(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(s(0), g(z0)) → f(z0, g(z0))
g(s(z0)) → g(z0)
Tuples:
F(s(0), g(z0)) → c(F(z0, g(z0)), G(z0))
G(s(z0)) → c1(G(z0))
S tuples:
F(s(0), g(z0)) → c(F(z0, g(z0)), G(z0))
G(s(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(s(0), g(z0)) → c(F(z0, g(z0)), G(z0))
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(s(0), g(z0)) → f(z0, g(z0))
g(s(z0)) → g(z0)
Tuples:
G(s(z0)) → c1(G(z0))
S tuples:
G(s(z0)) → c1(G(z0))
K tuples:none
Defined Rule Symbols:
f, g
Defined Pair Symbols:
G
Compound Symbols:
c1
(5) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
f(s(0), g(z0)) → f(z0, g(z0))
g(s(z0)) → g(z0)
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
G(s(z0)) → c1(G(z0))
S tuples:
G(s(z0)) → c1(G(z0))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
G
Compound Symbols:
c1
(7) CdtRuleRemovalProof (UPPER BOUND(ADD(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
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
G(s(z0)) → c1(G(z0))
S tuples:none
K tuples:
G(s(z0)) → c1(G(z0))
Defined Rule Symbols:none
Defined Pair Symbols:
G
Compound Symbols:
c1
(9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(10) BOUNDS(1, 1)