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

0 CpxTRS
↳1 NestedDefinedSymbolProof (BOTH BOUNDS(ID, ID), 18 ms)
↳2 CpxTRS
    ↳3 CpxTrsMatchBoundsTAProof (⇔, 81 ms)
    ↳4 BOUNDS(1, n^1)
    ↳5 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
    ↳6 CdtProblem
        ↳7 CdtLeafRemovalProof (ComplexityIfPolyImplication, 0 ms)
        ↳8 CdtProblem
        ↳9 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳10 CdtProblem
        ↳11 CdtUsableRulesProof (⇔, 0 ms)
        ↳12 CdtProblem
        ↳13 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 140 ms)
        ↳14 CdtProblem
        ↳15 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 25 ms)
        ↳16 CdtProblem
        ↳17 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 30 ms)
        ↳18 CdtProblem
        ↳19 CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)), 43 ms)
        ↳20 CdtProblem
        ↳21 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 0 ms)
        ↳22 CdtProblem
        ↳23 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 11 ms)
        ↳24 CdtProblem
        ↳25 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 0 ms)
        ↳26 CdtProblem
        ↳27 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 12 ms)
        ↳28 CdtProblem
        ↳29 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳30 BOUNDS(1, 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(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(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(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(length(X)) → length(proper(X))
any(proper(X)) → any(any(any(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:

proper(true) → ok(true)
top(ok(X)) → top(active(X))
proper(nil) → ok(nil)
take(X1, mark(X2)) → mark(take(X1, X2))
take(mark(X1), X2) → mark(take(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
any(X) → s(X)
inf(mark(X)) → mark(inf(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
s(ok(X)) → ok(s(X))
proper(false) → ok(false)
length(mark(X)) → mark(length(X))
proper(0) → ok(0)
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
inf(ok(X)) → ok(inf(X))
length(ok(X)) → ok(length(X))
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, 8, 9]
transitions:
true0() → 0
ok0(0) → 0
active0(0) → 0
nil0() → 0
mark0(0) → 0
false0() → 0
00() → 0
proper0(0) → 1
top0(0) → 2
take0(0, 0) → 3
cons0(0, 0) → 4
any0(0) → 5
inf0(0) → 6
s0(0) → 7
length0(0) → 8
eq0(0, 0) → 9
true1() → 10
ok1(10) → 1
active1(0) → 11
top1(11) → 2
nil1() → 12
ok1(12) → 1
take1(0, 0) → 13
mark1(13) → 3
cons1(0, 0) → 14
ok1(14) → 4
s1(0) → 5
inf1(0) → 15
mark1(15) → 6
take1(0, 0) → 16
ok1(16) → 3
s1(0) → 17
ok1(17) → 7
false1() → 18
ok1(18) → 1
length1(0) → 19
mark1(19) → 8
01() → 20
ok1(20) → 1
eq1(0, 0) → 21
ok1(21) → 9
inf1(0) → 22
ok1(22) → 6
length1(0) → 23
ok1(23) → 8
proper1(0) → 24
top1(24) → 2
ok1(10) → 24
ok1(12) → 24
mark1(13) → 13
mark1(13) → 16
ok1(14) → 14
mark1(15) → 15
mark1(15) → 22
ok1(16) → 13
ok1(16) → 16
ok1(17) → 5
ok1(17) → 17
ok1(18) → 24
mark1(19) → 19
mark1(19) → 23
ok1(20) → 24
ok1(21) → 21
ok1(22) → 15
ok1(22) → 22
ok1(23) → 19
ok1(23) → 23
active2(10) → 25
top2(25) → 2
active2(12) → 25
active2(18) → 25
active2(20) → 25

