YES(O(1), O(n^2)) 0.00/0.99 YES(O(1), O(n^2)) 2.44/1.01 2.44/1.01 2.44/1.01 2.44/1.01 2.44/1.02 2.44/1.02 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 2.44/1.02 2.44/1.02 2.44/1.02
2.44/1.02 2.44/1.02 2.44/1.02
2.44/1.02
2.44/1.02

(0) Obligation:

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

naiverev(Cons(x, xs)) → app(naiverev(xs), Cons(x, Nil)) 2.44/1.02
app(Cons(x, xs), ys) → Cons(x, app(xs, ys)) 2.44/1.02
notEmpty(Cons(x, xs)) → True 2.44/1.02
notEmpty(Nil) → False 2.44/1.02
naiverev(Nil) → Nil 2.44/1.02
app(Nil, ys) → ys 2.44/1.02
goal(xs) → naiverev(xs)

Rewrite Strategy: INNERMOST
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2.44/1.02

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

Converted CpxTRS to CDT
2.44/1.02
2.44/1.02

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

naiverev(Cons(z0, z1)) → app(naiverev(z1), Cons(z0, Nil)) 2.44/1.02
naiverev(Nil) → Nil 2.44/1.02
app(Cons(z0, z1), z2) → Cons(z0, app(z1, z2)) 2.44/1.02
app(Nil, z0) → z0 2.44/1.02
notEmpty(Cons(z0, z1)) → True 2.44/1.02
notEmpty(Nil) → False 2.44/1.02
goal(z0) → naiverev(z0)
Tuples:

NAIVEREV(Cons(z0, z1)) → c(APP(naiverev(z1), Cons(z0, Nil)), NAIVEREV(z1)) 2.44/1.02
APP(Cons(z0, z1), z2) → c2(APP(z1, z2)) 2.44/1.02
GOAL(z0) → c6(NAIVEREV(z0))
S tuples:

NAIVEREV(Cons(z0, z1)) → c(APP(naiverev(z1), Cons(z0, Nil)), NAIVEREV(z1)) 2.44/1.02
APP(Cons(z0, z1), z2) → c2(APP(z1, z2)) 2.44/1.02
GOAL(z0) → c6(NAIVEREV(z0))
K tuples:none
Defined Rule Symbols:

naiverev, app, notEmpty, goal

Defined Pair Symbols:

NAIVEREV, APP, GOAL

Compound Symbols:

c, c2, c6

2.44/1.02
2.44/1.02

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

GOAL(z0) → c6(NAIVEREV(z0))
2.44/1.02
2.44/1.02

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

naiverev(Cons(z0, z1)) → app(naiverev(z1), Cons(z0, Nil)) 2.44/1.02
naiverev(Nil) → Nil 2.44/1.02
app(Cons(z0, z1), z2) → Cons(z0, app(z1, z2)) 2.44/1.02
app(Nil, z0) → z0 2.44/1.02
notEmpty(Cons(z0, z1)) → True 2.44/1.02
notEmpty(Nil) → False 2.44/1.02
goal(z0) → naiverev(z0)
Tuples:

NAIVEREV(Cons(z0, z1)) → c(APP(naiverev(z1), Cons(z0, Nil)), NAIVEREV(z1)) 2.44/1.02
APP(Cons(z0, z1), z2) → c2(APP(z1, z2))
S tuples:

NAIVEREV(Cons(z0, z1)) → c(APP(naiverev(z1), Cons(z0, Nil)), NAIVEREV(z1)) 2.44/1.02
APP(Cons(z0, z1), z2) → c2(APP(z1, z2))
K tuples:none
Defined Rule Symbols:

naiverev, app, notEmpty, goal

Defined Pair Symbols:

NAIVEREV, APP

Compound Symbols:

c, c2

2.44/1.02
2.44/1.02

(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.

