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:

minus(minus(x)) → x
minus(h(x)) → h(minus(x))
minus(f(x, y)) → f(minus(y), minus(x))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(minus(z0)) → z0
minus(h(z0)) → h(minus(z0))
minus(f(z0, z1)) → f(minus(z1), minus(z0))

Tuples:

MINUS(h(z0)) → c1(MINUS(z0))
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))

S tuples:

MINUS(h(z0)) → c1(MINUS(z0))
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))

K tuples:none
Defined Rule Symbols:

minus

Defined Pair Symbols:

MINUS

Compound Symbols:

c1, c2

(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.

MINUS(h(z0)) → c1(MINUS(z0))
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))

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

MINUS(h(z0)) → c1(MINUS(z0))
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))

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

POL(MINUS(x₁)) = [4] + [4]x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(f(x₁, x₂)) = [4] + x₁ + x₂   
POL(h(x₁)) = [5] + x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(minus(z0)) → z0
minus(h(z0)) → h(minus(z0))
minus(f(z0, z1)) → f(minus(z1), minus(z0))

Tuples:

MINUS(h(z0)) → c1(MINUS(z0))
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))

S tuples:none
K tuples:

MINUS(h(z0)) → c1(MINUS(z0))
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))

Defined Rule Symbols:

minus

Defined Pair Symbols:

MINUS

Compound Symbols:

c1, c2

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(6) BOUNDS(O(1), O(1))