(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:

proper(true) → ok(true)
proper(nil) → ok(nil)
proper(false) → ok(false)
proper(0) → ok(0)
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
take(z0, mark(z1)) → mark(take(z0, z1))
take(mark(z0), z1) → mark(take(z0, z1))
take(ok(z0), ok(z1)) → ok(take(z0, z1))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
any(z0) → s(z0)
inf(mark(z0)) → mark(inf(z0))
inf(ok(z0)) → ok(inf(z0))
s(ok(z0)) → ok(s(z0))
length(mark(z0)) → mark(length(z0))
length(ok(z0)) → ok(length(z0))
eq(ok(z0), ok(z1)) → ok(eq(z0, z1))
Tuples:

PROPER(true) → c
PROPER(nil) → c1
PROPER(false) → c2
PROPER(0) → c3
TOP(ok(z0)) → c4(TOP(active(z0)))
TOP(mark(z0)) → c5(TOP(proper(z0)), PROPER(z0))
TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
ANY(z0) → c10(S(z0))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
S tuples:

PROPER(true) → c
PROPER(nil) → c1
PROPER(false) → c2
PROPER(0) → c3
TOP(ok(z0)) → c4(TOP(active(z0)))
TOP(mark(z0)) → c5(TOP(proper(z0)), PROPER(z0))
TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
ANY(z0) → c10(S(z0))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
K tuples:none
Defined Rule Symbols:

proper, top, take, cons, any, inf, s, length, eq

Defined Pair Symbols:

PROPER, TOP, TAKE, CONS, ANY, INF, S, LENGTH, EQ

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12, c13, c14, c15, c16

(7) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

ANY(z0) → c10(S(z0))
Removed 5 trailing nodes:

TOP(ok(z0)) → c4(TOP(active(z0)))
PROPER(true) → c
PROPER(false) → c2
PROPER(nil) → c1
PROPER(0) → c3

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(true) → ok(true)
proper(nil) → ok(nil)
proper(false) → ok(false)
proper(0) → ok(0)
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
take(z0, mark(z1)) → mark(take(z0, z1))
take(mark(z0), z1) → mark(take(z0, z1))
take(ok(z0), ok(z1)) → ok(take(z0, z1))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
any(z0) → s(z0)
inf(mark(z0)) → mark(inf(z0))
inf(ok(z0)) → ok(inf(z0))
s(ok(z0)) → ok(s(z0))
length(mark(z0)) → mark(length(z0))
length(ok(z0)) → ok(length(z0))
eq(ok(z0), ok(z1)) → ok(eq(z0, z1))
Tuples:

TOP(mark(z0)) → c5(TOP(proper(z0)), PROPER(z0))
TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
S tuples:

TOP(mark(z0)) → c5(TOP(proper(z0)), PROPER(z0))
TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
K tuples:none
Defined Rule Symbols:

proper, top, take, cons, any, inf, s, length, eq

Defined Pair Symbols:

TOP, TAKE, CONS, INF, S, LENGTH, EQ

Compound Symbols:

c5, c6, c7, c8, c9, c11, c12, c13, c14, c15, c16

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

Removed 1 trailing tuple parts

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(true) → ok(true)
proper(nil) → ok(nil)
proper(false) → ok(false)
proper(0) → ok(0)
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
take(z0, mark(z1)) → mark(take(z0, z1))
take(mark(z0), z1) → mark(take(z0, z1))
take(ok(z0), ok(z1)) → ok(take(z0, z1))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
any(z0) → s(z0)
inf(mark(z0)) → mark(inf(z0))
inf(ok(z0)) → ok(inf(z0))
s(ok(z0)) → ok(s(z0))
length(mark(z0)) → mark(length(z0))
length(ok(z0)) → ok(length(z0))
eq(ok(z0), ok(z1)) → ok(eq(z0, z1))
Tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
S tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
K tuples:none
Defined Rule Symbols:

proper, top, take, cons, any, inf, s, length, eq

Defined Pair Symbols:

