YES(O(1), O(n^1)) 0.00/0.75 YES(O(1), O(n^1)) 0.00/0.77 0.00/0.77 0.00/0.77 0.00/0.77 0.00/0.77 0.00/0.77 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.77 0.00/0.77 0.00/0.77
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(0) Obligation:

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

admit(x, nil) → nil 0.00/0.77
admit(x, .(u, .(v, .(w, z)))) → cond(=(sum(x, u, v), w), .(u, .(v, .(w, admit(carry(x, u, v), z))))) 0.00/0.77
cond(true, y) → y

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:

admit(z0, nil) → nil 0.00/0.77
admit(z0, .(z1, .(z2, .(w, z3)))) → cond(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3))))) 0.00/0.77
cond(true, z0) → z0
Tuples:

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(COND(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3))))), ADMIT(carry(z0, z1, z2), z3))
S tuples:

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(COND(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3))))), ADMIT(carry(z0, z1, z2), z3))
K tuples:none
Defined Rule Symbols:

admit, cond

Defined Pair Symbols:

ADMIT

Compound Symbols:

c1

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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:

admit(z0, nil) → nil 0.00/0.77
admit(z0, .(z1, .(z2, .(w, z3)))) → cond(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3))))) 0.00/0.77
cond(true, z0) → z0
Tuples:

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))
S tuples:

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))
K tuples:none
Defined Rule Symbols:

admit, cond

Defined Pair Symbols:

ADMIT

Compound Symbols:

c1

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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.

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))
We considered the (Usable) Rules:none
And the Tuples:

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.77

POL(.(x1, x2)) = x1 + x2    0.00/0.77
POL(ADMIT(x1, x2)) = x1 + x2    0.00/0.77
POL(c1(x1)) = x1    0.00/0.77
POL(carry(x1, x2, x3)) = x1 + x2 + x3    0.00/0.77
POL(w) = [1]   
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

admit(z0, nil) → nil 0.00/0.77
admit(z0, .(z1, .(z2, .(w, z3)))) → cond(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3))))) 0.00/0.77
cond(true, z0) → z0
Tuples:

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))
S tuples:none
K tuples:

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))
Defined Rule Symbols:

admit, cond

Defined Pair Symbols:

ADMIT

Compound Symbols:

c1

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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