YES(O(1), O(n^1)) 0.00/0.73 YES(O(1), O(n^1)) 0.00/0.74 0.00/0.74 0.00/0.74 0.00/0.74 0.00/0.74 0.00/0.74 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.74 0.00/0.74 0.00/0.74
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(0) Obligation:

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

is_empty(nil) → true 0.00/0.74
is_empty(cons(x, l)) → false 0.00/0.74
hd(cons(x, l)) → x 0.00/0.74
tl(cons(x, l)) → l 0.00/0.74
append(l1, l2) → ifappend(l1, l2, l1) 0.00/0.74
ifappend(l1, l2, nil) → l2 0.00/0.74
ifappend(l1, l2, cons(x, l)) → cons(x, append(l, l2))

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:

is_empty(nil) → true 0.00/0.74
is_empty(cons(z0, z1)) → false 0.00/0.74
hd(cons(z0, z1)) → z0 0.00/0.74
tl(cons(z0, z1)) → z1 0.00/0.74
append(z0, z1) → ifappend(z0, z1, z0) 0.00/0.74
ifappend(z0, z1, nil) → z1 0.00/0.74
ifappend(z0, z1, cons(z2, z3)) → cons(z2, append(z3, z1))
Tuples:

APPEND(z0, z1) → c4(IFAPPEND(z0, z1, z0)) 0.00/0.74
IFAPPEND(z0, z1, cons(z2, z3)) → c6(APPEND(z3, z1))
S tuples:

APPEND(z0, z1) → c4(IFAPPEND(z0, z1, z0)) 0.00/0.74
IFAPPEND(z0, z1, cons(z2, z3)) → c6(APPEND(z3, z1))
K tuples:none
Defined Rule Symbols:

is_empty, hd, tl, append, ifappend

Defined Pair Symbols:

APPEND, IFAPPEND

Compound Symbols:

c4, c6

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(3) 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.

APPEND(z0, z1) → c4(IFAPPEND(z0, z1, z0)) 0.00/0.74
IFAPPEND(z0, z1, cons(z2, z3)) → c6(APPEND(z3, z1))
We considered the (Usable) Rules:none
And the Tuples:

APPEND(z0, z1) → c4(IFAPPEND(z0, z1, z0)) 0.00/0.74
IFAPPEND(z0, z1, cons(z2, z3)) → c6(APPEND(z3, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.74

POL(APPEND(x1, x2)) = [4] + [2]x1    0.00/0.74
POL(IFAPPEND(x1, x2, x3)) = [2]x3    0.00/0.74
POL(c4(x1)) = x1    0.00/0.74
POL(c6(x1)) = x1    0.00/0.74
POL(cons(x1, x2)) = [4] + x2   
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

is_empty(nil) → true 0.00/0.74
is_empty(cons(z0, z1)) → false 0.00/0.74
hd(cons(z0, z1)) → z0 0.00/0.74
tl(cons(z0, z1)) → z1 0.00/0.74
append(z0, z1) → ifappend(z0, z1, z0) 0.00/0.74
ifappend(z0, z1, nil) → z1 0.00/0.74
ifappend(z0, z1, cons(z2, z3)) → cons(z2, append(z3, z1))
Tuples:

APPEND(z0, z1) → c4(IFAPPEND(z0, z1, z0)) 0.00/0.74
IFAPPEND(z0, z1, cons(z2, z3)) → c6(APPEND(z3, z1))
S tuples:none
K tuples:

APPEND(z0, z1) → c4(IFAPPEND(z0, z1, z0)) 0.00/0.74
IFAPPEND(z0, z1, cons(z2, z3)) → c6(APPEND(z3, z1))
Defined Rule Symbols:

is_empty, hd, tl, append, ifappend

Defined Pair Symbols:

APPEND, IFAPPEND

Compound Symbols:

c4, c6

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

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

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0.00/0.78 EOF