### (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 Cpx (relative) TRS 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(z0, 0) → c
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(0, s(z0)) → c2
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(0, z0) → c4
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(z0, 0) → c
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(0, s(z0)) → c2
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(0, z0) → c4
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:

c, c1, c2, c3, c4, c5, c6, c7

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

Removed 3 trailing nodes:

MINUS(z0, 0) → c
QUOT(0, s(z0)) → c2
PLUS(0, z0) → c4

### (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(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

### (5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 2 trailing tuple parts

### (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))
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

### (7) CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID) transformation)

Split RHS of tuples not part of any SCC

### (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(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2))))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(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(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2))))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
K tuples:none
Defined Rule Symbols:

minus, quot, plus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c7, c

### (9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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)))

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
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(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2))))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(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(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2))))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
K tuples:none
Defined Rule Symbols:

minus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c7, c

### (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.

PLUS(s(z0), z1) → c5(PLUS(z0, z1))
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:none
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(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2))))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(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)) = [2]x1 + [2]x2
POL(QUOT(x1, x2)) = 0
POL(c(x1)) = x1
POL(c1(x1)) = x1
POL(c3(x1, x2)) = x1 + x2
POL(c5(x1)) = x1
POL(c7(x1, x2)) = x1 + x2
POL(minus(x1, x2)) = [2]x1
POL(plus(x1, x2)) = [1] + x1 + x2
POL(s(x1)) = [1] + x1

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
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(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2))))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(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(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2))))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
K tuples:

PLUS(s(z0), z1) → c5(PLUS(z0, z1))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
Defined Rule Symbols:

minus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c7, c

### (13) CdtKnowledgeProof (BOTH BOUNDS(ID, ID) transformation)

The following tuples could be moved from S to K by knowledge propagation:

PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2))))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))

### (14) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
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(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2))))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(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(s(z0), z1) → c5(PLUS(z0, z1))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2))))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
Defined Rule Symbols:

minus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c7, c

### (15) 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)) → c3(QUOT(minus(z0, z1), s(z1)), 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))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2))))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(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)) = 0
POL(QUOT(x1, x2)) = x1
POL(c(x1)) = x1
POL(c1(x1)) = x1
POL(c3(x1, x2)) = x1 + x2
POL(c5(x1)) = x1
POL(c7(x1, x2)) = x1 + x2
POL(minus(x1, x2)) = x1
POL(plus(x1, x2)) = [3]
POL(s(x1)) = [1] + x1

### (16) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
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(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2))))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
K tuples:

PLUS(s(z0), z1) → c5(PLUS(z0, z1))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2))))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:

minus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c7, c

### (17) 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))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2))))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

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

### (18) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
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(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2))))
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
S tuples:none
K tuples:

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

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c7, c

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

The set S is empty