(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(0) → s(0)
f(s(x)) → g(s(s(x)))
g(0) → s(0)
g(s(0)) → s(0)
g(s(s(x))) → f(x)
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → s(0)
f(s(z0)) → g(s(s(z0)))
g(0) → s(0)
g(s(0)) → s(0)
g(s(s(z0))) → f(z0)
Tuples:
F(s(z0)) → c1(G(s(s(z0))))
G(s(s(z0))) → c4(F(z0))
S tuples:
F(s(z0)) → c1(G(s(s(z0))))
G(s(s(z0))) → c4(F(z0))
K tuples:none
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F, G
Compound Symbols:
c1, c4
(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(s(z0)) → c1(G(s(s(z0))))
We considered the (Usable) Rules:none
And the Tuples:
F(s(z0)) → c1(G(s(s(z0))))
G(s(s(z0))) → c4(F(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(F(x1)) = [2] + x1
POL(G(x1)) = x1
POL(c1(x1)) = x1
POL(c4(x1)) = x1
POL(s(x1)) = [1] + x1
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → s(0)
f(s(z0)) → g(s(s(z0)))
g(0) → s(0)
g(s(0)) → s(0)
g(s(s(z0))) → f(z0)
Tuples:
F(s(z0)) → c1(G(s(s(z0))))
G(s(s(z0))) → c4(F(z0))
S tuples:
G(s(s(z0))) → c4(F(z0))
K tuples:
F(s(z0)) → c1(G(s(s(z0))))
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F, G
Compound Symbols:
c1, c4
(5) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
G(s(s(z0))) → c4(F(z0))
F(s(z0)) → c1(G(s(s(z0))))
Now S is empty
(6) BOUNDS(O(1), O(1))