(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0, 1, z0) → f(s(z0), z0, z0)
f(z0, z1, s(z2)) → s(f(0, 1, z2))
Tuples:
F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
S tuples:
F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c, c1
(3) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
f(0, 1, z0) → f(s(z0), z0, z0)
f(z0, z1, s(z2)) → s(f(0, 1, z2))
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
S tuples:
F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c, c1
(5) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
We considered the (Usable) Rules:none
And the Tuples:
F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(1) = 0
POL(F(x1, x2, x3)) = x3
POL(c(x1)) = x1
POL(c1(x1)) = x1
POL(s(x1)) = [2] + x1
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
S tuples:
F(0, 1, z0) → c(F(s(z0), z0, z0))
K tuples:
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c, c1
(7) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
Now S is empty
(8) BOUNDS(1, 1)