YES(O(1), O(n^2)) 3.15/1.25 YES(O(1), O(n^2)) 3.15/1.28 3.15/1.28 3.15/1.28 3.15/1.28 3.15/1.28 3.15/1.28 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 3.15/1.28 3.15/1.28 3.15/1.28
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The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

0 CpxTRS
3.15/1.28 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
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↳2 CdtProblem
3.15/1.28 
↳3 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
3.15/1.28 
↳4 CdtProblem
3.15/1.28 
↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
3.15/1.28 
↳6 CdtProblem
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↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
3.15/1.28 
↳8 CdtProblem
3.15/1.28 
↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
3.15/1.28 
↳10 CdtProblem
3.15/1.28 
↳11 SIsEmptyProof (BOTH BOUNDS(ID, ID))
3.15/1.28 
↳12 BOUNDS(O(1), O(1))
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3.15/1.28 3.15/1.28
3.15/1.28
3.15/1.28

(0) Obligation:

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

-(x, 0) → x 3.15/1.28
-(0, s(y)) → 0 3.15/1.28
-(s(x), s(y)) → -(x, y) 3.15/1.28
lt(x, 0) → false 3.15/1.28
lt(0, s(y)) → true 3.15/1.28
lt(s(x), s(y)) → lt(x, y) 3.15/1.28
if(true, x, y) → x 3.15/1.28
if(false, x, y) → y 3.15/1.28
div(x, 0) → 0 3.15/1.28
div(0, y) → 0 3.15/1.28
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

Rewrite Strategy: INNERMOST
3.15/1.28
3.15/1.28

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

Converted CpxTRS to CDT
3.15/1.28
3.15/1.28

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

-(z0, 0) → z0 3.15/1.28
-(0, s(z0)) → 0 3.15/1.28
-(s(z0), s(z1)) → -(z0, z1) 3.15/1.28
lt(z0, 0) → false 3.15/1.28
lt(0, s(z0)) → true 3.15/1.28
lt(s(z0), s(z1)) → lt(z0, z1) 3.15/1.28
if(true, z0, z1) → z0 3.15/1.28
if(false, z0, z1) → z1 3.15/1.28
div(z0, 0) → 0 3.15/1.28
div(0, z0) → 0 3.15/1.28
div(s(z0), s(z1)) → if(lt(z0, z1), 0, s(div(-(z0, z1), s(z1))))
Tuples:

-'(s(z0), s(z1)) → c2(-'(z0, z1)) 3.15/1.28
LT(s(z0), s(z1)) → c5(LT(z0, z1)) 3.15/1.28
DIV(s(z0), s(z1)) → c10(IF(lt(z0, z1), 0, s(div(-(z0, z1), s(z1)))), LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
S tuples:

-'(s(z0), s(z1)) → c2(-'(z0, z1)) 3.15/1.28
LT(s(z0), s(z1)) → c5(LT(z0, z1)) 3.15/1.28
DIV(s(z0), s(z1)) → c10(IF(lt(z0, z1), 0, s(div(-(z0, z1), s(z1)))), LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
K tuples:none
Defined Rule Symbols:

-, lt, if, div

Defined Pair Symbols:

-', LT, DIV

Compound Symbols:

c2, c5, c10

3.15/1.28
3.15/1.28

(3) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts
3.15/1.28
3.15/1.28

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

-(z0, 0) → z0 3.15/1.28
-(0, s(z0)) → 0 3.15/1.28
-(s(z0), s(z1)) → -(z0, z1) 3.15/1.28
lt(z0, 0) → false 3.15/1.28
lt(0, s(z0)) → true 3.15/1.28
lt(s(z0), s(z1)) → lt(z0, z1) 3.15/1.28
if(true, z0, z1) → z0 3.15/1.28
if(false, z0, z1) → z1 3.15/1.28
div(z0, 0) → 0 3.15/1.28
div(0, z0) → 0 3.15/1.28
div(s(z0), s(z1)) → if(lt(z0, z1), 0, s(div(-(z0, z1), s(z1))))
Tuples:

