YES(O(1), O(n^1)) 0.00/0.83 YES(O(1), O(n^1)) 0.00/0.86 0.00/0.86 0.00/0.86 0.00/0.86 0.00/0.86 0.00/0.86 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.86 0.00/0.86 0.00/0.86
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The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

0 CpxTRS
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↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
0.00/0.86 
↳2 CdtProblem
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↳3 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
0.00/0.86 
↳4 CdtProblem
0.00/0.86 
↳5 CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID))
0.00/0.86 
↳6 CdtProblem
0.00/0.86 
↳7 CdtLeafRemovalProof (ComplexityIfPolyImplication)
0.00/0.86 
↳8 CdtProblem
0.00/0.86 
↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
0.00/0.86 
↳10 CdtProblem
0.00/0.86 
↳11 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
0.00/0.86 
↳12 CdtProblem
0.00/0.86 
↳13 SIsEmptyProof (BOTH BOUNDS(ID, ID))
0.00/0.86 
↳14 BOUNDS(O(1), O(1))
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0.00/0.86 0.00/0.86
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0.00/0.86

(0) Obligation:

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

foldl(x, Cons(S(0), xs)) → foldl(S(x), xs) 0.00/0.86
foldl(S(0), Cons(x, xs)) → foldl(S(x), xs) 0.00/0.86
foldr(a, Cons(x, xs)) → op(x, foldr(a, xs)) 0.00/0.86
foldr(a, Nil) → a 0.00/0.86
foldl(a, Nil) → a 0.00/0.86
notEmpty(Cons(x, xs)) → True 0.00/0.86
notEmpty(Nil) → False 0.00/0.86
op(x, S(0)) → S(x) 0.00/0.86
op(S(0), y) → S(y) 0.00/0.86
fold(a, xs) → Cons(foldl(a, xs), Cons(foldr(a, xs), Nil))

Rewrite Strategy: INNERMOST
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0.00/0.86

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

Converted CpxTRS to CDT
0.00/0.86
0.00/0.86

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

foldl(z0, Cons(S(0), z1)) → foldl(S(z0), z1) 0.00/0.86
foldl(S(0), Cons(z0, z1)) → foldl(S(z0), z1) 0.00/0.86
foldl(z0, Nil) → z0 0.00/0.86
foldr(z0, Cons(z1, z2)) → op(z1, foldr(z0, z2)) 0.00/0.86
foldr(z0, Nil) → z0 0.00/0.86
notEmpty(Cons(z0, z1)) → True 0.00/0.86
notEmpty(Nil) → False 0.00/0.86
op(z0, S(0)) → S(z0) 0.00/0.86
op(S(0), z0) → S(z0) 0.00/0.86
fold(z0, z1) → Cons(foldl(z0, z1), Cons(foldr(z0, z1), Nil))
Tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1)) 0.00/0.86
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1)) 0.00/0.86
FOLDR(z0, Cons(z1, z2)) → c3(OP(z1, foldr(z0, z2)), FOLDR(z0, z2)) 0.00/0.86
FOLD(z0, z1) → c9(FOLDL(z0, z1), FOLDR(z0, z1))
S tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1)) 0.00/0.86
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1)) 0.00/0.86
FOLDR(z0, Cons(z1, z2)) → c3(OP(z1, foldr(z0, z2)), FOLDR(z0, z2)) 0.00/0.86
FOLD(z0, z1) → c9(FOLDL(z0, z1), FOLDR(z0, z1))
K tuples:none
Defined Rule Symbols:

foldl, foldr, notEmpty, op, fold

Defined Pair Symbols:

FOLDL, FOLDR, FOLD

Compound Symbols:

c, c1, c3, c9

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0.00/0.86

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

Removed 1 trailing tuple parts
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0.00/0.86