NAIVEREV(Cons(z0, z1)) → c(APP(naiverev(z1), Cons(z0, Nil)), NAIVEREV(z1))
We considered the (Usable) Rules:

naiverev(Cons(z0, z1)) → app(naiverev(z1), Cons(z0, Nil)) 2.44/1.02
naiverev(Nil) → Nil 2.44/1.02
app(Cons(z0, z1), z2) → Cons(z0, app(z1, z2)) 2.44/1.02
app(Nil, z0) → z0
And the Tuples:

NAIVEREV(Cons(z0, z1)) → c(APP(naiverev(z1), Cons(z0, Nil)), NAIVEREV(z1)) 2.44/1.02
APP(Cons(z0, z1), z2) → c2(APP(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation : 2.44/1.02

POL(APP(x1, x2)) = [3]    2.44/1.02
POL(Cons(x1, x2)) = [4] + x1 + x2    2.44/1.02
POL(NAIVEREV(x1)) = [4]x1    2.44/1.02
POL(Nil) = 0    2.44/1.02
POL(app(x1, x2)) = [5]    2.44/1.02
POL(c(x1, x2)) = x1 + x2    2.44/1.02
POL(c2(x1)) = x1    2.44/1.02
POL(naiverev(x1)) = 0   
2.44/1.02
2.44/1.02

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

naiverev(Cons(z0, z1)) → app(naiverev(z1), Cons(z0, Nil)) 2.44/1.02
naiverev(Nil) → Nil 2.44/1.02
app(Cons(z0, z1), z2) → Cons(z0, app(z1, z2)) 2.44/1.02
app(Nil, z0) → z0 2.44/1.02
notEmpty(Cons(z0, z1)) → True 2.44/1.02
notEmpty(Nil) → False 2.44/1.02
goal(z0) → naiverev(z0)
Tuples:

NAIVEREV(Cons(z0, z1)) → c(APP(naiverev(z1), Cons(z0, Nil)), NAIVEREV(z1)) 2.44/1.02
APP(Cons(z0, z1), z2) → c2(APP(z1, z2))
S tuples:

APP(Cons(z0, z1), z2) → c2(APP(z1, z2))
K tuples:

NAIVEREV(Cons(z0, z1)) → c(APP(naiverev(z1), Cons(z0, Nil)), NAIVEREV(z1))
Defined Rule Symbols:

naiverev, app, notEmpty, goal

Defined Pair Symbols:

NAIVEREV, APP

Compound Symbols:

c, c2

2.44/1.02
2.44/1.02

(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.

APP(Cons(z0, z1), z2) → c2(APP(z1, z2))
We considered the (Usable) Rules:

naiverev(Cons(z0, z1)) → app(naiverev(z1), Cons(z0, Nil)) 2.44/1.02
naiverev(Nil) → Nil 2.44/1.02
app(Cons(z0, z1), z2) → Cons(z0, app(z1, z2)) 2.44/1.02
app(Nil, z0) → z0
And the Tuples:

NAIVEREV(Cons(z0, z1)) → c(APP(naiverev(z1), Cons(z0, Nil)), NAIVEREV(z1)) 2.44/1.02
APP(Cons(z0, z1), z2) → c2(APP(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation : 2.44/1.02

POL(APP(x1, x2)) = x1    2.44/1.02
POL(Cons(x1, x2)) = [1] + x2    2.44/1.02
POL(NAIVEREV(x1)) = x12    2.44/1.02
POL(Nil) = 0    2.44/1.02
POL(app(x1, x2)) = x1 + x2    2.44/1.02
POL(c(x1, x2)) = x1 + x2    2.44/1.02
POL(c2(x1)) = x1    2.44/1.02
POL(naiverev(x1)) = x1   
2.44/1.02
2.44/1.02

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

naiverev(Cons(z0, z1)) → app(naiverev(z1), Cons(z0, Nil)) 2.44/1.02
naiverev(Nil) → Nil 2.44/1.02
app(Cons(z0, z1), z2) → Cons(z0, app(z1, z2)) 2.44/1.02
app(Nil, z0) → z0 2.44/1.02
notEmpty(Cons(z0, z1)) → True 2.44/1.02
notEmpty(Nil) → False 2.44/1.02
goal(z0) → naiverev(z0)
Tuples:

NAIVEREV(Cons(z0, z1)) → c(APP(naiverev(z1), Cons(z0, Nil)), NAIVEREV(z1)) 2.44/1.02
APP(Cons(z0, z1), z2) → c2(APP(z1, z2))
S tuples:none
K tuples:

NAIVEREV(Cons(z0, z1)) → c(APP(naiverev(z1), Cons(z0, Nil)), NAIVEREV(z1)) 2.44/1.02
APP(Cons(z0, z1), z2) → c2(APP(z1, z2))
Defined Rule Symbols:

naiverev, app, notEmpty, goal

Defined Pair Symbols:

NAIVEREV, APP

Compound Symbols:

c, c2

2.44/1.02
2.44/1.02

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

The set S is empty
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2.44/1.02

(10) BOUNDS(O(1), O(1))

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2.44/1.02
2.44/1.06 EOF