YES(O(1), O(n^2)) 3.51/1.37 YES(O(1), O(n^2)) 3.90/1.43 3.90/1.43 3.90/1.43 3.90/1.43 3.90/1.43 3.90/1.43 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 3.90/1.43 3.90/1.43 3.90/1.43
3.90/1.43
3.90/1.43
3.90/1.43
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

0 CpxTRS
3.90/1.43 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
3.90/1.43 
↳2 CdtProblem
3.90/1.43 
↳3 CdtUnreachableProof (⇔)
3.90/1.43 
↳4 CdtProblem
3.90/1.43 
↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
3.90/1.43 
↳6 CdtProblem
3.90/1.43 
↳7 CdtKnowledgeProof (BOTH BOUNDS(ID, ID))
3.90/1.43 
↳8 CdtProblem
3.90/1.43 
↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
3.90/1.43 
↳10 CdtProblem
3.90/1.43 
↳11 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
3.90/1.43 
↳12 CdtProblem
3.90/1.43 
↳13 SIsEmptyProof (BOTH BOUNDS(ID, ID))
3.90/1.43 
↳14 BOUNDS(O(1), O(1))
3.90/1.43
3.90/1.43 3.90/1.43
3.90/1.43
3.90/1.43

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

plus(x, 0) → x 3.90/1.43
plus(0, y) → y 3.90/1.43
plus(s(x), y) → s(plus(x, y)) 3.90/1.43
times(0, y) → 0 3.90/1.43
times(s(0), y) → y 3.90/1.43
times(s(x), y) → plus(y, times(x, y)) 3.90/1.43
div(0, y) → 0 3.90/1.43
div(x, y) → quot(x, y, y) 3.90/1.43
quot(0, s(y), z) → 0 3.90/1.43
quot(s(x), s(y), z) → quot(x, y, z) 3.90/1.43
quot(x, 0, s(z)) → s(div(x, s(z))) 3.90/1.43
div(div(x, y), z) → div(x, times(y, z))

Rewrite Strategy: INNERMOST
3.90/1.43
3.90/1.43

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

Converted CpxTRS to CDT
3.90/1.43
3.90/1.43

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

plus(z0, 0) → z0 3.90/1.43
plus(0, z0) → z0 3.90/1.43
plus(s(z0), z1) → s(plus(z0, z1)) 3.90/1.43
times(0, z0) → 0 3.90/1.43
times(s(0), z0) → z0 3.90/1.43
times(s(z0), z1) → plus(z1, times(z0, z1)) 3.90/1.43
div(0, z0) → 0 3.90/1.43
div(z0, z1) → quot(z0, z1, z1) 3.90/1.43
div(div(z0, z1), z2) → div(z0, times(z1, z2)) 3.90/1.43
quot(0, s(z0), z1) → 0 3.90/1.43
quot(s(z0), s(z1), z2) → quot(z0, z1, z2) 3.90/1.43
quot(z0, 0, s(z1)) → s(div(z0, s(z1)))
Tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1)) 3.90/1.43
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1)) 3.90/1.43
DIV(z0, z1) → c7(QUOT(z0, z1, z1)) 3.90/1.43
DIV(div(z0, z1), z2) → c8(DIV(z0, times(z1, z2)), TIMES(z1, z2)) 3.90/1.43
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2)) 3.90/1.43
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))
S tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1)) 3.90/1.43
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1)) 3.90/1.43
DIV(z0, z1) → c7(QUOT(z0, z1, z1)) 3.90/1.43
DIV(div(z0, z1), z2) → c8(DIV(z0, times(z1, z2)), TIMES(z1, z2)) 3.90/1.43
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2)) 3.90/1.43
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

3.90/1.43
3.90/1.43

(3) CdtUnreachableProof (EQUIVALENT transformation)

The following tuples could be removed as they are not reachable from basic start terms:

DIV(div(z0, z1), z2) → c8(DIV(z0, times(z1, z2)), TIMES(z1, z2))
3.90/1.43
3.90/1.43

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

plus(z0, 0) → z0 3.90/1.43
plus(0, z0) → z0 3.90/1.43
plus(s(z0), z1) → s(plus(z0, z1)) 3.90/1.43
times(0, z0) → 0 3.90/1.43
times(s(0), z0) → z0 3.90/1.43
times(s(z0), z1) → plus(z1, times(z0, z1)) 3.90/1.43
div(0, z0) → 0 3.90/1.43
div(z0, z1) → quot(z0, z1, z1) 3.90/1.43
div(div(z0, z1), z2) → div(z0, times(z1, z2)) 3.90/1.43
quot(0, s(z0), z1) → 0 3.90/1.43
quot(s(z0), s(z1), z2) → quot(z0, z1, z2) 3.90/1.43
quot(z0, 0, s(z1)) → s(div(z0, s(z1)))
Tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1)) 3.90/1.43
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1)) 3.90/1.43
DIV(z0, z1) → c7(QUOT(z0, z1, z1)) 3.90/1.43
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2)) 3.90/1.43
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))
S tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1)) 3.90/1.43
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1)) 3.90/1.43
DIV(z0, z1) → c7(QUOT(z0, z1, z1)) 3.90/1.43
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2)) 3.90/1.43
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

3.90/1.43
3.90/1.43

(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.

QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2))
We considered the (Usable) Rules:

times(0, z0) → 0 3.90/1.43
times(s(0), z0) → z0 3.90/1.43
times(s(z0), z1) → plus(z1, times(z0, z1)) 3.90/1.43
plus(z0, 0) → z0 3.90/1.43
plus(0, z0) → z0 3.90/1.43
plus(s(z0), z1) → s(plus(z0, z1))
And the Tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1)) 3.90/1.43
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1)) 3.90/1.43
DIV(z0, z1) → c7(QUOT(z0, z1, z1)) 3.90/1.43
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2)) 3.90/1.43
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation : 3.90/1.43

POL(0) = 0    3.90/1.43
POL(DIV(x1, x2)) = x1    3.90/1.43
POL(PLUS(x1, x2)) = 0    3.90/1.43
POL(QUOT(x1, x2, x3)) = x1    3.90/1.43
POL(TIMES(x1, x2)) = 0    3.90/1.43
POL(c10(x1)) = x1    3.90/1.43
POL(c11(x1)) = x1    3.90/1.43
POL(c2(x1)) = x1    3.90/1.43
POL(c5(x1, x2)) = x1 + x2    3.90/1.43
POL(c7(x1)) = x1    3.90/1.43
POL(plus(x1, x2)) = [3] + [3]x1 + [5]x2    3.90/1.43
POL(s(x1)) = [1] + x1    3.90/1.43
POL(times(x1, x2)) = 0   
3.90/1.43
3.90/1.43

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

plus(z0, 0) → z0 3.90/1.43
plus(0, z0) → z0 3.90/1.43
plus(s(z0), z1) → s(plus(z0, z1)) 3.90/1.43
times(0, z0) → 0 3.90/1.43
times(s(0), z0) → z0 3.90/1.43
times(s(z0), z1) → plus(z1, times(z0, z1)) 3.90/1.43
div(0, z0) → 0 3.90/1.43
div(z0, z1) → quot(z0, z1, z1) 3.90/1.43
div(div(z0, z1), z2) → div(z0, times(z1, z2)) 3.90/1.43
quot(0, s(z0), z1) → 0 3.90/1.43
quot(s(z0), s(z1), z2) → quot(z0, z1, z2) 3.90/1.43
quot(z0, 0, s(z1)) → s(div(z0, s(z1)))
Tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1)) 3.90/1.43
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1)) 3.90/1.43
DIV(z0, z1) → c7(QUOT(z0, z1, z1)) 3.90/1.43
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2)) 3.90/1.43
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))
S tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1)) 3.90/1.43
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1)) 3.90/1.44
DIV(z0, z1) → c7(QUOT(z0, z1, z1)) 3.90/1.44
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))
K tuples:

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

3.90/1.44
3.90/1.44

(7) CdtKnowledgeProof (BOTH BOUNDS(ID, ID) transformation)

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

QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1))) 3.90/1.44
DIV(z0, z1) → c7(QUOT(z0, z1, z1)) 3.90/1.44
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2))
3.90/1.44
3.90/1.44

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

plus(z0, 0) → z0 3.90/1.44
plus(0, z0) → z0 3.90/1.44
plus(s(z0), z1) → s(plus(z0, z1)) 3.90/1.44
times(0, z0) → 0 3.90/1.44
times(s(0), z0) → z0 3.90/1.44
times(s(z0), z1) → plus(z1, times(z0, z1)) 3.90/1.44
div(0, z0) → 0 3.90/1.44
div(z0, z1) → quot(z0, z1, z1) 3.90/1.44
div(div(z0, z1), z2) → div(z0, times(z1, z2)) 3.90/1.44
quot(0, s(z0), z1) → 0 3.90/1.44
quot(s(z0), s(z1), z2) → quot(z0, z1, z2) 3.90/1.44
quot(z0, 0, s(z1)) → s(div(z0, s(z1)))
Tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1)) 3.90/1.44
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1)) 3.90/1.44
DIV(z0, z1) → c7(QUOT(z0, z1, z1)) 3.90/1.44
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2)) 3.90/1.44
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))
S tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1)) 3.90/1.44
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
K tuples:

QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2)) 3.90/1.44
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1))) 3.90/1.44
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

3.90/1.44
3.90/1.44

(9) 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 3.90/1.44
times(s(0), z0) → z0 3.90/1.44
times(s(z0), z1) → plus(z1, times(z0, z1)) 3.90/1.44
plus(z0, 0) → z0 3.90/1.44
plus(0, z0) → z0 3.90/1.44
plus(s(z0), z1) → s(plus(z0, z1))
And the Tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1)) 3.90/1.44
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1)) 3.90/1.44
DIV(z0, z1) → c7(QUOT(z0, z1, z1)) 3.90/1.44
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2)) 3.90/1.44
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation : 3.90/1.44

