The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^1).

(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:

anchored(Cons(x, xs), y) → anchored(xs, Cons(Cons(Nil, Nil), y))
anchored(Nil, y) → y
goal(x, y) → anchored(x, y)

Rewrite Strategy: INNERMOST

(1) CpxTrsMatchBoundsTAProof (EQUIVALENT transformation)

A linear upper bound on the runtime complexity of the TRS R could be shown with a Match-Bound[TAB_LEFTLINEAR,TAB_NONLEFTLINEAR] (for contructor-based start-terms) of 1.

The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2]
transitions:
Cons0(0, 0) → 0
Nil0() → 0
anchored0(0, 0) → 1
goal0(0, 0) → 2
Nil1() → 5
Nil1() → 6
Cons1(5, 6) → 4
Cons1(4, 0) → 3
anchored1(0, 3) → 1
anchored1(0, 0) → 2
Cons1(4, 3) → 3
anchored1(0, 3) → 2
0 → 1
0 → 2
3 → 1
3 → 2

(2) BOUNDS(1, n^1)

(3) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

anchored(Cons(z0, z1), z2) → anchored(z1, Cons(Cons(Nil, Nil), z2))
anchored(Nil, z0) → z0
goal(z0, z1) → anchored(z0, z1)

Tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))
ANCHORED(Nil, z0) → c1
GOAL(z0, z1) → c2(ANCHORED(z0, z1))

S tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))
ANCHORED(Nil, z0) → c1
GOAL(z0, z1) → c2(ANCHORED(z0, z1))

K tuples:none
Defined Rule Symbols:

anchored, goal

Defined Pair Symbols:

ANCHORED, GOAL

Compound Symbols:

c, c1, c2

(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

GOAL(z0, z1) → c2(ANCHORED(z0, z1))

Removed 1 trailing nodes:

ANCHORED(Nil, z0) → c1

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

anchored(Cons(z0, z1), z2) → anchored(z1, Cons(Cons(Nil, Nil), z2))
anchored(Nil, z0) → z0
goal(z0, z1) → anchored(z0, z1)

Tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))

S tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))

K tuples:none
Defined Rule Symbols:

anchored, goal

Defined Pair Symbols:

ANCHORED

Compound Symbols:

c

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

anchored(Cons(z0, z1), z2) → anchored(z1, Cons(Cons(Nil, Nil), z2))
anchored(Nil, z0) → z0
goal(z0, z1) → anchored(z0, z1)

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))

S tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

ANCHORED

Compound Symbols:

c

(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))

We considered the (Usable) Rules:none
And the Tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(ANCHORED(x₁, x₂)) = [3]x₁ + x₂   
POL(Cons(x₁, x₂)) = [3] + x₂   
POL(Nil) = 0   
POL(c(x₁)) = x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))

S tuples:none
K tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))

Defined Rule Symbols:none

Defined Pair Symbols:

ANCHORED

Compound Symbols:

c

(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(12) BOUNDS(1, 1)