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(0) → 0
+(x, 0) → x
+(0, y) → y
+(minus(1), 1) → 0
minus(minus(x)) → x
+(x, minus(y)) → minus(+(minus(x), y))
+(x, +(y, z)) → +(+(x, y), z)
+(minus(+(x, 1)), 1) → minus(x)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(0) → 0
minus(minus(z0)) → z0
+(z0, 0) → z0
+(0, z0) → z0
+(minus(1), 1) → 0
+(z0, minus(z1)) → minus(+(minus(z0), z1))
+(z0, +(z1, z2)) → +(+(z0, z1), z2)
+(minus(+(z0, 1)), 1) → minus(z0)

Tuples:

+'(z0, minus(z1)) → c5(MINUS(+(minus(z0), z1)), +'(minus(z0), z1), MINUS(z0))
+'(z0, +(z1, z2)) → c6(+'(+(z0, z1), z2), +'(z0, z1))
+'(minus(+(z0, 1)), 1) → c7(MINUS(z0))

S tuples:

+'(z0, minus(z1)) → c5(MINUS(+(minus(z0), z1)), +'(minus(z0), z1), MINUS(z0))
+'(z0, +(z1, z2)) → c6(+'(+(z0, z1), z2), +'(z0, z1))
+'(minus(+(z0, 1)), 1) → c7(MINUS(z0))

K tuples:none
Defined Rule Symbols:

minus, +

Defined Pair Symbols:

+'

Compound Symbols:

c5, c6, c7

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

Removed 1 trailing nodes:

+'(minus(+(z0, 1)), 1) → c7(MINUS(z0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(0) → 0
minus(minus(z0)) → z0
+(z0, 0) → z0
+(0, z0) → z0
+(minus(1), 1) → 0
+(z0, minus(z1)) → minus(+(minus(z0), z1))
+(z0, +(z1, z2)) → +(+(z0, z1), z2)
+(minus(+(z0, 1)), 1) → minus(z0)

Tuples:

+'(z0, minus(z1)) → c5(MINUS(+(minus(z0), z1)), +'(minus(z0), z1), MINUS(z0))
+'(z0, +(z1, z2)) → c6(+'(+(z0, z1), z2), +'(z0, z1))

S tuples:

+'(z0, minus(z1)) → c5(MINUS(+(minus(z0), z1)), +'(minus(z0), z1), MINUS(z0))
+'(z0, +(z1, z2)) → c6(+'(+(z0, z1), z2), +'(z0, z1))

K tuples:none
Defined Rule Symbols:

minus, +

Defined Pair Symbols:

+'

Compound Symbols:

c5, c6

(5) 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.

+'(z0, minus(z1)) → c5(MINUS(+(minus(z0), z1)), +'(minus(z0), z1), MINUS(z0))
+'(z0, +(z1, z2)) → c6(+'(+(z0, z1), z2), +'(z0, z1))

We considered the (Usable) Rules:

+(0, z0) → z0
+(minus(1), 1) → 0
+(z0, minus(z1)) → minus(+(minus(z0), z1))
+(z0, +(z1, z2)) → +(+(z0, z1), z2)
+(minus(+(z0, 1)), 1) → minus(z0)
minus(minus(z0)) → z0
minus(0) → 0

And the Tuples:

+'(z0, minus(z1)) → c5(MINUS(+(minus(z0), z1)), +'(minus(z0), z1), MINUS(z0))
+'(z0, +(z1, z2)) → c6(+'(+(z0, z1), z2), +'(z0, z1))

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

POL(+(x₁, x₂)) = [4] + [2]x₁ + [4]x₂   
POL(+'(x₁, x₂)) = [2] + [4]x₂   
POL(0) = [5]   
POL(1) = [3]   
POL(MINUS(x₁)) = [2]   
POL(c5(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(minus(x₁)) = [3] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(0) → 0
minus(minus(z0)) → z0
+(z0, 0) → z0
+(0, z0) → z0
+(minus(1), 1) → 0
+(z0, minus(z1)) → minus(+(minus(z0), z1))
+(z0, +(z1, z2)) → +(+(z0, z1), z2)
+(minus(+(z0, 1)), 1) → minus(z0)

Tuples:

+'(z0, minus(z1)) → c5(MINUS(+(minus(z0), z1)), +'(minus(z0), z1), MINUS(z0))
+'(z0, +(z1, z2)) → c6(+'(+(z0, z1), z2), +'(z0, z1))

S tuples:none
K tuples:

+'(z0, minus(z1)) → c5(MINUS(+(minus(z0), z1)), +'(minus(z0), z1), MINUS(z0))
+'(z0, +(z1, z2)) → c6(+'(+(z0, z1), z2), +'(z0, z1))

Defined Rule Symbols:

minus, +

Defined Pair Symbols:

+'

Compound Symbols:

c5, c6

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(8) BOUNDS(O(1), O(1))