(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(g(x), x, y) → f(y, y, g(y))
g(g(x)) → g(x)
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(g(z0), z0, z1) → f(z1, z1, g(z1))
g(g(z0)) → g(z0)
Tuples:
F(g(z0), z0, z1) → c(F(z1, z1, g(z1)), G(z1))
G(g(z0)) → c1(G(z0))
S tuples:
F(g(z0), z0, z1) → c(F(z1, z1, g(z1)), G(z1))
G(g(z0)) → c1(G(z0))
K tuples:none
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F, G
Compound Symbols:
c, c1
(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)
Removed 1 leading nodes:
F(g(z0), z0, z1) → c(F(z1, z1, g(z1)), G(z1))
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(g(z0), z0, z1) → f(z1, z1, g(z1))
g(g(z0)) → g(z0)
Tuples:
G(g(z0)) → c1(G(z0))
S tuples:
G(g(z0)) → c1(G(z0))
K tuples:none
Defined Rule Symbols:
f, g
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(g(z0)) → c1(G(z0))
We considered the (Usable) Rules:none
And the Tuples:
G(g(z0)) → c1(G(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(G(x1)) = x1
POL(c1(x1)) = x1
POL(g(x1)) = [1] + [4]x1
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(g(z0), z0, z1) → f(z1, z1, g(z1))
g(g(z0)) → g(z0)
Tuples:
G(g(z0)) → c1(G(z0))
S tuples:none
K tuples:
G(g(z0)) → c1(G(z0))
Defined Rule Symbols:
f, g
Defined Pair Symbols:
G
Compound Symbols:
c1
(7) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(8) BOUNDS(O(1), O(1))