(0) Obligation:
The Runtime Complexity (innermost) of the given
CpxTRS could be proven to be
BOUNDS(1, n^1).
The TRS R consists of the following rules:
duplicate(Cons(x, xs)) → Cons(x, Cons(x, duplicate(xs)))
duplicate(Nil) → Nil
goal(x) → duplicate(x)
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
duplicate(Cons(z0, z1)) → Cons(z0, Cons(z0, duplicate(z1)))
duplicate(Nil) → Nil
goal(z0) → duplicate(z0)
Tuples:
DUPLICATE(Cons(z0, z1)) → c(DUPLICATE(z1))
DUPLICATE(Nil) → c1
GOAL(z0) → c2(DUPLICATE(z0))
S tuples:
DUPLICATE(Cons(z0, z1)) → c(DUPLICATE(z1))
DUPLICATE(Nil) → c1
GOAL(z0) → c2(DUPLICATE(z0))
K tuples:none
Defined Rule Symbols:
duplicate, goal
Defined Pair Symbols:
DUPLICATE, GOAL
Compound Symbols:
c, c1, c2
(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)
Removed 1 leading nodes:
GOAL(z0) → c2(DUPLICATE(z0))
Removed 1 trailing nodes:
DUPLICATE(Nil) → c1
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
duplicate(Cons(z0, z1)) → Cons(z0, Cons(z0, duplicate(z1)))
duplicate(Nil) → Nil
goal(z0) → duplicate(z0)
Tuples:
DUPLICATE(Cons(z0, z1)) → c(DUPLICATE(z1))
S tuples:
DUPLICATE(Cons(z0, z1)) → c(DUPLICATE(z1))
K tuples:none
Defined Rule Symbols:
duplicate, goal
Defined Pair Symbols:
DUPLICATE
Compound Symbols:
c
(5) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
duplicate(Cons(z0, z1)) → Cons(z0, Cons(z0, duplicate(z1)))
duplicate(Nil) → Nil
goal(z0) → duplicate(z0)
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
DUPLICATE(Cons(z0, z1)) → c(DUPLICATE(z1))
S tuples:
DUPLICATE(Cons(z0, z1)) → c(DUPLICATE(z1))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
DUPLICATE
Compound Symbols:
c
(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.
DUPLICATE(Cons(z0, z1)) → c(DUPLICATE(z1))
We considered the (Usable) Rules:none
And the Tuples:
DUPLICATE(Cons(z0, z1)) → c(DUPLICATE(z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(Cons(x1, x2)) = [1] + x2
POL(DUPLICATE(x1)) = x1
POL(c(x1)) = x1
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
DUPLICATE(Cons(z0, z1)) → c(DUPLICATE(z1))
S tuples:none
K tuples:
DUPLICATE(Cons(z0, z1)) → c(DUPLICATE(z1))
Defined Rule Symbols:none
Defined Pair Symbols:
DUPLICATE
Compound Symbols:
c
(9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(10) BOUNDS(1, 1)