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

(0) Obligation:

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


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:

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

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

*'(s(z0), z1) → c3(*'(z0, z1))

K tuples:

EXP(z0, s(z1)) → c1(*'(z0, exp(z0, z1)), EXP(z0, z1))
-'(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(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]x₁ + x₁·x₂ + [2]x₁²   
POL(*'(x₁, x₂)) = x₁   
POL(+(x₁, x₂)) = [1]   
POL(-'(x₁, x₂)) = [2]x₁ + [2]x₂ + x₂² + x₁²   
POL(0) = [2]   
POL(EXP(x₁, x₂)) = [2]x₂ + x₂² + x₁·x₂   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(exp(x₁, x₂)) = [1] + [2]x₂ + [2]x₁²   
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:none
K tuples:

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

Defined Rule Symbols:

exp, *

Defined Pair Symbols:

EXP, *', -'

Compound Symbols:

c1, c3, c6

(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(12) BOUNDS(1, 1)