(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), s(y)) → f(x, s(c(s(y))))
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), s(z1)) → f(z0, s(c(s(z1))))
Tuples:
F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
S tuples:
F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c1, c2
(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.
F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
We considered the (Usable) Rules:
f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), s(z1)) → f(z0, s(c(s(z1))))
And the Tuples:
F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(F(x1, x2)) = x2
POL(c(x1)) = [4] + x1
POL(c1(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(f(x1, x2)) = [3] + [2]x1
POL(s(x1)) = 0
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), s(z1)) → f(z0, s(c(s(z1))))
Tuples:
F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
S tuples:
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
K tuples:
F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c1, c2
(5) CdtInstantiationProof (BOTH BOUNDS(ID, ID) transformation)
Use instantiation to replace
F(
z0,
c(
z1)) →
c1(
F(
z0,
s(
f(
z1,
z1))),
F(
z1,
z1)) by
F(c(z1), c(z1)) → c1(F(c(z1), s(f(z1, z1))), F(z1, z1))
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), s(z1)) → f(z0, s(c(s(z1))))
Tuples:
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
F(c(z1), c(z1)) → c1(F(c(z1), s(f(z1, z1))), F(z1, z1))
S tuples:
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
K tuples:
F(c(z1), c(z1)) → c1(F(c(z1), s(f(z1, z1))), F(z1, z1))
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c2, c1
(7) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), s(z1)) → f(z0, s(c(s(z1))))
Tuples:
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
F(c(z1), c(z1)) → c1(F(z1, z1))
S tuples:
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
K tuples:
F(c(z1), c(z1)) → c1(F(z1, z1))
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c2, c1
(9) 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.
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
We considered the (Usable) Rules:none
And the Tuples:
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
F(c(z1), c(z1)) → c1(F(z1, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(F(x1, x2)) = [4]x1 + x2
POL(c(x1)) = x1
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(s(x1)) = [5] + x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), s(z1)) → f(z0, s(c(s(z1))))
Tuples:
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
F(c(z1), c(z1)) → c1(F(z1, z1))
S tuples:none
K tuples:
F(c(z1), c(z1)) → c1(F(z1, z1))
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c2, c1
(11) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(12) BOUNDS(O(1), O(1))