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

(0) Obligation:

Runtime Complexity TRS:
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: INNERMOST

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
DIV(z0, z1) → c7(QUOT(z0, z1, z1))
DIV(div(z0, z1), z2) → c8(DIV(z0, times(z1, z2)), TIMES(z1, z2))
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))

S tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
DIV(z0, z1) → c7(QUOT(z0, z1, z1))
DIV(div(z0, z1), z2) → c8(DIV(z0, times(z1, z2)), TIMES(z1, z2))
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))

K tuples:none
Defined Rule Symbols:

plus, times, div, quot

Defined Pair Symbols:

PLUS, TIMES, DIV, QUOT

Compound Symbols:

c2, c5, c7, c8, c10, c11

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

DIV(div(z0, z1), z2) → c8(DIV(z0, times(z1, z2)), TIMES(z1, z2))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
DIV(z0, z1) → c7(QUOT(z0, z1, z1))
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))

S tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
DIV(z0, z1) → c7(QUOT(z0, z1, z1))
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))

K tuples:none
Defined Rule Symbols:

plus, times, div, quot

Defined Pair Symbols:

PLUS, TIMES, DIV, QUOT

Compound Symbols:

c2, c5, c7, c10, c11

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

We considered the (Usable) Rules:

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

And the Tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
DIV(z0, z1) → c7(QUOT(z0, z1, z1))
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [3]   
POL(DIV(x₁, x₂)) = x₂   
POL(PLUS(x₁, x₂)) = 0   
POL(QUOT(x₁, x₂, x₃)) = x₃   
POL(TIMES(x₁, x₂)) = [4]x₁   
POL(c10(x₁)) = x₁   
POL(c11(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(c5(x₁, x₂)) = x₁ + x₂   
POL(c7(x₁)) = x₁   
POL(plus(x₁, x₂)) = [2] + [4]x₁ + [2]x₂   
POL(s(x₁)) = [5] + x₁   
POL(times(x₁, x₂)) = [4] + [3]x₂   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
DIV(z0, z1) → c7(QUOT(z0, z1, z1))
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))

S tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
DIV(z0, z1) → c7(QUOT(z0, z1, z1))
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))

K tuples:

TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

Defined Rule Symbols:

plus, times, div, quot

Defined Pair Symbols:

PLUS, TIMES, DIV, QUOT

Compound Symbols:

c2, c5, c7, c10, c11

(7) CdtPolyRedPairProof (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), z2) → c10(QUOT(z0, z1, z2))

We considered the (Usable) Rules:

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

And the Tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
DIV(z0, z1) → c7(QUOT(z0, z1, z1))
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [3]   
POL(DIV(x₁, x₂)) = [2]x₁   
POL(PLUS(x₁, x₂)) = [1]   
POL(QUOT(x₁, x₂, x₃)) = [2]x₁   
POL(TIMES(x₁, x₂)) = x₁   
POL(c10(x₁)) = x₁   
POL(c11(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(c5(x₁, x₂)) = x₁ + x₂   
POL(c7(x₁)) = x₁   
POL(plus(x₁, x₂)) = [3] + [2]x₁   
POL(s(x₁)) = [2] + x₁   
POL(times(x₁, x₂)) = [2]x₂   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
DIV(z0, z1) → c7(QUOT(z0, z1, z1))
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))

S tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
DIV(z0, z1) → c7(QUOT(z0, z1, z1))
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))

K tuples:

TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2))

Defined Rule Symbols:

plus, times, div, quot

Defined Pair Symbols:

PLUS, TIMES, DIV, QUOT

Compound Symbols:

c2, c5, c7, c10, c11

(9) CdtKnowledgeProof (EQUIVALENT transformation)

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

QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))
DIV(z0, z1) → c7(QUOT(z0, z1, z1))
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
DIV(z0, z1) → c7(QUOT(z0, z1, z1))
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))

S tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))

K tuples:

TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))
DIV(z0, z1) → c7(QUOT(z0, z1, z1))

Defined Rule Symbols:

plus, times, div, quot

Defined Pair Symbols:

PLUS, TIMES, DIV, QUOT

Compound Symbols:

c2, c5, c7, c10, c11

(11) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) 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:

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

And the Tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
DIV(z0, z1) → c7(QUOT(z0, z1, z1))
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(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)) = x₁   
POL(QUOT(x₁, x₂, x₃)) = 0   
POL(TIMES(x₁, x₂)) = [2]x₁ + [2]x₁·x₂ + [3]x₁²   
POL(c10(x₁)) = x₁   
POL(c11(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(c5(x₁, x₂)) = x₁ + x₂   
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:

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

Tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
DIV(z0, z1) → c7(QUOT(z0, z1, z1))
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))

S tuples:none
K tuples:

TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))
DIV(z0, z1) → c7(QUOT(z0, z1, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))

Defined Rule Symbols:

plus, times, div, quot

Defined Pair Symbols:

PLUS, TIMES, DIV, QUOT

Compound Symbols:

c2, c5, c7, c10, c11

(13) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(14) BOUNDS(O(1), O(1))