TAKE, CONS, INF, S, LENGTH, EQ, TOP

Compound Symbols:

c6, c7, c8, c9, c11, c12, c13, c14, c15, c16, c5

(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))
take(z0, mark(z1)) → mark(take(z0, z1))
take(mark(z0), z1) → mark(take(z0, z1))
take(ok(z0), ok(z1)) → ok(take(z0, z1))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
any(z0) → s(z0)
inf(mark(z0)) → mark(inf(z0))
inf(ok(z0)) → ok(inf(z0))
s(ok(z0)) → ok(s(z0))
length(mark(z0)) → mark(length(z0))
length(ok(z0)) → ok(length(z0))
eq(ok(z0), ok(z1)) → ok(eq(z0, z1))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(true) → ok(true)
proper(nil) → ok(nil)
proper(false) → ok(false)
proper(0) → ok(0)
Tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
S tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
K tuples:none
Defined Rule Symbols:

proper

Defined Pair Symbols:

TAKE, CONS, INF, S, LENGTH, EQ, TOP

Compound Symbols:

c6, c7, c8, c9, c11, c12, c13, c14, c15, c16, c5

(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.

CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [2]   
POL(CONS(x1, x2)) = [2]x1   
POL(EQ(x1, x2)) = [3]x1 + [3]x2   
POL(INF(x1)) = 0   
POL(LENGTH(x1)) = 0   
POL(S(x1)) = 0   
POL(TAKE(x1, x2)) = 0   
POL(TOP(x1)) = 0   
POL(c11(x1)) = x1   
POL(c12(x1)) = x1   
POL(c13(x1)) = x1   
POL(c14(x1)) = x1   
POL(c15(x1)) = x1   
POL(c16(x1)) = x1   
POL(c5(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(false) = [2]   
POL(mark(x1)) = 0   
POL(nil) = [2]   
POL(ok(x1)) = [3] + x1   
POL(proper(x1)) = [2] + [3]x1   
POL(true) = [2]   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(true) → ok(true)
proper(nil) → ok(nil)
proper(false) → ok(false)
proper(0) → ok(0)
Tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
S tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
TOP(mark(z0)) → c5(TOP(proper(z0)))
K tuples:

CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
Defined Rule Symbols:

proper

Defined Pair Symbols:

TAKE, CONS, INF, S, LENGTH, EQ, TOP

Compound Symbols:

c6, c7, c8, c9, c11, c12, c13, c14, c15, c16, c5

(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.

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(CONS(x1, x2)) = [2]x1 + x2   
POL(EQ(x1, x2)) = [3]x2   
POL(INF(x1)) = 0   
POL(LENGTH(x1)) = 0   
POL(S(x1)) = x1   
POL(TAKE(x1, x2)) = x2   
POL(TOP(x1)) = 0   
POL(c11(x1)) = x1   
POL(c12(x1)) = x1   
POL(c13(x1)) = x1   
POL(c14(x1)) = x1   
POL(c15(x1)) = x1   
POL(c16(x1)) = x1   
POL(c5(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(false) = 0   
POL(mark(x1)) = [1] + x1   
POL(nil) = 0   
POL(ok(x1)) = x1   
POL(proper(x1)) = 0   
POL(true) = 0   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(true) → ok(true)
proper(nil) → ok(nil)
proper(false) → ok(false)
proper(0) → ok(0)
Tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
S tuples:

TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
TOP(mark(z0)) → c5(TOP(proper(z0)))
K tuples:

CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
Defined Rule Symbols:

proper

Defined Pair Symbols:

TAKE, CONS, INF, S, LENGTH, EQ, TOP

Compound Symbols:

c6, c7, c8, c9, c11, c12, c13, c14, c15, c16, c5

(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.

