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

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

-(0, y) → 0
-(x, 0) → x
-(x, s(y)) → if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) → 0
p(s(x)) → x

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

-(0, z0) → 0
-(z0, 0) → z0
-(z0, s(z1)) → if(greater(z0, s(z1)), s(-(z0, p(s(z1)))), 0)
p(0) → 0
p(s(z0)) → z0

Tuples:

-'(z0, s(z1)) → c2(-'(z0, p(s(z1))), P(s(z1)))

S tuples:

-'(z0, s(z1)) → c2(-'(z0, p(s(z1))), P(s(z1)))

K tuples:none
Defined Rule Symbols:

-, p

Defined Pair Symbols:

-'

Compound Symbols:

c2

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

Removed 1 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

-(0, z0) → 0
-(z0, 0) → z0
-(z0, s(z1)) → if(greater(z0, s(z1)), s(-(z0, p(s(z1)))), 0)
p(0) → 0
p(s(z0)) → z0

Tuples:

-'(z0, s(z1)) → c2(-'(z0, p(s(z1))))

S tuples:

-'(z0, s(z1)) → c2(-'(z0, p(s(z1))))

K tuples:none
Defined Rule Symbols:

-, p

Defined Pair Symbols:

-'

Compound Symbols:

c2

(5) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace -'(z0, s(z1)) → c2(-'(z0, p(s(z1)))) by

-'(x0, s(z0)) → c2(-'(x0, z0))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

-(0, z0) → 0
-(z0, 0) → z0
-(z0, s(z1)) → if(greater(z0, s(z1)), s(-(z0, p(s(z1)))), 0)
p(0) → 0
p(s(z0)) → z0

Tuples:

-'(x0, s(z0)) → c2(-'(x0, z0))

S tuples:

-'(x0, s(z0)) → c2(-'(x0, z0))

K tuples:none
Defined Rule Symbols:

-, p

Defined Pair Symbols:

-'

Compound Symbols:

c2

(7) 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.

-'(x0, s(z0)) → c2(-'(x0, z0))

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

-'(x0, s(z0)) → c2(-'(x0, z0))

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

POL(-'(x₁, x₂)) = [5]x₂   
POL(c2(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

-(0, z0) → 0
-(z0, 0) → z0
-(z0, s(z1)) → if(greater(z0, s(z1)), s(-(z0, p(s(z1)))), 0)
p(0) → 0
p(s(z0)) → z0

Tuples:

-'(x0, s(z0)) → c2(-'(x0, z0))

S tuples:none
K tuples:

-'(x0, s(z0)) → c2(-'(x0, z0))

Defined Rule Symbols:

-, p

Defined Pair Symbols:

-'

Compound Symbols:

c2

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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