YES(O(1), O(n^1)) 0.00/0.82 YES(O(1), O(n^1)) 0.00/0.84 0.00/0.84 0.00/0.84 0.00/0.84 0.00/0.84 0.00/0.84 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.84 0.00/0.84 0.00/0.84
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(0) Obligation:

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

c(c(c(y))) → c(c(a(y, 0))) 0.00/0.84
c(a(a(0, x), y)) → a(c(c(c(0))), y) 0.00/0.84
c(y) → y

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

Converted CpxTRS to CDT
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0.00/0.84

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(c(z0))) → c(c(a(z0, 0))) 0.00/0.84
c(a(a(0, z0), z1)) → a(c(c(c(0))), z1) 0.00/0.84
c(z0) → z0
Tuples:

C(c(c(z0))) → c1(C(c(a(z0, 0))), C(a(z0, 0))) 0.00/0.84
C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)), C(0))
S tuples:

C(c(c(z0))) → c1(C(c(a(z0, 0))), C(a(z0, 0))) 0.00/0.84
C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)), C(0))
K tuples:none
Defined Rule Symbols:

c

Defined Pair Symbols:

C

Compound Symbols:

c1, c2

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0.00/0.84

(3) CdtUnreachableProof (EQUIVALENT transformation)

The following tuples could be removed as they are not reachable from basic start terms:

C(c(c(z0))) → c1(C(c(a(z0, 0))), C(a(z0, 0)))
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(c(z0))) → c(c(a(z0, 0))) 0.00/0.84
c(a(a(0, z0), z1)) → a(c(c(c(0))), z1) 0.00/0.84
c(z0) → z0
Tuples:

C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)), C(0))
S tuples:

C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)), C(0))
K tuples:none
Defined Rule Symbols:

c

Defined Pair Symbols:

C

Compound Symbols:

c2

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0.00/0.84

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

Removed 1 trailing tuple parts
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0.00/0.84

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(c(z0))) → c(c(a(z0, 0))) 0.00/0.84
c(a(a(0, z0), z1)) → a(c(c(c(0))), z1) 0.00/0.84
c(z0) → z0
Tuples:

C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))
S tuples:

C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))
K tuples:none
Defined Rule Symbols:

c

Defined Pair Symbols:

C

Compound Symbols:

c2

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0.00/0.84

(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.

C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))
We considered the (Usable) Rules:

c(z0) → z0 0.00/0.84
c(a(a(0, z0), z1)) → a(c(c(c(0))), z1)
And the Tuples:

C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.84

POL(0) = 0    0.00/0.84
POL(C(x1)) = [5] + [4]x1    0.00/0.84
POL(a(x1, x2)) = [4] + x1 + x2    0.00/0.84
POL(c(x1)) = [1] + x1    0.00/0.84
POL(c2(x1, x2)) = x1 + x2   
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(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(c(z0))) → c(c(a(z0, 0))) 0.00/0.84
c(a(a(0, z0), z1)) → a(c(c(c(0))), z1) 0.00/0.84
c(z0) → z0
Tuples:

C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))
S tuples:none
K tuples:

C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))
Defined Rule Symbols:

c

Defined Pair Symbols:

C

Compound Symbols:

c2

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0.00/0.84

(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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0.00/0.90 EOF