### (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 Cpx (relative) TRS 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(z0, 0) → c
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(0, z0) → c2
LE(s(z0), 0) → c3
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(0, s(z0)) → c5
QUOT(s(z0), s(z1)) → c6(QUOT(minus(s(z0), s(z1)), s(z1)), MINUS(s(z0), s(z1)))
S tuples:

MINUS(z0, 0) → c
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(0, z0) → c2
LE(s(z0), 0) → c3
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(0, s(z0)) → c5
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:

c, c1, c2, c3, c4, c5, c6

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

Removed 4 trailing nodes:

MINUS(z0, 0) → c
QUOT(0, s(z0)) → c5
LE(0, z0) → c2
LE(s(z0), 0) → c3

### (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))
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

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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)))

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Defined Pair Symbols:

MINUS, LE, QUOT

Compound Symbols:

c1, c4, c6

### (7) CdtRuleRemovalProof (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:none
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

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Defined Pair Symbols:

MINUS, LE, QUOT

Compound Symbols:

c1, c4, c6

### (9) 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)))

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Defined Pair Symbols:

MINUS, LE, QUOT

Compound Symbols:

c1, c4, c6

### (11) CdtRuleRemovalProof (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(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(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
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

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Defined Pair Symbols:

MINUS, LE, QUOT

Compound Symbols:

c1, c4, c6

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

### (14) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Defined Pair Symbols:

MINUS, LE, QUOT

Compound Symbols:

c1, c4, c6

### (15) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty