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(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
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times(X, s(Y)) → plus(X, times(Y, X))
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:
plus(plus(z0, z1), z2) → plus(z0, plus(z1, z2))
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times(z0, s(z1)) → plus(z0, times(z1, z0))
Tuples:
PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
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TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
S tuples:
PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
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TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
K tuples:none
Defined Rule Symbols:
plus, times
Defined Pair Symbols:
PLUS, TIMES
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.
TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
We considered the (Usable) Rules:
times(z0, s(z1)) → plus(z0, times(z1, z0))
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plus(plus(z0, z1), z2) → plus(z0, plus(z1, z2))
And the Tuples:
PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
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TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(PLUS(x1, x2)) = 0
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POL(TIMES(x1, x2)) = [5]x1 + [5]x2
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POL(c(x1, x2)) = x1 + x2
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POL(c1(x1, x2)) = x1 + x2
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POL(plus(x1, x2)) = [3] + x2
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POL(s(x1)) = [3] + x1
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POL(times(x1, x2)) = [5] + [5]x1 + [5]x2
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(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
plus(plus(z0, z1), z2) → plus(z0, plus(z1, z2))
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times(z0, s(z1)) → plus(z0, times(z1, z0))
Tuples:
PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
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TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
S tuples:
PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
K tuples:
TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
Defined Rule Symbols:
plus, times
Defined Pair Symbols:
PLUS, TIMES
Compound Symbols:
c, c1
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(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
We considered the (Usable) Rules:
times(z0, s(z1)) → plus(z0, times(z1, z0))
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plus(plus(z0, z1), z2) → plus(z0, plus(z1, z2))
And the Tuples:
PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
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TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(PLUS(x1, x2)) = x1
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POL(TIMES(x1, x2)) = [2]x1 + [3]x2 + x1·x2
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POL(c(x1, x2)) = x1 + x2
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POL(c1(x1, x2)) = x1 + x2
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POL(plus(x1, x2)) = [1] + [2]x1 + x2
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POL(s(x1)) = [3] + x1
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POL(times(x1, x2)) = 0
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(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
plus(plus(z0, z1), z2) → plus(z0, plus(z1, z2))
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times(z0, s(z1)) → plus(z0, times(z1, z0))
Tuples:
PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
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TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
S tuples:none
K tuples:
TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
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PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
Defined Rule Symbols:
plus, times
Defined Pair Symbols:
PLUS, TIMES
Compound Symbols:
c, 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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