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)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))

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)
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(s(z0), s(z1)), s(z1)))

Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z0), s(z1)) → c6(QUOT(minus(s(z0), s(z1)), s(z1)), MINUS(s(z0), s(z1)))

S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z0), s(z1)) → c6(QUOT(minus(s(z0), s(z1)), s(z1)), MINUS(s(z0), s(z1)))

K tuples:none
Defined Rule Symbols:

minus, le, quot

Defined Pair Symbols:

MINUS, LE, QUOT

Compound Symbols:

c1, c4, c6

(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.

LE(s(z0), s(z1)) → c4(LE(z0, z1))

We considered the (Usable) Rules:

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

And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z0), s(z1)) → c6(QUOT(minus(s(z0), s(z1)), s(z1)), MINUS(s(z0), s(z1)))

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

POL(0) = 0   
POL(LE(x₁, x₂)) = [2]x₁   
POL(MINUS(x₁, x₂)) = 0   
POL(QUOT(x₁, x₂)) = 0   
POL(c1(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(minus(x₁, x₂)) = [3]   
POL(s(x₁)) = [4] + x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(s(z0), s(z1)), s(z1)))

Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z0), s(z1)) → c6(QUOT(minus(s(z0), s(z1)), s(z1)), MINUS(s(z0), s(z1)))

S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c6(QUOT(minus(s(z0), s(z1)), s(z1)), MINUS(s(z0), s(z1)))

K tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))

Defined Rule Symbols:

minus, le, quot

Defined Pair Symbols:

MINUS, LE, QUOT

Compound Symbols:

c1, c4, c6

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

Use narrowing to replace QUOT(s(z0), s(z1)) → c6(QUOT(minus(s(z0), s(z1)), s(z1)), MINUS(s(z0), s(z1))) by

QUOT(s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
QUOT(s(x0), s(x1)) → c6

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(s(z0), s(z1)), s(z1)))

Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
QUOT(s(x0), s(x1)) → c6

S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
QUOT(s(x0), s(x1)) → c6

K tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))

Defined Rule Symbols:

minus, le, quot

Defined Pair Symbols:

MINUS, LE, QUOT

Compound Symbols:

c1, c4, c6, c6

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

Removed 1 trailing nodes:

QUOT(s(x0), s(x1)) → c6

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(s(z0), s(z1)), s(z1)))

Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))

S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))

K tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))

Defined Rule Symbols:

minus, le, quot

Defined Pair Symbols:

MINUS, LE, QUOT

Compound Symbols:

c1, c4, c6

(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)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(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))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))

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

POL(0) = 0   
POL(LE(x₁, x₂)) = [3]x₂   
POL(MINUS(x₁, x₂)) = 0   
POL(QUOT(x₁, x₂)) = [2]x₁   
POL(c1(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(minus(x₁, x₂)) = 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)
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(s(z0), s(z1)), s(z1)))

Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))

S tuples:

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

K tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))

Defined Rule Symbols:

minus, le, quot

Defined Pair Symbols:

MINUS, LE, QUOT

Compound Symbols:

c1, c4, c6

(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))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))

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

POL(0) = 0   
POL(LE(x₁, x₂)) = [2]x₁ + [3]x₂ + x₁·x₂ + x₁²   
POL(MINUS(x₁, x₂)) = [1] + x₁   
POL(QUOT(x₁, x₂)) = [2]x₁²   
POL(c1(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(minus(x₁, x₂)) = x₁   
POL(s(x₁)) = [2] + x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(s(z0), s(z1)), s(z1)))

Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))

S tuples:none
K tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))

Defined Rule Symbols:

minus, le, quot

Defined Pair Symbols:

MINUS, LE, QUOT

Compound Symbols:

c1, c4, c6

(13) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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