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

foldl(z0, Cons(S(0), z1)) → foldl(S(z0), z1) 0.00/0.86
foldl(S(0), Cons(z0, z1)) → foldl(S(z0), z1) 0.00/0.86
foldl(z0, Nil) → z0 0.00/0.86
foldr(z0, Cons(z1, z2)) → op(z1, foldr(z0, z2)) 0.00/0.86
foldr(z0, Nil) → z0 0.00/0.86
notEmpty(Cons(z0, z1)) → True 0.00/0.86
notEmpty(Nil) → False 0.00/0.86
op(z0, S(0)) → S(z0) 0.00/0.86
op(S(0), z0) → S(z0) 0.00/0.86
fold(z0, z1) → Cons(foldl(z0, z1), Cons(foldr(z0, z1), Nil))
Tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1)) 0.00/0.86
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1)) 0.00/0.86
FOLD(z0, z1) → c9(FOLDL(z0, z1), FOLDR(z0, z1)) 0.00/0.86
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
S tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1)) 0.00/0.86
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1)) 0.00/0.86
FOLD(z0, z1) → c9(FOLDL(z0, z1), FOLDR(z0, z1)) 0.00/0.86
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
K tuples:none
Defined Rule Symbols:

foldl, foldr, notEmpty, op, fold

Defined Pair Symbols:

FOLDL, FOLD, FOLDR

Compound Symbols:

c, c1, c9, c3

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0.00/0.86

(5) CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID) transformation)

Split RHS of tuples not part of any SCC
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0.00/0.86

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

foldl(z0, Cons(S(0), z1)) → foldl(S(z0), z1) 0.00/0.86
foldl(S(0), Cons(z0, z1)) → foldl(S(z0), z1) 0.00/0.86
foldl(z0, Nil) → z0 0.00/0.86
foldr(z0, Cons(z1, z2)) → op(z1, foldr(z0, z2)) 0.00/0.86
foldr(z0, Nil) → z0 0.00/0.86
notEmpty(Cons(z0, z1)) → True 0.00/0.86
notEmpty(Nil) → False 0.00/0.86
op(z0, S(0)) → S(z0) 0.00/0.86
op(S(0), z0) → S(z0) 0.00/0.86
fold(z0, z1) → Cons(foldl(z0, z1), Cons(foldr(z0, z1), Nil))
Tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1)) 0.00/0.86
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1)) 0.00/0.86
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2)) 0.00/0.86
FOLD(z0, z1) → c2(FOLDL(z0, z1)) 0.00/0.86
FOLD(z0, z1) → c2(FOLDR(z0, z1))
S tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1)) 0.00/0.86
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1)) 0.00/0.86
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2)) 0.00/0.86
FOLD(z0, z1) → c2(FOLDL(z0, z1)) 0.00/0.86
FOLD(z0, z1) → c2(FOLDR(z0, z1))
K tuples:none
Defined Rule Symbols:

foldl, foldr, notEmpty, op, fold

Defined Pair Symbols:

FOLDL, FOLDR, FOLD

Compound Symbols:

c, c1, c3, c2

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0.00/0.86

(7) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 2 leading nodes:

FOLD(z0, z1) → c2(FOLDL(z0, z1)) 0.00/0.86
FOLD(z0, z1) → c2(FOLDR(z0, z1))
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0.00/0.86

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

foldl(z0, Cons(S(0), z1)) → foldl(S(z0), z1) 0.00/0.86
foldl(S(0), Cons(z0, z1)) → foldl(S(z0), z1) 0.00/0.86
foldl(z0, Nil) → z0 0.00/0.86
foldr(z0, Cons(z1, z2)) → op(z1, foldr(z0, z2)) 0.00/0.86
foldr(z0, Nil) → z0 0.00/0.86
notEmpty(Cons(z0, z1)) → True 0.00/0.86
notEmpty(Nil) → False 0.00/0.86
op(z0, S(0)) → S(z0) 0.00/0.86
op(S(0), z0) → S(z0) 0.00/0.86
fold(z0, z1) → Cons(foldl(z0, z1), Cons(foldr(z0, z1), Nil))
Tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1)) 0.00/0.86
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1)) 0.00/0.86
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
S tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1)) 0.00/0.86
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1)) 0.00/0.86
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
K tuples:none
Defined Rule Symbols:

foldl, foldr, notEmpty, op, fold

Defined Pair Symbols:

FOLDL, FOLDR

Compound Symbols:

c, c1, c3

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0.00/0.86

(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.

