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:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
length1(ok(X)) → ok(length1(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Rewrite Strategy: INNERMOST

(1) NestedDefinedSymbolProof (BOTH BOUNDS(ID, ID) transformation)

The following defined symbols can occur below the 0th argument of top: proper, active
The following defined symbols can occur below the 0th argument of proper: proper, active
The following defined symbols can occur below the 0th argument of active: proper, active

Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(length1(X)) → length1(proper(X))

(2) 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:

top(ok(X)) → top(active(X))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
from(mark(X)) → mark(from(X))
s(ok(X)) → ok(s(X))
s(mark(X)) → mark(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
proper(0) → ok(0)
length1(ok(X)) → ok(length1(X))
length(ok(X)) → ok(length(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
top(mark(X)) → top(proper(X))

Rewrite Strategy: INNERMOST

(3) 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, 3, 4, 5, 6, 7]
transitions:
ok0(0) → 0
active0(0) → 0
nil0() → 0
mark0(0) → 0
00() → 0
top0(0) → 1
proper0(0) → 2
from0(0) → 3
s0(0) → 4
cons0(0, 0) → 5
length10(0) → 6
length0(0) → 7
active1(0) → 8
top1(8) → 1
nil1() → 9
ok1(9) → 2
from1(0) → 10
ok1(10) → 3
from1(0) → 11
mark1(11) → 3
s1(0) → 12
ok1(12) → 4
s1(0) → 13
mark1(13) → 4
cons1(0, 0) → 14
ok1(14) → 5
01() → 15
ok1(15) → 2
length11(0) → 16
ok1(16) → 6
length1(0) → 17
ok1(17) → 7
cons1(0, 0) → 18
mark1(18) → 5
proper1(0) → 19
top1(19) → 1
ok1(9) → 19
ok1(10) → 10
ok1(10) → 11
mark1(11) → 10
mark1(11) → 11
ok1(12) → 12
ok1(12) → 13
mark1(13) → 12
mark1(13) → 13
ok1(14) → 14
ok1(14) → 18
ok1(15) → 19
ok1(16) → 16
ok1(17) → 17
mark1(18) → 14
mark1(18) → 18
active2(9) → 20
top2(20) → 1
active2(15) → 20

(4) BOUNDS(1, n^1)

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

Converted Cpx (relative) TRS to CDT

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
proper(nil) → ok(nil)
proper(0) → ok(0)
from(ok(z0)) → ok(from(z0))
from(mark(z0)) → mark(from(z0))
s(ok(z0)) → ok(s(z0))
s(mark(z0)) → mark(s(z0))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
cons(mark(z0), z1) → mark(cons(z0, z1))
length1(ok(z0)) → ok(length1(z0))
length(ok(z0)) → ok(length(z0))

Tuples:

TOP(ok(z0)) → c(TOP(active(z0)))
TOP(mark(z0)) → c1(TOP(proper(z0)), PROPER(z0))
PROPER(nil) → c2
PROPER(0) → c3
FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))

S tuples:

TOP(ok(z0)) → c(TOP(active(z0)))
TOP(mark(z0)) → c1(TOP(proper(z0)), PROPER(z0))
PROPER(nil) → c2
PROPER(0) → c3
FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))

K tuples:none
Defined Rule Symbols:

top, proper, from, s, cons, length1, length

Defined Pair Symbols:

TOP, PROPER, FROM, S, CONS, LENGTH1, LENGTH

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11

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

Removed 3 trailing nodes:

PROPER(nil) → c2
PROPER(0) → c3
TOP(ok(z0)) → c(TOP(active(z0)))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
proper(nil) → ok(nil)
proper(0) → ok(0)
from(ok(z0)) → ok(from(z0))
from(mark(z0)) → mark(from(z0))
s(ok(z0)) → ok(s(z0))
s(mark(z0)) → mark(s(z0))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
cons(mark(z0), z1) → mark(cons(z0, z1))
length1(ok(z0)) → ok(length1(z0))
length(ok(z0)) → ok(length(z0))

Tuples:

TOP(mark(z0)) → c1(TOP(proper(z0)), PROPER(z0))
FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))

S tuples:

TOP(mark(z0)) → c1(TOP(proper(z0)), PROPER(z0))
FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))

K tuples:none
Defined Rule Symbols:

top, proper, from, s, cons, length1, length

Defined Pair Symbols:

