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(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
eq0(S(x'), S(x)) → eq0(x', x)
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eq0(S(x), 0) → 0
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eq0(0, S(x)) → 0
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eq0(0, 0) → S(0)
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:
eq0(S(z0), S(z1)) → eq0(z0, z1)
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eq0(S(z0), 0) → 0
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eq0(0, S(z0)) → 0
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eq0(0, 0) → S(0)
Tuples:
EQ0(S(z0), S(z1)) → c(EQ0(z0, z1))
S tuples:
EQ0(S(z0), S(z1)) → c(EQ0(z0, z1))
K tuples:none
Defined Rule Symbols:
eq0
Defined Pair Symbols:
EQ0
Compound Symbols:
c
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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.
EQ0(S(z0), S(z1)) → c(EQ0(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:
EQ0(S(z0), S(z1)) → c(EQ0(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
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POL(EQ0(x1, x2)) = x1
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POL(S(x1)) = [2] + x1
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POL(c(x1)) = x1
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(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
eq0(S(z0), S(z1)) → eq0(z0, z1)
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eq0(S(z0), 0) → 0
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eq0(0, S(z0)) → 0
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eq0(0, 0) → S(0)
Tuples:
EQ0(S(z0), S(z1)) → c(EQ0(z0, z1))
S tuples:none
K tuples:
EQ0(S(z0), S(z1)) → c(EQ0(z0, z1))
Defined Rule Symbols:
eq0
Defined Pair Symbols:
EQ0
Compound Symbols:
c
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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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