YES(O(1), O(n^1)) 0.00/0.75 YES(O(1), O(n^1)) 0.00/0.76 0.00/0.76 0.00/0.76 0.00/0.76 0.00/0.76 0.00/0.76 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.76 0.00/0.76 0.00/0.76
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The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(O(1), O(n^1)).

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 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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↳6 CdtProblem
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↳7 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳8 BOUNDS(O(1), O(1))
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(0) Obligation:

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

f(C(x1, x2)) → C(f(x1), f(x2)) 0.00/0.76
f(Z) → Z 0.00/0.76
eqZList(C(x1, x2), C(y1, y2)) → and(eqZList(x1, y1), eqZList(x2, y2)) 0.00/0.76
eqZList(C(x1, x2), Z) → False 0.00/0.76
eqZList(Z, C(y1, y2)) → False 0.00/0.76
eqZList(Z, Z) → True 0.00/0.76
second(C(x1, x2)) → x2 0.00/0.76
first(C(x1, x2)) → x1 0.00/0.76
g(x) → x

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

and(False, False) → False 0.00/0.76
and(True, False) → False 0.00/0.76
and(False, True) → False 0.00/0.76
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 0.00/0.76
and(True, False) → False 0.00/0.76
and(False, True) → False 0.00/0.76
and(True, True) → True 0.00/0.76
f(C(z0, z1)) → C(f(z0), f(z1)) 0.00/0.76
f(Z) → Z 0.00/0.76
eqZList(C(z0, z1), C(z2, z3)) → and(eqZList(z0, z2), eqZList(z1, z3)) 0.00/0.76
eqZList(C(z0, z1), Z) → False 0.00/0.76
eqZList(Z, C(z0, z1)) → False 0.00/0.76
eqZList(Z, Z) → True 0.00/0.76
second(C(z0, z1)) → z1 0.00/0.76
first(C(z0, z1)) → z0 0.00/0.76
g(z0) → z0
Tuples:

F(C(z0, z1)) → c4(F(z0), F(z1)) 0.00/0.76
EQZLIST(C(z0, z1), C(z2, z3)) → c6(AND(eqZList(z0, z2), eqZList(z1, z3)), EQZLIST(z0, z2), EQZLIST(z1, z3))
S tuples:

F(C(z0, z1)) → c4(F(z0), F(z1)) 0.00/0.76
EQZLIST(C(z0, z1), C(z2, z3)) → c6(AND(eqZList(z0, z2), eqZList(z1, z3)), EQZLIST(z0, z2), EQZLIST(z1, z3))
K tuples:none
Defined Rule Symbols:

f, eqZList, second, first, g, and

Defined Pair Symbols:

F, EQZLIST

Compound Symbols:

c4, c6

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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 0.00/0.76
and(True, False) → False 0.00/0.76
and(False, True) → False 0.00/0.76
and(True, True) → True 0.00/0.76
f(C(z0, z1)) → C(f(z0), f(z1)) 0.00/0.76
f(Z) → Z 0.00/0.76
eqZList(C(z0, z1), C(z2, z3)) → and(eqZList(z0, z2), eqZList(z1, z3)) 0.00/0.76
eqZList(C(z0, z1), Z) → False 0.00/0.76
eqZList(Z, C(z0, z1)) → False 0.00/0.76
eqZList(Z, Z) → True 0.00/0.76
second(C(z0, z1)) → z1 0.00/0.76
first(C(z0, z1)) → z0 0.00/0.76
g(z0) → z0
Tuples:

F(C(z0, z1)) → c4(F(z0), F(z1)) 0.00/0.76
EQZLIST(C(z0, z1), C(z2, z3)) → c6(EQZLIST(z0, z2), EQZLIST(z1, z3))
S tuples:

F(C(z0, z1)) → c4(F(z0), F(z1)) 0.00/0.76
EQZLIST(C(z0, z1), C(z2, z3)) → c6(EQZLIST(z0, z2), EQZLIST(z1, z3))
K tuples:none
Defined Rule Symbols:

f, eqZList, second, first, g, and

Defined Pair Symbols:

F, EQZLIST

Compound Symbols:

c4, c6

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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.

F(C(z0, z1)) → c4(F(z0), F(z1)) 0.00/0.76
EQZLIST(C(z0, z1), C(z2, z3)) → c6(EQZLIST(z0, z2), EQZLIST(z1, z3))
We considered the (Usable) Rules:none
And the Tuples:

F(C(z0, z1)) → c4(F(z0), F(z1)) 0.00/0.76
EQZLIST(C(z0, z1), C(z2, z3)) → c6(EQZLIST(z0, z2), EQZLIST(z1, z3))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.76

POL(C(x1, x2)) = [2] + x1 + x2    0.00/0.76
POL(EQZLIST(x1, x2)) = [3]x1    0.00/0.76
POL(F(x1)) = [3] + [2]x1    0.00/0.76
POL(c4(x1, x2)) = x1 + x2    0.00/0.76
POL(c6(x1, x2)) = x1 + x2   
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(False, False) → False 0.00/0.76
and(True, False) → False 0.00/0.76
and(False, True) → False 0.00/0.76
and(True, True) → True 0.00/0.76
f(C(z0, z1)) → C(f(z0), f(z1)) 0.00/0.76
f(Z) → Z 0.00/0.76
eqZList(C(z0, z1), C(z2, z3)) → and(eqZList(z0, z2), eqZList(z1, z3)) 0.00/0.76
eqZList(C(z0, z1), Z) → False 0.00/0.76
eqZList(Z, C(z0, z1)) → False 0.00/0.76
eqZList(Z, Z) → True 0.00/0.76
second(C(z0, z1)) → z1 0.00/0.76
first(C(z0, z1)) → z0 0.00/0.76
g(z0) → z0
Tuples:

F(C(z0, z1)) → c4(F(z0), F(z1)) 0.00/0.76
EQZLIST(C(z0, z1), C(z2, z3)) → c6(EQZLIST(z0, z2), EQZLIST(z1, z3))
S tuples:none
K tuples:

F(C(z0, z1)) → c4(F(z0), F(z1)) 0.00/0.76
EQZLIST(C(z0, z1), C(z2, z3)) → c6(EQZLIST(z0, z2), EQZLIST(z1, z3))
Defined Rule Symbols:

f, eqZList, second, first, g, and

Defined Pair Symbols:

F, EQZLIST

Compound Symbols:

c4, c6

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

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

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0.00/0.79 EOF