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(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
minus(x, 0) → x
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minus(s(x), s(y)) → minus(x, y)
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quot(0, s(y)) → 0
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quot(s(x), s(y)) → s(quot(minus(x, y), s(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:
minus(z0, 0) → z0
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minus(s(z0), s(z1)) → minus(z0, z1)
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quot(0, s(z0)) → 0
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quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
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QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
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QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:
minus, quot
Defined Pair Symbols:
MINUS, QUOT
Compound Symbols:
c1, c3
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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.
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
We considered the (Usable) Rules:
minus(z0, 0) → z0
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minus(s(z0), s(z1)) → minus(z0, z1)
And the Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
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QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(0) = 0
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POL(MINUS(x1, x2)) = 0
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POL(QUOT(x1, x2)) = x1
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POL(c1(x1)) = x1
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POL(c3(x1, x2)) = x1 + x2
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POL(minus(x1, x2)) = x1
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POL(s(x1)) = [1] + x1
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(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
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minus(s(z0), s(z1)) → minus(z0, z1)
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quot(0, s(z0)) → 0
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quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
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QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
K tuples:
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:
minus, quot
Defined Pair Symbols:
MINUS, QUOT
Compound Symbols:
c1, c3
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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.
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
We considered the (Usable) Rules:
minus(z0, 0) → z0
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minus(s(z0), s(z1)) → minus(z0, z1)
And the Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
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QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(0) = [3]
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POL(MINUS(x1, x2)) = [3]x1 + x2
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POL(QUOT(x1, x2)) = [2]x1 + [2]x1·x2 + x12
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POL(c1(x1)) = x1
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POL(c3(x1, x2)) = x1 + x2
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POL(minus(x1, x2)) = x1
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POL(s(x1)) = [2] + x1
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(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
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minus(s(z0), s(z1)) → minus(z0, z1)
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quot(0, s(z0)) → 0
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quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
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QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:none
K tuples:
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
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MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
Defined Rule Symbols:
minus, quot
Defined Pair Symbols:
MINUS, QUOT
Compound Symbols:
c1, c3
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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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