TOP, FROM, S, CONS, LENGTH1, LENGTH

Compound Symbols:

c1, c4, c5, c6, c7, c8, c9, c10, c11

(9) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
proper(nil) → ok(nil)
proper(0) → ok(0)
from(ok(z0)) → ok(from(z0))
from(mark(z0)) → mark(from(z0))
s(ok(z0)) → ok(s(z0))
s(mark(z0)) → mark(s(z0))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
cons(mark(z0), z1) → mark(cons(z0, z1))
length1(ok(z0)) → ok(length1(z0))
length(ok(z0)) → ok(length(z0))

Tuples:

FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))
TOP(mark(z0)) → c1(TOP(proper(z0)))

S tuples:

FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))
TOP(mark(z0)) → c1(TOP(proper(z0)))

K tuples:none
Defined Rule Symbols:

top, proper, from, s, cons, length1, length

Defined Pair Symbols:

FROM, S, CONS, LENGTH1, LENGTH, TOP

Compound Symbols:

c4, c5, c6, c7, c8, c9, c10, c11, c1

(11) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
from(ok(z0)) → ok(from(z0))
from(mark(z0)) → mark(from(z0))
s(ok(z0)) → ok(s(z0))
s(mark(z0)) → mark(s(z0))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
cons(mark(z0), z1) → mark(cons(z0, z1))
length1(ok(z0)) → ok(length1(z0))
length(ok(z0)) → ok(length(z0))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(nil) → ok(nil)
proper(0) → ok(0)

Tuples:

FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))
TOP(mark(z0)) → c1(TOP(proper(z0)))

S tuples:

FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))
TOP(mark(z0)) → c1(TOP(proper(z0)))

K tuples:none
Defined Rule Symbols:

proper

Defined Pair Symbols:

FROM, S, CONS, LENGTH1, LENGTH, TOP

Compound Symbols:

c4, c5, c6, c7, c8, c9, c10, c11, c1

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

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

TOP(mark(z0)) → c1(TOP(proper(z0)))

We considered the (Usable) Rules:

proper(nil) → ok(nil)
proper(0) → ok(0)

And the Tuples:

FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))
TOP(mark(z0)) → c1(TOP(proper(z0)))

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

POL(0) = 0   
POL(CONS(x₁, x₂)) = 0   
POL(FROM(x₁)) = 0   
POL(LENGTH(x₁)) = 0   
POL(LENGTH1(x₁)) = 0   
POL(S(x₁)) = 0   
POL(TOP(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c10(x₁)) = x₁   
POL(c11(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(mark(x₁)) = [1] + x₁   
POL(nil) = 0   
POL(ok(x₁)) = 0   
POL(proper(x₁)) = 0   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(nil) → ok(nil)
proper(0) → ok(0)

Tuples:

FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))
TOP(mark(z0)) → c1(TOP(proper(z0)))

S tuples:

FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))

K tuples:

TOP(mark(z0)) → c1(TOP(proper(z0)))

Defined Rule Symbols:

proper

Defined Pair Symbols:

FROM, S, CONS, LENGTH1, LENGTH, TOP

Compound Symbols:

c4, c5, c6, c7, c8, c9, c10, c11, c1

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

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

S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))

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

FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))
TOP(mark(z0)) → c1(TOP(proper(z0)))

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

POL(0) = 0   
POL(CONS(x₁, x₂)) = 0   
POL(FROM(x₁)) = 0   
POL(LENGTH(x₁)) = 0   
POL(LENGTH1(x₁)) = [2]x₁   
POL(S(x₁)) = [2]x₁   
POL(TOP(x₁)) = 0   
POL(c1(x₁)) = x₁   
POL(c10(x₁)) = x₁   
POL(c11(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(mark(x₁)) = [1] + x₁   
POL(nil) = 0   
POL(ok(x₁)) = [1] + x₁   
POL(proper(x₁)) = 0   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(nil) → ok(nil)
proper(0) → ok(0)

Tuples:

FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))
TOP(mark(z0)) → c1(TOP(proper(z0)))

S tuples:

FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH(ok(z0)) → c11(LENGTH(z0))

K tuples:

TOP(mark(z0)) → c1(TOP(proper(z0)))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))

Defined Rule Symbols:

proper

Defined Pair Symbols:

FROM, S, CONS, LENGTH1, LENGTH, TOP

Compound Symbols:

c4, c5, c6, c7, c8, c9, c10, c11, c1

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

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

CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH(ok(z0)) → c11(LENGTH(z0))

