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:

rev(ls) → r1(ls, empty)
r1(empty, a) → a
r1(cons(x, k), a) → r1(k, cons(x, a))

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:
empty0() → 0
cons0(0, 0) → 0
rev0(0) → 1
r10(0, 0) → 2
empty1() → 3
r11(0, 3) → 1
cons1(0, 0) → 4
r11(0, 4) → 2
cons1(0, 3) → 4
r11(0, 4) → 1
cons1(0, 4) → 4
0 → 2
3 → 1
4 → 2
4 → 1

(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:

rev(z0) → r1(z0, empty)
r1(empty, z0) → z0
r1(cons(z0, z1), z2) → r1(z1, cons(z0, z2))

Tuples:

REV(z0) → c(R1(z0, empty))
R1(empty, z0) → c1
R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))

S tuples:

REV(z0) → c(R1(z0, empty))
R1(empty, z0) → c1
R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))

K tuples:none
Defined Rule Symbols:

rev, r1

Defined Pair Symbols:

REV, R1

Compound Symbols:

c, c1, c2

(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

REV(z0) → c(R1(z0, empty))

Removed 1 trailing nodes:

R1(empty, z0) → c1

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(z0) → r1(z0, empty)
r1(empty, z0) → z0
r1(cons(z0, z1), z2) → r1(z1, cons(z0, z2))

Tuples:

R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))

S tuples:

R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))

K tuples:none
Defined Rule Symbols:

rev, r1

Defined Pair Symbols:

R1

Compound Symbols:

c2

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

rev(z0) → r1(z0, empty)
r1(empty, z0) → z0
r1(cons(z0, z1), z2) → r1(z1, cons(z0, z2))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))

S tuples:

R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

R1

Compound Symbols:

c2

(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.

R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))

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

R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))

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

POL(R1(x₁, x₂)) = x₁   
POL(c2(x₁)) = x₁   
POL(cons(x₁, x₂)) = [1] + x₂   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))

S tuples:none
K tuples:

R1(cons(z0, z1), z2) → c2(R1(z1, cons(z0, z2)))

Defined Rule Symbols:none

Defined Pair Symbols:

R1

Compound Symbols:

c2

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

The set S is empty

(12) BOUNDS(1, 1)