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1))
We considered the (Usable) Rules:none
And the Tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1)) 0.00/0.86
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1)) 0.00/0.86
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.86

POL(0) = 0    0.00/0.86
POL(Cons(x1, x2)) = x1 + x2    0.00/0.86
POL(FOLDL(x1, x2)) = [2]x1 + [4]x2    0.00/0.86
POL(FOLDR(x1, x2)) = 0    0.00/0.86
POL(S(x1)) = [5] + x1    0.00/0.86
POL(c(x1)) = x1    0.00/0.86
POL(c1(x1)) = x1    0.00/0.86
POL(c3(x1)) = x1   
0.00/0.86
0.00/0.86

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

foldl(z0, Cons(S(0), z1)) → foldl(S(z0), z1) 0.00/0.86
foldl(S(0), Cons(z0, z1)) → foldl(S(z0), z1) 0.00/0.86
foldl(z0, Nil) → z0 0.00/0.86
foldr(z0, Cons(z1, z2)) → op(z1, foldr(z0, z2)) 0.00/0.86
foldr(z0, Nil) → z0 0.00/0.86
notEmpty(Cons(z0, z1)) → True 0.00/0.86
notEmpty(Nil) → False 0.00/0.86
op(z0, S(0)) → S(z0) 0.00/0.86
op(S(0), z0) → S(z0) 0.00/0.86
fold(z0, z1) → Cons(foldl(z0, z1), Cons(foldr(z0, z1), Nil))
Tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1)) 0.00/0.86
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1)) 0.00/0.86
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
S tuples:

FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1)) 0.00/0.86
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
K tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1))
Defined Rule Symbols:

foldl, foldr, notEmpty, op, fold

Defined Pair Symbols:

FOLDL, FOLDR

Compound Symbols:

c, c1, c3

0.00/0.86
0.00/0.86

(11) 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.

FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1)) 0.00/0.86
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
We considered the (Usable) Rules:none
And the Tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1)) 0.00/0.86
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1)) 0.00/0.86
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.86

POL(0) = [3]    0.00/0.86
POL(Cons(x1, x2)) = [1] + x2    0.00/0.86
POL(FOLDL(x1, x2)) = x2    0.00/0.86
POL(FOLDR(x1, x2)) = x2    0.00/0.86
POL(S(x1)) = 0    0.00/0.86
POL(c(x1)) = x1    0.00/0.86
POL(c1(x1)) = x1    0.00/0.86
POL(c3(x1)) = x1   
0.00/0.86
0.00/0.86

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

foldl(z0, Cons(S(0), z1)) → foldl(S(z0), z1) 0.00/0.86
foldl(S(0), Cons(z0, z1)) → foldl(S(z0), z1) 0.00/0.86
foldl(z0, Nil) → z0 0.00/0.86
foldr(z0, Cons(z1, z2)) → op(z1, foldr(z0, z2)) 0.00/0.86
foldr(z0, Nil) → z0 0.00/0.86
notEmpty(Cons(z0, z1)) → True 0.00/0.86
notEmpty(Nil) → False 0.00/0.86
op(z0, S(0)) → S(z0) 0.00/0.86
op(S(0), z0) → S(z0) 0.00/0.86
fold(z0, z1) → Cons(foldl(z0, z1), Cons(foldr(z0, z1), Nil))
Tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1)) 0.00/0.86
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1)) 0.00/0.86
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
S tuples:none
K tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1)) 0.00/0.86
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1)) 0.00/0.86
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
Defined Rule Symbols:

foldl, foldr, notEmpty, op, fold

Defined Pair Symbols:

FOLDL, FOLDR

Compound Symbols:

c, c1, c3

0.00/0.86
0.00/0.86

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

The set S is empty
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0.00/0.86

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

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0.00/0.86
0.00/0.91 EOF