### (0) Obligation:

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

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: FULL

### (1) NestedDefinedSymbolProof (BOTH BOUNDS(ID, ID) transformation)

The following defined symbols can occur below the 0th argument of quot: minus
The following defined symbols can occur below the 0th argument of minus: minus

Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
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)))

### (2) Obligation:

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

The TRS R consists of the following rules:

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

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
quot(s(x), s([]))

The defined contexts are:
quot([], s(x1))
minus([], x1)

[] just represents basic- or constructor-terms in the following defined contexts:
quot([], s(x1))

As the TRS is an overlay system and the defined contexts and the duplicating contexts don't overlap, we have rc = irc.

### (4) Obligation:

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

The TRS R consists of the following rules:

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

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
quot(0, s(z0)) → 0
plus(s(z0), z1) → s(plus(z0, z1))
plus(0, z0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
minus(z0, 0) → z0
Tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
QUOT(0, s(z0)) → c1
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
PLUS(0, z0) → c3
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
MINUS(z0, 0) → c5
S tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
QUOT(0, s(z0)) → c1
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
PLUS(0, z0) → c3
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
MINUS(z0, 0) → c5
K tuples:none
Defined Rule Symbols:

quot, plus, minus

Defined Pair Symbols:

QUOT, PLUS, MINUS

Compound Symbols:

c, c1, c2, c3, c4, c5

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

Removed 3 trailing nodes:

MINUS(z0, 0) → c5
PLUS(0, z0) → c3
QUOT(0, s(z0)) → c1

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
quot(0, s(z0)) → 0
plus(s(z0), z1) → s(plus(z0, z1))
plus(0, z0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
minus(z0, 0) → z0
Tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
S tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

quot, plus, minus

Defined Pair Symbols:

QUOT, PLUS, MINUS

Compound Symbols:

c, c2, c4

### (9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
S tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

minus

Defined Pair Symbols:

QUOT, PLUS, MINUS

Compound Symbols:

c, c2, c4

### (11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

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

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
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
POL(QUOT(x1, x2)) = 0
POL(c(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c4(x1)) = x1
POL(minus(x1, x2)) = 0
POL(s(x1)) = [2] + x1

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
S tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
K tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
Defined Rule Symbols:

minus

Defined Pair Symbols:

QUOT, PLUS, MINUS

Compound Symbols:

c, c2, c4

### (13) CdtRuleRemovalProof (UPPER BOUND(ADD(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)) → c(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:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(MINUS(x1, x2)) = [3]
POL(PLUS(x1, x2)) = x1
POL(QUOT(x1, x2)) = [2]x1
POL(c(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c4(x1)) = x1
POL(minus(x1, x2)) = x1
POL(s(x1)) = [2] + x1

### (14) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
S tuples:

MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
K tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:

minus

Defined Pair Symbols:

QUOT, PLUS, MINUS

Compound Symbols:

c, c2, c4

### (15) CdtRuleRemovalProof (UPPER BOUND(ADD(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)) → c4(MINUS(z0, z1))
We considered the (Usable) Rules:

minus(s(z0), s(z1)) → minus(z0, z1)
minus(z0, 0) → z0
And the Tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [2]
POL(MINUS(x1, x2)) = [2] + x2
POL(PLUS(x1, x2)) = [2]x1
POL(QUOT(x1, x2)) = [2]x1·x2
POL(c(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c4(x1)) = x1
POL(minus(x1, x2)) = x1
POL(s(x1)) = [2] + x1

### (16) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
S tuples:none
K tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
Defined Rule Symbols:

minus

Defined Pair Symbols:

QUOT, PLUS, MINUS

Compound Symbols:

c, c2, c4

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

The set S is empty