YES(O(1), O(n^2)) 0.00/0.82 YES(O(1), O(n^2)) 0.00/0.85 0.00/0.85 0.00/0.85 0.00/0.85 0.00/0.85 0.00/0.85 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.85 0.00/0.85 0.00/0.85
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The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(O(1), O(n^2)).

0 CpxRelTRS
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↳1 CpxRelTrsToCDT (UPPER BOUND (ID))
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↳2 CdtProblem
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↳3 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
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↳4 CdtProblem
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↳5 CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID))
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↳6 CdtProblem
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↳7 CdtLeafRemovalProof (ComplexityIfPolyImplication)
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↳8 CdtProblem
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↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
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↳10 CdtProblem
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↳11 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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↳12 CdtProblem
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↳13 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳14 BOUNDS(O(1), O(1))
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(0) Obligation:

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

lte(Cons(x', xs'), Cons(x, xs)) → lte(xs', xs) 0.00/0.85
lte(Cons(x, xs), Nil) → False 0.00/0.85
even(Cons(x, Nil)) → False 0.00/0.85
even(Cons(x', Cons(x, xs))) → even(xs) 0.00/0.85
notEmpty(Cons(x, xs)) → True 0.00/0.85
notEmpty(Nil) → False 0.00/0.85
lte(Nil, y) → True 0.00/0.85
even(Nil) → True 0.00/0.85
goal(x, y) → and(lte(x, y), even(x))

The (relative) TRS S consists of the following rules:

and(False, False) → False 0.00/0.85
and(True, False) → False 0.00/0.85
and(False, True) → False 0.00/0.85
and(True, True) → True

Rewrite Strategy: INNERMOST
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(1) CpxRelTrsToCDT (UPPER BOUND (ID) transformation)

Relative innermost TRS to CDT Problem.
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(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(False, False) → False 0.00/0.85
and(True, False) → False 0.00/0.85
and(False, True) → False 0.00/0.85
and(True, True) → True 0.00/0.85
lte(Cons(z0, z1), Cons(z2, z3)) → lte(z1, z3) 0.00/0.85
lte(Cons(z0, z1), Nil) → False 0.00/0.85
lte(Nil, z0) → True 0.00/0.85
even(Cons(z0, Nil)) → False 0.00/0.85
even(Cons(z0, Cons(z1, z2))) → even(z2) 0.00/0.85
even(Nil) → True 0.00/0.85
notEmpty(Cons(z0, z1)) → True 0.00/0.85
notEmpty(Nil) → False 0.00/0.85
goal(z0, z1) → and(lte(z0, z1), even(z0))
Tuples:

LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3)) 0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2)) 0.00/0.85
GOAL(z0, z1) → c12(AND(lte(z0, z1), even(z0)), LTE(z0, z1), EVEN(z0))
S tuples:

LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3)) 0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2)) 0.00/0.85
GOAL(z0, z1) → c12(AND(lte(z0, z1), even(z0)), LTE(z0, z1), EVEN(z0))
K tuples:none
Defined Rule Symbols:

lte, even, notEmpty, goal, and

Defined Pair Symbols:

LTE, EVEN, GOAL

Compound Symbols:

c4, c8, c12

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

Removed 1 trailing tuple parts
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(False, False) → False 0.00/0.85
and(True, False) → False 0.00/0.85
and(False, True) → False 0.00/0.85
and(True, True) → True 0.00/0.85
lte(Cons(z0, z1), Cons(z2, z3)) → lte(z1, z3) 0.00/0.85
lte(Cons(z0, z1), Nil) → False 0.00/0.85
lte(Nil, z0) → True 0.00/0.85
even(Cons(z0, Nil)) → False 0.00/0.85
even(Cons(z0, Cons(z1, z2))) → even(z2) 0.00/0.85
even(Nil) → True 0.00/0.85
notEmpty(Cons(z0, z1)) → True 0.00/0.85
notEmpty(Nil) → False 0.00/0.85
goal(z0, z1) → and(lte(z0, z1), even(z0))
Tuples:

LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3)) 0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2)) 0.00/0.85
GOAL(z0, z1) → c12(LTE(z0, z1), EVEN(z0))
S tuples:

LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3)) 0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2)) 0.00/0.85
GOAL(z0, z1) → c12(LTE(z0, z1), EVEN(z0))
K tuples:none
Defined Rule Symbols:

lte, even, notEmpty, goal, and

Defined Pair Symbols:

LTE, EVEN, GOAL

Compound Symbols:

c4, c8, c12

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

Split RHS of tuples not part of any SCC
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(False, False) → False 0.00/0.85
and(True, False) → False 0.00/0.85
and(False, True) → False 0.00/0.85
and(True, True) → True 0.00/0.85
lte(Cons(z0, z1), Cons(z2, z3)) → lte(z1, z3) 0.00/0.85
lte(Cons(z0, z1), Nil) → False 0.00/0.85
lte(Nil, z0) → True 0.00/0.85
even(Cons(z0, Nil)) → False 0.00/0.85
even(Cons(z0, Cons(z1, z2))) → even(z2) 0.00/0.85
even(Nil) → True 0.00/0.85
notEmpty(Cons(z0, z1)) → True 0.00/0.85
notEmpty(Nil) → False 0.00/0.85
goal(z0, z1) → and(lte(z0, z1), even(z0))
Tuples:

LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3)) 0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2)) 0.00/0.85
GOAL(z0, z1) → c(LTE(z0, z1)) 0.00/0.85
GOAL(z0, z1) → c(EVEN(z0))
S tuples:

LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3)) 0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2)) 0.00/0.85
GOAL(z0, z1) → c(LTE(z0, z1)) 0.00/0.85
GOAL(z0, z1) → c(EVEN(z0))
K tuples:none
Defined Rule Symbols:

lte, even, notEmpty, goal, and

Defined Pair Symbols:

LTE, EVEN, GOAL

Compound Symbols:

c4, c8, c

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(7) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 2 leading nodes:

GOAL(z0, z1) → c(LTE(z0, z1)) 0.00/0.85
GOAL(z0, z1) → c(EVEN(z0))
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(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(False, False) → False 0.00/0.85
and(True, False) → False 0.00/0.85
and(False, True) → False 0.00/0.85
and(True, True) → True 0.00/0.85
lte(Cons(z0, z1), Cons(z2, z3)) → lte(z1, z3) 0.00/0.85
lte(Cons(z0, z1), Nil) → False 0.00/0.85
lte(Nil, z0) → True 0.00/0.85
even(Cons(z0, Nil)) → False 0.00/0.85
even(Cons(z0, Cons(z1, z2))) → even(z2) 0.00/0.85
even(Nil) → True 0.00/0.85
notEmpty(Cons(z0, z1)) → True 0.00/0.85
notEmpty(Nil) → False 0.00/0.85
goal(z0, z1) → and(lte(z0, z1), even(z0))
Tuples:

LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3)) 0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
S tuples:

LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3)) 0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
K tuples:none
Defined Rule Symbols:

lte, even, notEmpty, goal, and

Defined Pair Symbols:

LTE, EVEN

Compound Symbols:

c4, c8

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(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.

LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3))
We considered the (Usable) Rules:none
And the Tuples:

LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3)) 0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.85

POL(Cons(x1, x2)) = [1] + x2    0.00/0.85
POL(EVEN(x1)) = 0    0.00/0.85
POL(LTE(x1, x2)) = x1·x2    0.00/0.85
POL(c4(x1)) = x1    0.00/0.85
POL(c8(x1)) = x1   
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(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(False, False) → False 0.00/0.85
and(True, False) → False 0.00/0.85
and(False, True) → False 0.00/0.85
and(True, True) → True 0.00/0.85
lte(Cons(z0, z1), Cons(z2, z3)) → lte(z1, z3) 0.00/0.85
lte(Cons(z0, z1), Nil) → False 0.00/0.85
lte(Nil, z0) → True 0.00/0.85
even(Cons(z0, Nil)) → False 0.00/0.85
even(Cons(z0, Cons(z1, z2))) → even(z2) 0.00/0.85
even(Nil) → True 0.00/0.85
notEmpty(Cons(z0, z1)) → True 0.00/0.85
notEmpty(Nil) → False 0.00/0.85
goal(z0, z1) → and(lte(z0, z1), even(z0))
Tuples:

LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3)) 0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
S tuples:

EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
K tuples:

LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3))
Defined Rule Symbols:

lte, even, notEmpty, goal, and

Defined Pair Symbols:

LTE, EVEN

Compound Symbols:

c4, c8

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(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.

EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
We considered the (Usable) Rules:none
And the Tuples:

LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3)) 0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.85

POL(Cons(x1, x2)) = [1] + x2    0.00/0.85
POL(EVEN(x1)) = [2]x1    0.00/0.85
POL(LTE(x1, x2)) = [3]x1 + [5]x2    0.00/0.85
POL(c4(x1)) = x1    0.00/0.85
POL(c8(x1)) = x1   
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(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(False, False) → False 0.00/0.85
and(True, False) → False 0.00/0.85
and(False, True) → False 0.00/0.85
and(True, True) → True 0.00/0.85
lte(Cons(z0, z1), Cons(z2, z3)) → lte(z1, z3) 0.00/0.85
lte(Cons(z0, z1), Nil) → False 0.00/0.85
lte(Nil, z0) → True 0.00/0.85
even(Cons(z0, Nil)) → False 0.00/0.85
even(Cons(z0, Cons(z1, z2))) → even(z2) 0.00/0.85
even(Nil) → True 0.00/0.85
notEmpty(Cons(z0, z1)) → True 0.00/0.85
notEmpty(Nil) → False 0.00/0.85
goal(z0, z1) → and(lte(z0, z1), even(z0))
Tuples:

LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3)) 0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
S tuples:none
K tuples:

LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3)) 0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
Defined Rule Symbols:

lte, even, notEmpty, goal, and

Defined Pair Symbols:

LTE, EVEN

Compound Symbols:

c4, c8

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

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

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0.00/0.89 EOF