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:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(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:

sum(0) → 0
sum(s(z0)) → +(sum(z0), s(z0))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

Tuples:

SUM(0) → c
SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, 0) → c2
+'(z0, s(z1)) → c3(+'(z0, z1))

S tuples:

SUM(0) → c
SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, 0) → c2
+'(z0, s(z1)) → c3(+'(z0, z1))

K tuples:none
Defined Rule Symbols:

sum, +

Defined Pair Symbols:

SUM, +'

Compound Symbols:

c, c1, c2, c3

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

Removed 2 trailing nodes:

SUM(0) → c
+'(z0, 0) → c2

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

sum(0) → 0
sum(s(z0)) → +(sum(z0), s(z0))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

Tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))

S tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))

K tuples:none
Defined Rule Symbols:

sum, +

Defined Pair Symbols:

SUM, +'

Compound Symbols:

c1, c3

(5) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))

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

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))

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

POL(+(x₁, x₂)) = 0   
POL(+'(x₁, x₂)) = 0   
POL(0) = 0   
POL(SUM(x₁)) = x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   
POL(sum(x₁)) = 0   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

sum(0) → 0
sum(s(z0)) → +(sum(z0), s(z0))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

Tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))

S tuples:

+'(z0, s(z1)) → c3(+'(z0, z1))

K tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))

Defined Rule Symbols:

sum, +

Defined Pair Symbols:

SUM, +'

Compound Symbols:

c1, c3

(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

+'(z0, s(z1)) → c3(+'(z0, z1))

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

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))

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

POL(+(x₁, x₂)) = x₂ + x₁·x₂   
POL(+'(x₁, x₂)) = [1] + x₂   
POL(0) = [1]   
POL(SUM(x₁)) = [2]x₁ + x₁²   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(s(x₁)) = [2] + x₁   
POL(sum(x₁)) = [2]   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

sum(0) → 0
sum(s(z0)) → +(sum(z0), s(z0))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

Tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))

S tuples:none
K tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))

Defined Rule Symbols:

sum, +

Defined Pair Symbols:

SUM, +'

Compound Symbols:

c1, c3

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

The set S is empty

(10) BOUNDS(1, 1)