TOP(mark(z0)) → c5(TOP(proper(z0)))
We considered the (Usable) Rules:

proper(true) → ok(true)
proper(nil) → ok(nil)
proper(false) → ok(false)
proper(0) → ok(0)
And the Tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(CONS(x1, x2)) = 0   
POL(EQ(x1, x2)) = 0   
POL(INF(x1)) = 0   
POL(LENGTH(x1)) = 0   
POL(S(x1)) = 0   
POL(TAKE(x1, x2)) = 0   
POL(TOP(x1)) = x1   
POL(c11(x1)) = x1   
POL(c12(x1)) = x1   
POL(c13(x1)) = x1   
POL(c14(x1)) = x1   
POL(c15(x1)) = x1   
POL(c16(x1)) = x1   
POL(c5(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(false) = 0   
POL(mark(x1)) = [2]   
POL(nil) = 0   
POL(ok(x1)) = [1]   
POL(proper(x1)) = [1]   
POL(true) = 0   

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(true) → ok(true)
proper(nil) → ok(nil)
proper(false) → ok(false)
proper(0) → ok(0)
Tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
S tuples:

TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
K tuples:

CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
Defined Rule Symbols:

proper

Defined Pair Symbols:

TAKE, CONS, INF, S, LENGTH, EQ, TOP

Compound Symbols:

c6, c7, c8, c9, c11, c12, c13, c14, c15, c16, c5

(19) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

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

TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(CONS(x1, x2)) = x2 + [2]x22   
POL(EQ(x1, x2)) = x1 + x12   
POL(INF(x1)) = 0   
POL(LENGTH(x1)) = 0   
POL(S(x1)) = 0   
POL(TAKE(x1, x2)) = x1   
POL(TOP(x1)) = 0   
POL(c11(x1)) = x1   
POL(c12(x1)) = x1   
POL(c13(x1)) = x1   
POL(c14(x1)) = x1   
POL(c15(x1)) = x1   
POL(c16(x1)) = x1   
POL(c5(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(false) = [1]   
POL(mark(x1)) = x1   
POL(nil) = [2]   
POL(ok(x1)) = [1] + x1   
POL(proper(x1)) = [2] + [2]x1 + [2]x12   
POL(true) = [1]   

(20) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(true) → ok(true)
proper(nil) → ok(nil)
proper(false) → ok(false)
proper(0) → ok(0)
Tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
S tuples:

TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
K tuples:

CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
Defined Rule Symbols:

proper

Defined Pair Symbols:

TAKE, CONS, INF, S, LENGTH, EQ, TOP

Compound Symbols:

c6, c7, c8, c9, c11, c12, c13, c14, c15, c16, c5

(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.

TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(CONS(x1, x2)) = 0   
POL(EQ(x1, x2)) = 0   
POL(INF(x1)) = 0   
POL(LENGTH(x1)) = 0   
POL(S(x1)) = 0   
POL(TAKE(x1, x2)) = x1   
POL(TOP(x1)) = 0   
POL(c11(x1)) = x1   
POL(c12(x1)) = x1   
POL(c13(x1)) = x1   
POL(c14(x1)) = x1   
POL(c15(x1)) = x1   
POL(c16(x1)) = x1   
POL(c5(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(false) = 0   
POL(mark(x1)) = [1] + x1   
POL(nil) = 0   
POL(ok(x1)) = x1   
POL(proper(x1)) = 0   
POL(true) = 0   

(22) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(true) → ok(true)
proper(nil) → ok(nil)
proper(false) → ok(false)
proper(0) → ok(0)
Tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
S tuples:

INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
K tuples:

CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
Defined Rule Symbols:

proper

Defined Pair Symbols:

TAKE, CONS, INF, S, LENGTH, EQ, TOP

Compound Symbols:

c6, c7, c8, c9, c11, c12, c13, c14, c15, c16, c5

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

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

INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
We considered the (Usable) Rules:none
And the Tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(CONS(x1, x2)) = x1   
POL(EQ(x1, x2)) = 0   
POL(INF(x1)) = x1   
POL(LENGTH(x1)) = 0   
POL(S(x1)) = x1   
POL(TAKE(x1, x2)) = 0   
POL(TOP(x1)) = 0   
POL(c11(x1)) = x1   
POL(c12(x1)) = x1   
POL(c13(x1)) = x1   
POL(c14(x1)) = x1   
POL(c15(x1)) = x1   
POL(c16(x1)) = x1   
POL(c5(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(false) = 0   
POL(mark(x1)) = [1] + x1   
POL(nil) = 0   
POL(ok(x1)) = [1] + x1   
POL(proper(x1)) = 0   
POL(true) = 0   

