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