POL(0) = 0    3.90/1.44
POL(DIV(x1, x2)) = [5]x1    3.90/1.44
POL(PLUS(x1, x2)) = [1]    3.90/1.44
POL(QUOT(x1, x2, x3)) = [5]x1    3.90/1.44
POL(TIMES(x1, x2)) = [4]x1    3.90/1.44
POL(c10(x1)) = x1    3.90/1.44
POL(c11(x1)) = x1    3.90/1.44
POL(c2(x1)) = x1    3.90/1.44
POL(c5(x1, x2)) = x1 + x2    3.90/1.44
POL(c7(x1)) = x1    3.90/1.44
POL(plus(x1, x2)) = [3] + [3]x1 + [5]x2    3.90/1.44
POL(s(x1)) = [4] + x1    3.90/1.44
POL(times(x1, x2)) = 0   
3.90/1.44
3.90/1.44

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

plus(z0, 0) → z0 3.90/1.44
plus(0, z0) → z0 3.90/1.44
plus(s(z0), z1) → s(plus(z0, z1)) 3.90/1.44
times(0, z0) → 0 3.90/1.44
times(s(0), z0) → z0 3.90/1.44
times(s(z0), z1) → plus(z1, times(z0, z1)) 3.90/1.44
div(0, z0) → 0 3.90/1.44
div(z0, z1) → quot(z0, z1, z1) 3.90/1.44
div(div(z0, z1), z2) → div(z0, times(z1, z2)) 3.90/1.44
quot(0, s(z0), z1) → 0 3.90/1.44
quot(s(z0), s(z1), z2) → quot(z0, z1, z2) 3.90/1.44
quot(z0, 0, s(z1)) → s(div(z0, s(z1)))
Tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1)) 3.90/1.44
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1)) 3.90/1.44
DIV(z0, z1) → c7(QUOT(z0, z1, z1)) 3.90/1.44
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2)) 3.90/1.44
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))
S tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
K tuples:

QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2)) 3.90/1.44
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1))) 3.90/1.44
DIV(z0, z1) → c7(QUOT(z0, z1, z1)) 3.90/1.44
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

3.90/1.44
3.90/1.44

(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 3.90/1.44
times(s(0), z0) → z0 3.90/1.44
times(s(z0), z1) → plus(z1, times(z0, z1)) 3.90/1.44
plus(z0, 0) → z0 3.90/1.44
plus(0, z0) → z0 3.90/1.44
plus(s(z0), z1) → s(plus(z0, z1))
And the Tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1)) 3.90/1.44
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1)) 3.90/1.44
DIV(z0, z1) → c7(QUOT(z0, z1, z1)) 3.90/1.44
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2)) 3.90/1.44
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation : 3.90/1.44

POL(0) = 0    3.90/1.44
POL(DIV(x1, x2)) = 0    3.90/1.44
POL(PLUS(x1, x2)) = [2]x1    3.90/1.44
POL(QUOT(x1, x2, x3)) = 0    3.90/1.44
POL(TIMES(x1, x2)) = x1 + [2]x1·x2    3.90/1.44
POL(c10(x1)) = x1    3.90/1.44
POL(c11(x1)) = x1    3.90/1.44
POL(c2(x1)) = x1    3.90/1.44
POL(c5(x1, x2)) = x1 + x2    3.90/1.44
POL(c7(x1)) = x1    3.90/1.44
POL(plus(x1, x2)) = [2] + [2]x1 + x2    3.90/1.44
POL(s(x1)) = [2] + x1    3.90/1.44
POL(times(x1, x2)) = [2]x2 + x1·x2 + [2]x12   
3.90/1.44
3.90/1.44

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

plus(z0, 0) → z0 3.90/1.44
plus(0, z0) → z0 3.90/1.44
plus(s(z0), z1) → s(plus(z0, z1)) 3.90/1.44
times(0, z0) → 0 3.90/1.44
times(s(0), z0) → z0 3.90/1.44
times(s(z0), z1) → plus(z1, times(z0, z1)) 3.90/1.44
div(0, z0) → 0 3.90/1.44
div(z0, z1) → quot(z0, z1, z1) 3.90/1.44
div(div(z0, z1), z2) → div(z0, times(z1, z2)) 3.90/1.44
quot(0, s(z0), z1) → 0 3.90/1.44
quot(s(z0), s(z1), z2) → quot(z0, z1, z2) 3.90/1.44
quot(z0, 0, s(z1)) → s(div(z0, s(z1)))
Tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1)) 3.90/1.44
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1)) 3.90/1.44
DIV(z0, z1) → c7(QUOT(z0, z1, z1)) 3.90/1.44
QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2)) 3.90/1.44
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1)))
S tuples:none
K tuples:

QUOT(s(z0), s(z1), z2) → c10(QUOT(z0, z1, z2)) 3.90/1.44
QUOT(z0, 0, s(z1)) → c11(DIV(z0, s(z1))) 3.90/1.44
DIV(z0, z1) → c7(QUOT(z0, z1, z1)) 3.90/1.44
TIMES(s(z0), z1) → c5(PLUS(z1, times(z0, z1)), TIMES(z0, z1)) 3.90/1.44
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

3.90/1.44
3.90/1.44

(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty
3.90/1.44
3.90/1.44

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

3.90/1.44
3.90/1.44
4.13/1.51 EOF