### (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)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
div(s(X), s([]))

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

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

### (2) 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:

minus(X, 0) → X
minus(s(X), s(Y)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

MINUS(z0, 0) → c
MINUS(s(z0), s(z1)) → c1(P(minus(z0, z1)), MINUS(z0, z1))
P(s(z0)) → c2
DIV(0, s(z0)) → c3
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:

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

minus, p, div

Defined Pair Symbols:

MINUS, P, DIV

Compound Symbols:

c, c1, c2, c3, c4

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

Removed 3 trailing nodes:

MINUS(z0, 0) → c
DIV(0, s(z0)) → c3
P(s(z0)) → c2

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

MINUS(s(z0), s(z1)) → c1(P(minus(z0, z1)), MINUS(z0, z1))
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:

MINUS(s(z0), s(z1)) → c1(P(minus(z0, z1)), MINUS(z0, z1))
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

minus, p, div

Defined Pair Symbols:

MINUS, DIV

Compound Symbols:

c1, c4

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

Removed 1 trailing tuple parts

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
S tuples:

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

minus, p, div

Defined Pair Symbols:

DIV, MINUS

Compound Symbols:

c4, c1

### (9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
S tuples:

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

minus, p

Defined Pair Symbols:

DIV, MINUS

Compound Symbols:

c4, c1

### (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.

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
We considered the (Usable) Rules:

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

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(DIV(x1, x2)) = [2]x1
POL(MINUS(x1, x2)) = [1]
POL(c1(x1)) = x1
POL(c4(x1, x2)) = x1 + x2
POL(minus(x1, x2)) = x1
POL(p(x1)) = [1] + x1
POL(s(x1)) = [2] + x1

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
S tuples:

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

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:

minus, p

Defined Pair Symbols:

DIV, MINUS

Compound Symbols:

c4, c1

### (13) 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)) → c1(MINUS(z0, z1))
We considered the (Usable) Rules:

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

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(DIV(x1, x2)) = x12
POL(MINUS(x1, x2)) = [2]x1
POL(c1(x1)) = x1
POL(c4(x1, x2)) = x1 + x2
POL(minus(x1, x2)) = x1
POL(p(x1)) = [1] + x1
POL(s(x1)) = [1] + x1

### (14) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
S tuples:none
K tuples:

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
Defined Rule Symbols:

minus, p

Defined Pair Symbols:

DIV, MINUS

Compound Symbols:

c4, c1

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

The set S is empty