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

(0) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0)))
plus(plus(z0, s(0)), plus(z1, s(s(z2)))) → plus(plus(z1, s(s(z2))), plus(z0, s(0)))

Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(PLUS(minus(z1, s(s(z2))), minus(z0, s(0))), MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(plus(z1, s(s(z2))), plus(z0, s(0))), PLUS(z1, s(s(z2))), PLUS(z0, s(0)))

S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(PLUS(minus(z1, s(s(z2))), minus(z0, s(0))), MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(plus(z1, s(s(z2))), plus(z0, s(0))), PLUS(z1, s(s(z2))), PLUS(z0, s(0)))

K tuples:none
Defined Rule Symbols:

minus, quot, plus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c6, c7

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

Removed 2 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0)))
plus(plus(z0, s(0)), plus(z1, s(s(z2)))) → plus(plus(z1, s(s(z2))), plus(z0, s(0)))

Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))

S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))

K tuples:none
Defined Rule Symbols:

minus, quot, plus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c6, c7

(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.

PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))

We considered the (Usable) Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)

And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))

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

POL(0) = [4]   
POL(MINUS(x₁, x₂)) = 0   
POL(PLUS(x₁, x₂)) = [4]x₂   
POL(QUOT(x₁, x₂)) = 0   
POL(c1(x₁)) = x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁)) = x₁   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(c7(x₁, x₂)) = x₁ + x₂   
POL(minus(x₁, x₂)) = [1]   
POL(plus(x₁, x₂)) = [4]x₂   
POL(s(x₁)) = [4]   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0)))
plus(plus(z0, s(0)), plus(z1, s(s(z2)))) → plus(plus(z1, s(s(z2))), plus(z0, s(0)))

Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))

S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))

K tuples:

PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))

Defined Rule Symbols:

minus, quot, plus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c6, c7

(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.

PLUS(s(z0), z1) → c5(PLUS(z0, z1))

We considered the (Usable) Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)

And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))

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

POL(0) = 0   
POL(MINUS(x₁, x₂)) = 0   
POL(PLUS(x₁, x₂)) = [4]x₁ + [4]x₂   
POL(QUOT(x₁, x₂)) = 0   
POL(c1(x₁)) = x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁)) = x₁   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(c7(x₁, x₂)) = x₁ + x₂   
POL(minus(x₁, x₂)) = [3]x₁   
POL(plus(x₁, x₂)) = [2]x₁ + x₂   
POL(s(x₁)) = [2] + x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0)))
plus(plus(z0, s(0)), plus(z1, s(s(z2)))) → plus(plus(z1, s(s(z2))), plus(z0, s(0)))

Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))

S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

K tuples:

PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))

Defined Rule Symbols:

minus, quot, plus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c6, c7

(9) 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.

QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

We considered the (Usable) Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)

And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))

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

POL(0) = [3]   
POL(MINUS(x₁, x₂)) = 0   
POL(PLUS(x₁, x₂)) = 0   
POL(QUOT(x₁, x₂)) = [2]x₁   
POL(c1(x₁)) = x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁)) = x₁   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(c7(x₁, x₂)) = x₁ + x₂   
POL(minus(x₁, x₂)) = x₁   
POL(plus(x₁, x₂)) = [3] + [3]x₁   
POL(s(x₁)) = [4] + x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0)))
plus(plus(z0, s(0)), plus(z1, s(s(z2)))) → plus(plus(z1, s(s(z2))), plus(z0, s(0)))

Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))

S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))

K tuples:

PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

Defined Rule Symbols:

minus, quot, plus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c6, c7

(11) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))

We considered the (Usable) Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)

And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))

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

POL(0) = [2]   
POL(MINUS(x₁, x₂)) = [2] + x₁   
POL(PLUS(x₁, x₂)) = [1] + [3]x₁ + [2]x₂   
POL(QUOT(x₁, x₂)) = [2]x₁²   
POL(c1(x₁)) = x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁)) = x₁   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(c7(x₁, x₂)) = x₁ + x₂   
POL(minus(x₁, x₂)) = [1] + x₁   
POL(plus(x₁, x₂)) = [2]x₁ + [2]x₂² + [3]x₁·x₂ + [3]x₁²   
POL(s(x₁)) = [2] + x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0)))
plus(plus(z0, s(0)), plus(z1, s(s(z2)))) → plus(plus(z1, s(s(z2))), plus(z0, s(0)))

Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))

S tuples:none
K tuples:

PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))

Defined Rule Symbols:

minus, quot, plus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c6, c7

(13) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(14) BOUNDS(O(1), O(1))