(24) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(true) → ok(true)
proper(nil) → ok(nil)
proper(false) → ok(false)
proper(0) → ok(0)
Tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
S tuples:

LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
K tuples:

CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
Defined Rule Symbols:

proper

Defined Pair Symbols:

TAKE, CONS, INF, S, LENGTH, EQ, TOP

Compound Symbols:

c6, c7, c8, c9, c11, c12, c13, c14, c15, c16, c5

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

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

LENGTH(mark(z0)) → c14(LENGTH(z0))
We considered the (Usable) Rules:none
And the Tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(CONS(x1, x2)) = 0   
POL(EQ(x1, x2)) = [2]x1   
POL(INF(x1)) = 0   
POL(LENGTH(x1)) = [2]x1   
POL(S(x1)) = 0   
POL(TAKE(x1, x2)) = [2]x1   
POL(TOP(x1)) = 0   
POL(c11(x1)) = x1   
POL(c12(x1)) = x1   
POL(c13(x1)) = x1   
POL(c14(x1)) = x1   
POL(c15(x1)) = x1   
POL(c16(x1)) = x1   
POL(c5(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(false) = 0   
POL(mark(x1)) = [2] + x1   
POL(nil) = 0   
POL(ok(x1)) = x1   
POL(proper(x1)) = 0   
POL(true) = 0   

(26) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(true) → ok(true)
proper(nil) → ok(nil)
proper(false) → ok(false)
proper(0) → ok(0)
Tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
S tuples:

LENGTH(ok(z0)) → c15(LENGTH(z0))
K tuples:

CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
Defined Rule Symbols:

proper

Defined Pair Symbols:

TAKE, CONS, INF, S, LENGTH, EQ, TOP

Compound Symbols:

c6, c7, c8, c9, c11, c12, c13, c14, c15, c16, c5

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

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

LENGTH(ok(z0)) → c15(LENGTH(z0))
We considered the (Usable) Rules:none
And the Tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(CONS(x1, x2)) = 0   
POL(EQ(x1, x2)) = 0   
POL(INF(x1)) = 0   
POL(LENGTH(x1)) = x1   
POL(S(x1)) = 0   
POL(TAKE(x1, x2)) = 0   
POL(TOP(x1)) = 0   
POL(c11(x1)) = x1   
POL(c12(x1)) = x1   
POL(c13(x1)) = x1   
POL(c14(x1)) = x1   
POL(c15(x1)) = x1   
POL(c16(x1)) = x1   
POL(c5(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(false) = 0   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(ok(x1)) = [1] + x1   
POL(proper(x1)) = 0   
POL(true) = 0   

(28) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(true) → ok(true)
proper(nil) → ok(nil)
proper(false) → ok(false)
proper(0) → ok(0)
Tuples:

TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
S tuples:none
K tuples:

CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
EQ(ok(z0), ok(z1)) → c16(EQ(z0, z1))
TAKE(z0, mark(z1)) → c6(TAKE(z0, z1))
TOP(mark(z0)) → c5(TOP(proper(z0)))
TAKE(ok(z0), ok(z1)) → c8(TAKE(z0, z1))
TAKE(mark(z0), z1) → c7(TAKE(z0, z1))
INF(mark(z0)) → c11(INF(z0))
INF(ok(z0)) → c12(INF(z0))
S(ok(z0)) → c13(S(z0))
LENGTH(mark(z0)) → c14(LENGTH(z0))
LENGTH(ok(z0)) → c15(LENGTH(z0))
Defined Rule Symbols:

proper

Defined Pair Symbols:

TAKE, CONS, INF, S, LENGTH, EQ, TOP

Compound Symbols:

c6, c7, c8, c9, c11, c12, c13, c14, c15, c16, c5

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

The set S is empty

(30) BOUNDS(1, 1)