-'(s(z0), s(z1)) → c2(-'(z0, z1)) 3.15/1.28
LT(s(z0), s(z1)) → c5(LT(z0, z1)) 3.15/1.28
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
S tuples:

-'(s(z0), s(z1)) → c2(-'(z0, z1)) 3.15/1.28
LT(s(z0), s(z1)) → c5(LT(z0, z1)) 3.15/1.28
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
K tuples:none
Defined Rule Symbols:

-, lt, if, div

Defined Pair Symbols:

-', LT, DIV

Compound Symbols:

c2, c5, c10

3.15/1.28
3.15/1.28

(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.

DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
We considered the (Usable) Rules:

-(z0, 0) → z0 3.15/1.28
-(0, s(z0)) → 0 3.15/1.28
-(s(z0), s(z1)) → -(z0, z1)
And the Tuples:

-'(s(z0), s(z1)) → c2(-'(z0, z1)) 3.15/1.28
LT(s(z0), s(z1)) → c5(LT(z0, z1)) 3.15/1.28
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 3.15/1.28

POL(-(x1, x2)) = x1    3.15/1.28
POL(-'(x1, x2)) = 0    3.15/1.28
POL(0) = 0    3.15/1.28
POL(DIV(x1, x2)) = [4]x1    3.15/1.28
POL(LT(x1, x2)) = 0    3.15/1.28
POL(c10(x1, x2, x3)) = x1 + x2 + x3    3.15/1.28
POL(c2(x1)) = x1    3.15/1.28
POL(c5(x1)) = x1    3.15/1.28
POL(s(x1)) = [2] + x1   
3.15/1.28
3.15/1.28

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

-(z0, 0) → z0 3.15/1.28
-(0, s(z0)) → 0 3.15/1.28
-(s(z0), s(z1)) → -(z0, z1) 3.15/1.28
lt(z0, 0) → false 3.15/1.28
lt(0, s(z0)) → true 3.15/1.28
lt(s(z0), s(z1)) → lt(z0, z1) 3.15/1.28
if(true, z0, z1) → z0 3.15/1.28
if(false, z0, z1) → z1 3.15/1.28
div(z0, 0) → 0 3.15/1.28
div(0, z0) → 0 3.15/1.28
div(s(z0), s(z1)) → if(lt(z0, z1), 0, s(div(-(z0, z1), s(z1))))
Tuples:

-'(s(z0), s(z1)) → c2(-'(z0, z1)) 3.15/1.28
LT(s(z0), s(z1)) → c5(LT(z0, z1)) 3.15/1.28
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
S tuples:

-'(s(z0), s(z1)) → c2(-'(z0, z1)) 3.15/1.28
LT(s(z0), s(z1)) → c5(LT(z0, z1))
K tuples:

DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
Defined Rule Symbols:

-, lt, if, div

Defined Pair Symbols:

-', LT, DIV

Compound Symbols:

c2, c5, c10

3.15/1.28
3.15/1.28

(7) 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.

LT(s(z0), s(z1)) → c5(LT(z0, z1))
We considered the (Usable) Rules:

-(z0, 0) → z0 3.15/1.28
-(0, s(z0)) → 0 3.15/1.28
-(s(z0), s(z1)) → -(z0, z1)
And the Tuples:

-'(s(z0), s(z1)) → c2(-'(z0, z1)) 3.15/1.28
LT(s(z0), s(z1)) → c5(LT(z0, z1)) 3.15/1.28
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 3.15/1.28

POL(-(x1, x2)) = x1    3.15/1.28
POL(-'(x1, x2)) = 0    3.15/1.28
POL(0) = 0    3.15/1.28
POL(DIV(x1, x2)) = x12    3.15/1.28
POL(LT(x1, x2)) = x1    3.15/1.28
POL(c10(x1, x2, x3)) = x1 + x2 + x3    3.15/1.28
POL(c2(x1)) = x1    3.15/1.28
POL(c5(x1)) = x1    3.15/1.28
POL(s(x1)) = [1] + x1   
3.15/1.28
3.15/1.28

