YES(O(1), O(n^1)) 0.00/0.84 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))
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↳2 CdtProblem
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↳3 CdtLeafRemovalProof (ComplexityIfPolyImplication)
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↳4 CdtProblem
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↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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↳6 CdtProblem
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↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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↳8 CdtProblem
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↳9 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳10 BOUNDS(O(1), O(1))
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(0) Obligation:

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

addlist(Cons(x, xs'), Cons(S(0), xs)) → Cons(S(x), addlist(xs', xs)) 0.00/0.86
addlist(Cons(S(0), xs'), Cons(x, xs)) → Cons(S(x), addlist(xs', xs)) 0.00/0.86
addlist(Nil, ys) → Nil 0.00/0.86
notEmpty(Cons(x, xs)) → True 0.00/0.86
notEmpty(Nil) → False 0.00/0.86
goal(xs, ys) → addlist(xs, ys)

Rewrite Strategy: INNERMOST
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(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT
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(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

addlist(Cons(z0, z1), Cons(S(0), z2)) → Cons(S(z0), addlist(z1, z2)) 0.00/0.86
addlist(Cons(S(0), z0), Cons(z1, z2)) → Cons(S(z1), addlist(z0, z2)) 0.00/0.86
addlist(Nil, z0) → Nil 0.00/0.86
notEmpty(Cons(z0, z1)) → True 0.00/0.86
notEmpty(Nil) → False 0.00/0.86
goal(z0, z1) → addlist(z0, z1)
Tuples:

ADDLIST(Cons(z0, z1), Cons(S(0), z2)) → c(ADDLIST(z1, z2)) 0.00/0.86
ADDLIST(Cons(S(0), z0), Cons(z1, z2)) → c1(ADDLIST(z0, z2)) 0.00/0.86
GOAL(z0, z1) → c5(ADDLIST(z0, z1))
S tuples:

ADDLIST(Cons(z0, z1), Cons(S(0), z2)) → c(ADDLIST(z1, z2)) 0.00/0.86
ADDLIST(Cons(S(0), z0), Cons(z1, z2)) → c1(ADDLIST(z0, z2)) 0.00/0.86
GOAL(z0, z1) → c5(ADDLIST(z0, z1))
K tuples:none
Defined Rule Symbols:

addlist, notEmpty, goal

Defined Pair Symbols:

ADDLIST, GOAL

Compound Symbols:

c, c1, c5

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(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

GOAL(z0, z1) → c5(ADDLIST(z0, z1))
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

addlist(Cons(z0, z1), Cons(S(0), z2)) → Cons(S(z0), addlist(z1, z2)) 0.00/0.86
addlist(Cons(S(0), z0), Cons(z1, z2)) → Cons(S(z1), addlist(z0, z2)) 0.00/0.86
addlist(Nil, z0) → Nil 0.00/0.86
notEmpty(Cons(z0, z1)) → True 0.00/0.86
notEmpty(Nil) → False 0.00/0.86
goal(z0, z1) → addlist(z0, z1)
Tuples:

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

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

addlist, notEmpty, goal

Defined Pair Symbols:

ADDLIST

Compound Symbols:

c, c1

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(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.

ADDLIST(Cons(z0, z1), Cons(S(0), z2)) → c(ADDLIST(z1, z2))
We considered the (Usable) Rules:none
And the Tuples:

ADDLIST(Cons(z0, z1), Cons(S(0), z2)) → c(ADDLIST(z1, z2)) 0.00/0.86
ADDLIST(Cons(S(0), z0), Cons(z1, z2)) → c1(ADDLIST(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(ADDLIST(x1, x2)) = x2    0.00/0.86
POL(Cons(x1, x2)) = x1 + x2    0.00/0.86
POL(S(x1)) = [2]    0.00/0.86
POL(c(x1)) = x1    0.00/0.86
POL(c1(x1)) = x1   
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

addlist(Cons(z0, z1), Cons(S(0), z2)) → Cons(S(z0), addlist(z1, z2)) 0.00/0.86
addlist(Cons(S(0), z0), Cons(z1, z2)) → Cons(S(z1), addlist(z0, z2)) 0.00/0.86
addlist(Nil, z0) → Nil 0.00/0.86
notEmpty(Cons(z0, z1)) → True 0.00/0.86
notEmpty(Nil) → False 0.00/0.86
goal(z0, z1) → addlist(z0, z1)
Tuples:

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

ADDLIST(Cons(S(0), z0), Cons(z1, z2)) → c1(ADDLIST(z0, z2))
K tuples:

ADDLIST(Cons(z0, z1), Cons(S(0), z2)) → c(ADDLIST(z1, z2))
Defined Rule Symbols:

addlist, notEmpty, goal

Defined Pair Symbols:

ADDLIST

Compound Symbols:

c, c1

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(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.

ADDLIST(Cons(S(0), z0), Cons(z1, z2)) → c1(ADDLIST(z0, z2))
We considered the (Usable) Rules:none
And the Tuples:

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

POL(0) = [1]    0.00/0.86
POL(ADDLIST(x1, x2)) = [2]x1    0.00/0.86
POL(Cons(x1, x2)) = [1] + x2    0.00/0.86
POL(S(x1)) = [1] + x1    0.00/0.86
POL(c(x1)) = x1    0.00/0.86
POL(c1(x1)) = x1   
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(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

addlist(Cons(z0, z1), Cons(S(0), z2)) → Cons(S(z0), addlist(z1, z2)) 0.00/0.86
addlist(Cons(S(0), z0), Cons(z1, z2)) → Cons(S(z1), addlist(z0, z2)) 0.00/0.86
addlist(Nil, z0) → Nil 0.00/0.86
notEmpty(Cons(z0, z1)) → True 0.00/0.86
notEmpty(Nil) → False 0.00/0.86
goal(z0, z1) → addlist(z0, z1)
Tuples:

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

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

addlist, notEmpty, goal

Defined Pair Symbols:

ADDLIST

Compound Symbols:

c, c1

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

The set S is empty
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(10) BOUNDS(O(1), O(1))

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0.00/0.89 EOF