(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
a__f(f(a)) → c(f(g(f(a))))
mark(f(X)) → a__f(mark(X))
mark(a) → a
mark(c(X)) → c(X)
mark(g(X)) → g(mark(X))
a__f(X) → f(X)
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
a__f(f(a)) → c(f(g(f(a))))
a__f(z0) → f(z0)
mark(f(z0)) → a__f(mark(z0))
mark(a) → a
mark(c(z0)) → c(z0)
mark(g(z0)) → g(mark(z0))
Tuples:
MARK(f(z0)) → c3(A__F(mark(z0)), MARK(z0))
MARK(g(z0)) → c6(MARK(z0))
S tuples:
MARK(f(z0)) → c3(A__F(mark(z0)), MARK(z0))
MARK(g(z0)) → c6(MARK(z0))
K tuples:none
Defined Rule Symbols:
a__f, mark
Defined Pair Symbols:
MARK
Compound Symbols:
c3, c6
(3) 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.
MARK(f(z0)) → c3(A__F(mark(z0)), MARK(z0))
We considered the (Usable) Rules:
mark(f(z0)) → a__f(mark(z0))
mark(a) → a
mark(c(z0)) → c(z0)
mark(g(z0)) → g(mark(z0))
a__f(f(a)) → c(f(g(f(a))))
a__f(z0) → f(z0)
And the Tuples:
MARK(f(z0)) → c3(A__F(mark(z0)), MARK(z0))
MARK(g(z0)) → c6(MARK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(A__F(x1)) = [3]
POL(MARK(x1)) = x12
POL(a) = 0
POL(a__f(x1)) = [3] + x1
POL(c(x1)) = x1
POL(c3(x1, x2)) = x1 + x2
POL(c6(x1)) = x1
POL(f(x1)) = [3] + x1
POL(g(x1)) = x1
POL(mark(x1)) = [2]x1
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
a__f(f(a)) → c(f(g(f(a))))
a__f(z0) → f(z0)
mark(f(z0)) → a__f(mark(z0))
mark(a) → a
mark(c(z0)) → c(z0)
mark(g(z0)) → g(mark(z0))
Tuples:
MARK(f(z0)) → c3(A__F(mark(z0)), MARK(z0))
MARK(g(z0)) → c6(MARK(z0))
S tuples:
MARK(g(z0)) → c6(MARK(z0))
K tuples:
MARK(f(z0)) → c3(A__F(mark(z0)), MARK(z0))
Defined Rule Symbols:
a__f, mark
Defined Pair Symbols:
MARK
Compound Symbols:
c3, c6
(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.
MARK(g(z0)) → c6(MARK(z0))
We considered the (Usable) Rules:
mark(f(z0)) → a__f(mark(z0))
mark(a) → a
mark(c(z0)) → c(z0)
mark(g(z0)) → g(mark(z0))
a__f(f(a)) → c(f(g(f(a))))
a__f(z0) → f(z0)
And the Tuples:
MARK(f(z0)) → c3(A__F(mark(z0)), MARK(z0))
MARK(g(z0)) → c6(MARK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(A__F(x1)) = [5]
POL(MARK(x1)) = [4]x1
POL(a) = [2]
POL(a__f(x1)) = [3] + [3]x1
POL(c(x1)) = [2]
POL(c3(x1, x2)) = x1 + x2
POL(c6(x1)) = x1
POL(f(x1)) = [4] + x1
POL(g(x1)) = [3] + x1
POL(mark(x1)) = [2]
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
a__f(f(a)) → c(f(g(f(a))))
a__f(z0) → f(z0)
mark(f(z0)) → a__f(mark(z0))
mark(a) → a
mark(c(z0)) → c(z0)
mark(g(z0)) → g(mark(z0))
Tuples:
MARK(f(z0)) → c3(A__F(mark(z0)), MARK(z0))
MARK(g(z0)) → c6(MARK(z0))
S tuples:none
K tuples:
MARK(f(z0)) → c3(A__F(mark(z0)), MARK(z0))
MARK(g(z0)) → c6(MARK(z0))
Defined Rule Symbols:
a__f, mark
Defined Pair Symbols:
MARK
Compound Symbols:
c3, c6
(7) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(8) BOUNDS(O(1), O(1))