### (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:

plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))

Rewrite Strategy: FULL

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

The following defined symbols can occur below the 1th argument of plus: plus, times

Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
div(div(x, y), z) → div(x, times(y, z))

### (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:

plus(s(x), y) → s(plus(x, y))
times(s(x), y) → plus(y, times(x, y))
quot(x, 0, s(z)) → s(div(x, s(z)))
plus(x, 0) → x
plus(0, y) → y
div(0, y) → 0
div(x, y) → quot(x, y, y)
times(s(0), y) → y
quot(s(x), s(y), z) → quot(x, y, z)
times(0, y) → 0
quot(0, s(y), z) → 0

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

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

The defined contexts are:
plus(x0, [])

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:

plus(s(x), y) → s(plus(x, y))
times(s(x), y) → plus(y, times(x, y))
quot(x, 0, s(z)) → s(div(x, s(z)))
plus(x, 0) → x
plus(0, y) → y
div(0, y) → 0
div(x, y) → quot(x, y, y)
times(s(0), y) → y
quot(s(x), s(y), z) → quot(x, y, z)
times(0, y) → 0
quot(0, s(y), z) → 0

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

plus(s(z0), z1) → s(plus(z0, z1))
plus(z0, 0) → z0
plus(0, z0) → z0
times(s(z0), z1) → plus(z1, times(z0, z1))
times(s(0), z0) → z0
times(0, z0) → 0
quot(z0, 0, s(z1)) → s(div(z0, s(z1)))
quot(s(z0), s(z1), z2) → quot(z0, z1, z2)
quot(0, s(z0), z1) → 0
div(0, z0) → 0
div(z0, z1) → quot(z0, z1, z1)
Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
PLUS(z0, 0) → c1
PLUS(0, z0) → c2
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
TIMES(s(0), z0) → c4
TIMES(0, z0) → c5
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
QUOT(0, s(z0), z1) → c8
DIV(0, z0) → c9
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
S tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
PLUS(z0, 0) → c1
PLUS(0, z0) → c2
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
TIMES(s(0), z0) → c4
TIMES(0, z0) → c5
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
QUOT(0, s(z0), z1) → c8
DIV(0, z0) → c9
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
K tuples:none
Defined Rule Symbols:

plus, times, quot, div

Defined Pair Symbols:

PLUS, TIMES, QUOT, DIV

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10

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

Removed 6 trailing nodes:

DIV(0, z0) → c9
PLUS(z0, 0) → c1
PLUS(0, z0) → c2
TIMES(s(0), z0) → c4
TIMES(0, z0) → c5
QUOT(0, s(z0), z1) → c8

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

plus(s(z0), z1) → s(plus(z0, z1))
plus(z0, 0) → z0
plus(0, z0) → z0
times(s(z0), z1) → plus(z1, times(z0, z1))
times(s(0), z0) → z0
times(0, z0) → 0
quot(z0, 0, s(z1)) → s(div(z0, s(z1)))
quot(s(z0), s(z1), z2) → quot(z0, z1, z2)
quot(0, s(z0), z1) → 0
div(0, z0) → 0
div(z0, z1) → quot(z0, z1, z1)
Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
S tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
K tuples:none
Defined Rule Symbols:

plus, times, quot, div

Defined Pair Symbols:

PLUS, TIMES, QUOT, DIV

Compound Symbols:

c, c3, c6, c7, c10

### (9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

quot(z0, 0, s(z1)) → s(div(z0, s(z1)))
quot(s(z0), s(z1), z2) → quot(z0, z1, z2)
quot(0, s(z0), z1) → 0
div(0, z0) → 0
div(z0, z1) → quot(z0, z1, z1)

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(s(z0), z1) → plus(z1, times(z0, z1))
times(s(0), z0) → z0
times(0, z0) → 0
plus(s(z0), z1) → s(plus(z0, z1))
plus(z0, 0) → z0
plus(0, z0) → z0
Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
S tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
K tuples:none
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

PLUS, TIMES, QUOT, DIV

Compound Symbols:

c, c3, c6, c7, c10

### (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.

QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
We considered the (Usable) Rules:none
And the Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(DIV(x1, x2)) = x1
POL(PLUS(x1, x2)) = 0
POL(QUOT(x1, x2, x3)) = x1
POL(TIMES(x1, x2)) = 0
POL(c(x1)) = x1
POL(c10(x1)) = x1
POL(c3(x1, x2)) = x1 + x2
POL(c6(x1)) = x1
POL(c7(x1)) = x1
POL(plus(x1, x2)) = 0
POL(s(x1)) = [1] + x1
POL(times(x1, x2)) = 0

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(s(z0), z1) → plus(z1, times(z0, z1))
times(s(0), z0) → z0
times(0, z0) → 0
plus(s(z0), z1) → s(plus(z0, z1))
plus(z0, 0) → z0
plus(0, z0) → z0
Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
S tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
K tuples:

QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

PLUS, TIMES, QUOT, DIV

Compound Symbols:

c, c3, c6, c7, c10

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

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

QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))

### (14) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(s(z0), z1) → plus(z1, times(z0, z1))
times(s(0), z0) → z0
times(0, z0) → 0
plus(s(z0), z1) → s(plus(z0, z1))
plus(z0, 0) → z0
plus(0, z0) → z0
Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
S tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
K tuples:

QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

PLUS, TIMES, QUOT, DIV

Compound Symbols:

c, c3, c6, c7, c10

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

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

TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(DIV(x1, x2)) = 0
POL(PLUS(x1, x2)) = 0
POL(QUOT(x1, x2, x3)) = 0
POL(TIMES(x1, x2)) = x1
POL(c(x1)) = x1
POL(c10(x1)) = x1
POL(c3(x1, x2)) = x1 + x2
POL(c6(x1)) = x1
POL(c7(x1)) = x1
POL(plus(x1, x2)) = 0
POL(s(x1)) = [1] + x1
POL(times(x1, x2)) = 0

### (16) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(s(z0), z1) → plus(z1, times(z0, z1))
times(s(0), z0) → z0
times(0, z0) → 0
plus(s(z0), z1) → s(plus(z0, z1))
plus(z0, 0) → z0
plus(0, z0) → z0
Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
S tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
K tuples:

QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

PLUS, TIMES, QUOT, DIV

Compound Symbols:

c, c3, c6, c7, c10

### (17) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

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

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

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [1]
POL(DIV(x1, x2)) = 0
POL(PLUS(x1, x2)) = x1
POL(QUOT(x1, x2, x3)) = 0
POL(TIMES(x1, x2)) = [2]x1·x2
POL(c(x1)) = x1
POL(c10(x1)) = x1
POL(c3(x1, x2)) = x1 + x2
POL(c6(x1)) = x1
POL(c7(x1)) = x1
POL(plus(x1, x2)) = [2] + [2]x1 + [2]x2 + x22 + [2]x1·x2 + [2]x12
POL(s(x1)) = [2] + x1
POL(times(x1, x2)) = [1] + x1 + x2 + x22 + x1·x2 + x12

### (18) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(s(z0), z1) → plus(z1, times(z0, z1))
times(s(0), z0) → z0
times(0, z0) → 0
plus(s(z0), z1) → s(plus(z0, z1))
plus(z0, 0) → z0
plus(0, z0) → z0
Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
S tuples:none
K tuples:

QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
PLUS(s(z0), z1) → c(PLUS(z0, z1))
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

PLUS, TIMES, QUOT, DIV

Compound Symbols:

c, c3, c6, c7, c10

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

The set S is empty