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

-(z0, 0) → z0 3.15/1.28
-(0, s(z0)) → 0 3.15/1.28
-(s(z0), s(z1)) → -(z0, z1) 3.15/1.28
lt(z0, 0) → false 3.15/1.28
lt(0, s(z0)) → true 3.15/1.28
lt(s(z0), s(z1)) → lt(z0, z1) 3.15/1.28
if(true, z0, z1) → z0 3.15/1.28
if(false, z0, z1) → z1 3.15/1.28
div(z0, 0) → 0 3.15/1.28
div(0, z0) → 0 3.15/1.28
div(s(z0), s(z1)) → if(lt(z0, z1), 0, s(div(-(z0, z1), s(z1))))
Tuples:

-'(s(z0), s(z1)) → c2(-'(z0, z1)) 3.15/1.28
LT(s(z0), s(z1)) → c5(LT(z0, z1)) 3.15/1.28
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
S tuples:

-'(s(z0), s(z1)) → c2(-'(z0, z1))
K tuples:

DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1)) 3.15/1.28
LT(s(z0), s(z1)) → c5(LT(z0, z1))
Defined Rule Symbols:

-, lt, if, div

Defined Pair Symbols:

-', LT, DIV

Compound Symbols:

c2, c5, c10

3.15/1.28
3.15/1.28

(9) 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.

-'(s(z0), s(z1)) → c2(-'(z0, z1))
We considered the (Usable) Rules:

-(z0, 0) → z0 3.15/1.28
-(0, s(z0)) → 0 3.15/1.28
-(s(z0), s(z1)) → -(z0, z1)
And the Tuples:

-'(s(z0), s(z1)) → c2(-'(z0, z1)) 3.15/1.28
LT(s(z0), s(z1)) → c5(LT(z0, z1)) 3.15/1.28
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 3.15/1.28

POL(-(x1, x2)) = x1    3.15/1.28
POL(-'(x1, x2)) = x1    3.15/1.28
POL(0) = 0    3.15/1.28
POL(DIV(x1, x2)) = x12    3.15/1.28
POL(LT(x1, x2)) = 0    3.15/1.28
POL(c10(x1, x2, x3)) = x1 + x2 + x3    3.15/1.28
POL(c2(x1)) = x1    3.15/1.28
POL(c5(x1)) = x1    3.15/1.28
POL(s(x1)) = [2] + x1   
3.15/1.28
3.15/1.28

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

-(z0, 0) → z0 3.15/1.28
-(0, s(z0)) → 0 3.15/1.28
-(s(z0), s(z1)) → -(z0, z1) 3.15/1.28
lt(z0, 0) → false 3.15/1.28
lt(0, s(z0)) → true 3.15/1.28
lt(s(z0), s(z1)) → lt(z0, z1) 3.15/1.28
if(true, z0, z1) → z0 3.15/1.28
if(false, z0, z1) → z1 3.15/1.28
div(z0, 0) → 0 3.15/1.28
div(0, z0) → 0 3.15/1.28
div(s(z0), s(z1)) → if(lt(z0, z1), 0, s(div(-(z0, z1), s(z1))))
Tuples:

-'(s(z0), s(z1)) → c2(-'(z0, z1)) 3.15/1.28
LT(s(z0), s(z1)) → c5(LT(z0, z1)) 3.15/1.28
DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1))
S tuples:none
K tuples:

DIV(s(z0), s(z1)) → c10(LT(z0, z1), DIV(-(z0, z1), s(z1)), -'(z0, z1)) 3.15/1.28
LT(s(z0), s(z1)) → c5(LT(z0, z1)) 3.15/1.28
-'(s(z0), s(z1)) → c2(-'(z0, z1))
Defined Rule Symbols:

-, lt, if, div

Defined Pair Symbols:

-', LT, DIV

Compound Symbols:

c2, c5, c10

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3.15/1.28

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

The set S is empty
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3.15/1.28

(12) BOUNDS(O(1), O(1))

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3.48/1.32 EOF