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:

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

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 2.

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
f0(0, 0) → 1
g0(0, 0) → 2
empty1() → 3
g1(0, 3) → 1
cons1(0, 0) → 4
f1(4, 0) → 1
cons1(0, 0) → 5
g1(0, 5) → 2
g1(4, 3) → 1
cons1(0, 4) → 4
cons1(0, 3) → 5
g1(0, 5) → 1
cons1(0, 5) → 5
cons2(0, 3) → 6
g2(0, 6) → 1
g2(4, 6) → 1
cons1(0, 6) → 5
cons2(0, 6) → 6
0 → 2
3 → 1
5 → 2
5 → 1
6 → 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:

f(z0, empty) → g(z0, empty)
f(z0, cons(z1, z2)) → f(cons(z1, z0), z2)
g(empty, z0) → z0
g(cons(z0, z1), z2) → g(z1, cons(z0, z2))

Tuples:

F(z0, empty) → c(G(z0, empty))
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(empty, z0) → c2
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

S tuples:

F(z0, empty) → c(G(z0, empty))
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(empty, z0) → c2
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, c2, c3

(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing nodes:

G(empty, z0) → c2

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, empty) → g(z0, empty)
f(z0, cons(z1, z2)) → f(cons(z1, z0), z2)
g(empty, z0) → z0
g(cons(z0, z1), z2) → g(z1, cons(z0, z2))

Tuples:

F(z0, empty) → c(G(z0, empty))
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

S tuples:

F(z0, empty) → c(G(z0, empty))
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, c3

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(z0, empty) → g(z0, empty)
f(z0, cons(z1, z2)) → f(cons(z1, z0), z2)
g(empty, z0) → z0
g(cons(z0, z1), z2) → g(z1, cons(z0, z2))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(z0, empty) → c(G(z0, empty))
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

S tuples:

F(z0, empty) → c(G(z0, empty))
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, c3

(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.

F(z0, empty) → c(G(z0, empty))
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

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

F(z0, empty) → c(G(z0, empty))
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

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

POL(F(x₁, x₂)) = [2] + x₁ + [2]x₂   
POL(G(x₁, x₂)) = [2] + x₁   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c3(x₁)) = x₁   
POL(cons(x₁, x₂)) = [2] + x₁ + x₂   
POL(empty) = [3]   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(z0, empty) → c(G(z0, empty))
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

S tuples:none
K tuples:

F(z0, empty) → c(G(z0, empty))
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

Defined Rule Symbols:none

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, c3

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

The set S is empty

(12) BOUNDS(1, 1)