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(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
p(m, n, s(r)) → p(m, r, n)
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p(m, s(n), 0) → p(0, n, m)
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p(m, 0, 0) → m
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:
p(z0, z1, s(z2)) → p(z0, z2, z1)
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p(z0, s(z1), 0) → p(0, z1, z0)
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p(z0, 0, 0) → z0
Tuples:
P(z0, z1, s(z2)) → c(P(z0, z2, z1))
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P(z0, s(z1), 0) → c1(P(0, z1, z0))
S tuples:
P(z0, z1, s(z2)) → c(P(z0, z2, z1))
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P(z0, s(z1), 0) → c1(P(0, z1, z0))
K tuples:none
Defined Rule Symbols:
p
Defined Pair Symbols:
P
Compound Symbols:
c, c1
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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.
P(z0, z1, s(z2)) → c(P(z0, z2, z1))
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P(z0, s(z1), 0) → c1(P(0, z1, z0))
We considered the (Usable) Rules:none
And the Tuples:
P(z0, z1, s(z2)) → c(P(z0, z2, z1))
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P(z0, s(z1), 0) → c1(P(0, z1, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(0) = 0
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POL(P(x1, x2, x3)) = x1 + x2 + x3
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POL(c(x1)) = x1
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POL(c1(x1)) = x1
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POL(s(x1)) = [1] + x1
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(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
p(z0, z1, s(z2)) → p(z0, z2, z1)
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p(z0, s(z1), 0) → p(0, z1, z0)
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p(z0, 0, 0) → z0
Tuples:
P(z0, z1, s(z2)) → c(P(z0, z2, z1))
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P(z0, s(z1), 0) → c1(P(0, z1, z0))
S tuples:none
K tuples:
P(z0, z1, s(z2)) → c(P(z0, z2, z1))
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P(z0, s(z1), 0) → c1(P(0, z1, z0))
Defined Rule Symbols:
p
Defined Pair Symbols:
P
Compound Symbols:
c, c1
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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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