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

FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))
TOP(mark(z0)) → c1(TOP(proper(z0)))

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

POL(0) = 0   
POL(CONS(x₁, x₂)) = x₁   
POL(FROM(x₁)) = 0   
POL(LENGTH(x₁)) = x₁   
POL(LENGTH1(x₁)) = 0   
POL(S(x₁)) = x₁   
POL(TOP(x₁)) = 0   
POL(c1(x₁)) = x₁   
POL(c10(x₁)) = x₁   
POL(c11(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(mark(x₁)) = [1] + x₁   
POL(nil) = 0   
POL(ok(x₁)) = [1] + x₁   
POL(proper(x₁)) = 0   

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(nil) → ok(nil)
proper(0) → ok(0)

Tuples:

FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))
TOP(mark(z0)) → c1(TOP(proper(z0)))

S tuples:

FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))

K tuples:

TOP(mark(z0)) → c1(TOP(proper(z0)))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH(ok(z0)) → c11(LENGTH(z0))

Defined Rule Symbols:

proper

Defined Pair Symbols:

FROM, S, CONS, LENGTH1, LENGTH, TOP

Compound Symbols:

c4, c5, c6, c7, c8, c9, c10, c11, c1

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

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

FROM(ok(z0)) → c4(FROM(z0))

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

FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))
TOP(mark(z0)) → c1(TOP(proper(z0)))

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

POL(0) = [3]   
POL(CONS(x₁, x₂)) = [2]x₁   
POL(FROM(x₁)) = [2]x₁   
POL(LENGTH(x₁)) = 0   
POL(LENGTH1(x₁)) = 0   
POL(S(x₁)) = 0   
POL(TOP(x₁)) = 0   
POL(c1(x₁)) = x₁   
POL(c10(x₁)) = x₁   
POL(c11(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(mark(x₁)) = x₁   
POL(nil) = [3]   
POL(ok(x₁)) = [1] + x₁   
POL(proper(x₁)) = [2] + [3]x₁   

(20) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(nil) → ok(nil)
proper(0) → ok(0)

Tuples:

FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))
TOP(mark(z0)) → c1(TOP(proper(z0)))

S tuples:

FROM(mark(z0)) → c5(FROM(z0))

K tuples:

TOP(mark(z0)) → c1(TOP(proper(z0)))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH(ok(z0)) → c11(LENGTH(z0))
FROM(ok(z0)) → c4(FROM(z0))

Defined Rule Symbols:

proper

Defined Pair Symbols:

FROM, S, CONS, LENGTH1, LENGTH, TOP

Compound Symbols:

c4, c5, c6, c7, c8, c9, c10, c11, c1

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

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

FROM(mark(z0)) → c5(FROM(z0))

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

FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))
TOP(mark(z0)) → c1(TOP(proper(z0)))

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

POL(0) = 0   
POL(CONS(x₁, x₂)) = 0   
POL(FROM(x₁)) = x₁   
POL(LENGTH(x₁)) = 0   
POL(LENGTH1(x₁)) = 0   
POL(S(x₁)) = 0   
POL(TOP(x₁)) = 0   
POL(c1(x₁)) = x₁   
POL(c10(x₁)) = x₁   
POL(c11(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(mark(x₁)) = [1] + x₁   
POL(nil) = 0   
POL(ok(x₁)) = x₁   
POL(proper(x₁)) = 0   

(22) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(nil) → ok(nil)
proper(0) → ok(0)

Tuples:

FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
LENGTH(ok(z0)) → c11(LENGTH(z0))
TOP(mark(z0)) → c1(TOP(proper(z0)))

S tuples:none
K tuples:

TOP(mark(z0)) → c1(TOP(proper(z0)))
S(ok(z0)) → c6(S(z0))
S(mark(z0)) → c7(S(z0))
LENGTH1(ok(z0)) → c10(LENGTH1(z0))
CONS(ok(z0), ok(z1)) → c8(CONS(z0, z1))
CONS(mark(z0), z1) → c9(CONS(z0, z1))
LENGTH(ok(z0)) → c11(LENGTH(z0))
FROM(ok(z0)) → c4(FROM(z0))
FROM(mark(z0)) → c5(FROM(z0))

Defined Rule Symbols:

proper

Defined Pair Symbols:

FROM, S, CONS, LENGTH1, LENGTH, TOP

Compound Symbols:

c4, c5, c6, c7, c8, c9, c10, c11, c1

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

The set S is empty

(24) BOUNDS(1, 1)