

### (13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty