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:

foldl#3(x2, Nil) → x2
foldl#3(x16, Cons(x24, x6)) → foldl#3(Cons(x24, x16), x6)
main(x1) → foldl#3(Nil, x1)

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:
Nil0() → 0
Cons0(0, 0) → 0
foldl#30(0, 0) → 1
main0(0) → 2
Cons1(0, 0) → 3
foldl#31(3, 0) → 1
Nil1() → 4
foldl#31(4, 0) → 2
Cons1(0, 3) → 3
Cons1(0, 4) → 3
foldl#31(3, 0) → 2
0 → 1
3 → 1
3 → 2
4 → 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:

foldl#3(z0, Nil) → z0
foldl#3(z0, Cons(z1, z2)) → foldl#3(Cons(z1, z0), z2)
main(z0) → foldl#3(Nil, z0)

Tuples:

FOLDL#3(z0, Nil) → c
FOLDL#3(z0, Cons(z1, z2)) → c1(FOLDL#3(Cons(z1, z0), z2))
MAIN(z0) → c2(FOLDL#3(Nil, z0))

S tuples:

FOLDL#3(z0, Nil) → c
FOLDL#3(z0, Cons(z1, z2)) → c1(FOLDL#3(Cons(z1, z0), z2))
MAIN(z0) → c2(FOLDL#3(Nil, z0))

K tuples:none
Defined Rule Symbols:

foldl#3, main

Defined Pair Symbols:

FOLDL#3, MAIN

Compound Symbols:

c, c1, c2

(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

MAIN(z0) → c2(FOLDL#3(Nil, z0))

Removed 1 trailing nodes:

FOLDL#3(z0, Nil) → c

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

foldl#3(z0, Nil) → z0
foldl#3(z0, Cons(z1, z2)) → foldl#3(Cons(z1, z0), z2)
main(z0) → foldl#3(Nil, z0)

Tuples:

FOLDL#3(z0, Cons(z1, z2)) → c1(FOLDL#3(Cons(z1, z0), z2))

S tuples:

FOLDL#3(z0, Cons(z1, z2)) → c1(FOLDL#3(Cons(z1, z0), z2))

K tuples:none
Defined Rule Symbols:

foldl#3, main

Defined Pair Symbols:

FOLDL#3

Compound Symbols:

c1

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

foldl#3(z0, Nil) → z0
foldl#3(z0, Cons(z1, z2)) → foldl#3(Cons(z1, z0), z2)
main(z0) → foldl#3(Nil, z0)

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

FOLDL#3(z0, Cons(z1, z2)) → c1(FOLDL#3(Cons(z1, z0), z2))

S tuples:

FOLDL#3(z0, Cons(z1, z2)) → c1(FOLDL#3(Cons(z1, z0), z2))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

FOLDL#3

Compound Symbols:

c1

(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.

FOLDL#3(z0, Cons(z1, z2)) → c1(FOLDL#3(Cons(z1, z0), z2))

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

FOLDL#3(z0, Cons(z1, z2)) → c1(FOLDL#3(Cons(z1, z0), z2))

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

POL(Cons(x₁, x₂)) = [3] + x₂   
POL(FOLDL#3(x₁, x₂)) = x₁ + [3]x₂   
POL(c1(x₁)) = x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

FOLDL#3(z0, Cons(z1, z2)) → c1(FOLDL#3(Cons(z1, z0), z2))

S tuples:none
K tuples:

FOLDL#3(z0, Cons(z1, z2)) → c1(FOLDL#3(Cons(z1, z0), z2))

Defined Rule Symbols:none

Defined Pair Symbols:

FOLDL#3

Compound Symbols:

c1

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

The set S is empty

(12) BOUNDS(1, 1)