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(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
implies(not(x), y) → or(x, y)
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implies(not(x), or(y, z)) → implies(y, or(x, z))
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implies(x, or(y, z)) → or(y, implies(x, z))
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:
implies(not(z0), z1) → or(z0, z1)
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implies(not(z0), or(z1, z2)) → implies(z1, or(z0, z2))
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implies(z0, or(z1, z2)) → or(z1, implies(z0, z2))
Tuples:
IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
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IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
S tuples:
IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
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IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
K tuples:none
Defined Rule Symbols:
implies
Defined Pair Symbols:
IMPLIES
Compound Symbols:
c1, c2
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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.
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
We considered the (Usable) Rules:none
And the Tuples:
IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
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IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(IMPLIES(x1, x2)) = x2
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POL(c1(x1)) = x1
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POL(c2(x1)) = x1
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POL(not(x1)) = 0
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POL(or(x1, x2)) = [1] + x2
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(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
implies(not(z0), z1) → or(z0, z1)
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implies(not(z0), or(z1, z2)) → implies(z1, or(z0, z2))
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implies(z0, or(z1, z2)) → or(z1, implies(z0, z2))
Tuples:
IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
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IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
S tuples:
IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
K tuples:
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
Defined Rule Symbols:
implies
Defined Pair Symbols:
IMPLIES
Compound Symbols:
c1, c2
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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.
IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
We considered the (Usable) Rules:none
And the Tuples:
IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
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IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(IMPLIES(x1, x2)) = [2]x1 + [2]x2
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POL(c1(x1)) = x1
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POL(c2(x1)) = x1
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POL(not(x1)) = [1] + x1
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POL(or(x1, x2)) = x1 + x2
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(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
implies(not(z0), z1) → or(z0, z1)
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implies(not(z0), or(z1, z2)) → implies(z1, or(z0, z2))
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implies(z0, or(z1, z2)) → or(z1, implies(z0, z2))
Tuples:
IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
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IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
S tuples:none
K tuples:
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
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IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
Defined Rule Symbols:
implies
Defined Pair Symbols:
IMPLIES
Compound Symbols:
c1, c2
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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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