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), 10 ms)
↳2 CpxTRS
    ↳3 CpxTrsMatchBoundsTAProof (⇔, 80 ms)
    ↳4 BOUNDS(1, n^1)
    ↳5 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
    ↳6 CdtProblem
        ↳7 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳8 CdtProblem
        ↳9 CdtUsableRulesProof (⇔, 0 ms)
        ↳10 CdtProblem
        ↳11 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 78 ms)
        ↳12 CdtProblem
        ↳13 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 3 ms)
        ↳14 CdtProblem
        ↳15 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 11 ms)
        ↳16 CdtProblem
        ↳17 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 23 ms)
        ↳18 CdtProblem
        ↳19 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 0 ms)
        ↳20 CdtProblem
        ↳21 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳22 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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))

(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))
from(ok(X)) → ok(from(X))
from(mark(X)) → mark(from(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
2nd(mark(X)) → mark(2nd(X))
2nd(ok(X)) → ok(2nd(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
s(ok(X)) → ok(s(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(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 1.

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]
transitions:
ok0(0) → 0
active0(0) → 0
mark0(0) → 0
proper0(0) → 0
top0(0) → 1
from0(0) → 2
cons10(0, 0) → 3
cons0(0, 0) → 4
2nd0(0) → 5
s0(0) → 6
active1(0) → 7
top1(7) → 1
from1(0) → 8
ok1(8) → 2
from1(0) → 9
mark1(9) → 2
cons11(0, 0) → 10
ok1(10) → 3
cons1(0, 0) → 11
ok1(11) → 4
2nd1(0) → 12
mark1(12) → 5
2nd1(0) → 13
ok1(13) → 5
cons11(0, 0) → 14
mark1(14) → 3
s1(0) → 15
ok1(15) → 6
s1(0) → 16
mark1(16) → 6
cons1(0, 0) → 17
mark1(17) → 4
proper1(0) → 18
top1(18) → 1
ok1(8) → 8
ok1(8) → 9
mark1(9) → 8
mark1(9) → 9
ok1(10) → 10
ok1(10) → 14
ok1(11) → 11
ok1(11) → 17
mark1(12) → 12
mark1(12) → 13
ok1(13) → 12
ok1(13) → 13
mark1(14) → 10
mark1(14) → 14
ok1(15) → 15
ok1(15) → 16
mark1(16) → 15
mark1(16) → 16
mark1(17) → 11
mark1(17) → 17

(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))
from(ok(z0)) → ok(from(z0))
from(mark(z0)) → mark(from(z0))
cons1(ok(z0), ok(z1)) → ok(cons1(z0, z1))
cons1(mark(z0), z1) → mark(cons1(z0, z1))
cons1(z0, mark(z1)) → mark(cons1(z0, z1))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
cons(mark(z0), z1) → mark(cons(z0, z1))
2nd(mark(z0)) → mark(2nd(z0))
2nd(ok(z0)) → ok(2nd(z0))
s(ok(z0)) → ok(s(z0))
s(mark(z0)) → mark(s(z0))
Tuples:

TOP(ok(z0)) → c(TOP(active(z0)))
TOP(mark(z0)) → c1(TOP(proper(z0)))
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
2ND(mark(z0)) → c9(2ND(z0))
2ND(ok(z0)) → c10(2ND(z0))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
S tuples:

TOP(ok(z0)) → c(TOP(active(z0)))
TOP(mark(z0)) → c1(TOP(proper(z0)))
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
2ND(mark(z0)) → c9(2ND(z0))
2ND(ok(z0)) → c10(2ND(z0))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
K tuples:none
Defined Rule Symbols:

top, from, cons1, cons, 2nd, s

Defined Pair Symbols:

TOP, FROM, CONS1, CONS, 2ND, S

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12

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

Removed 2 trailing nodes:

TOP(ok(z0)) → c(TOP(active(z0)))
TOP(mark(z0)) → c1(TOP(proper(z0)))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
from(ok(z0)) → ok(from(z0))
from(mark(z0)) → mark(from(z0))
cons1(ok(z0), ok(z1)) → ok(cons1(z0, z1))
cons1(mark(z0), z1) → mark(cons1(z0, z1))
cons1(z0, mark(z1)) → mark(cons1(z0, z1))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
cons(mark(z0), z1) → mark(cons(z0, z1))
2nd(mark(z0)) → mark(2nd(z0))
2nd(ok(z0)) → ok(2nd(z0))
s(ok(z0)) → ok(s(z0))
s(mark(z0)) → mark(s(z0))
Tuples:

FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
2ND(mark(z0)) → c9(2ND(z0))
2ND(ok(z0)) → c10(2ND(z0))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
S tuples:

FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
2ND(mark(z0)) → c9(2ND(z0))
2ND(ok(z0)) → c10(2ND(z0))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
K tuples:none
Defined Rule Symbols:

top, from, cons1, cons, 2nd, s

Defined Pair Symbols:

FROM, CONS1, CONS, 2ND, S

Compound Symbols:

c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12

(9) 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))
cons1(ok(z0), ok(z1)) → ok(cons1(z0, z1))
cons1(mark(z0), z1) → mark(cons1(z0, z1))
cons1(z0, mark(z1)) → mark(cons1(z0, z1))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
cons(mark(z0), z1) → mark(cons(z0, z1))
2nd(mark(z0)) → mark(2nd(z0))
2nd(ok(z0)) → ok(2nd(z0))
s(ok(z0)) → ok(s(z0))
s(mark(z0)) → mark(s(z0))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
2ND(mark(z0)) → c9(2ND(z0))
2ND(ok(z0)) → c10(2ND(z0))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
S tuples:

FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
2ND(mark(z0)) → c9(2ND(z0))
2ND(ok(z0)) → c10(2ND(z0))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

FROM, CONS1, CONS, 2ND, S

Compound Symbols:

c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12

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

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

CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
We considered the (Usable) Rules:none
And the Tuples:

FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
2ND(mark(z0)) → c9(2ND(z0))
2ND(ok(z0)) → c10(2ND(z0))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(2ND(x1)) = 0   
POL(CONS(x1, x2)) = 0   
POL(CONS1(x1, x2)) = x1   
POL(FROM(x1)) = 0   
POL(S(x1)) = [2]x1   
POL(c10(x1)) = x1   
POL(c11(x1)) = x1   
POL(c12(x1)) = x1   
POL(c2(x1)) = x1   
POL(c3(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)) = [2] + x1   
POL(ok(x1)) = [1] + x1   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
2ND(mark(z0)) → c9(2ND(z0))
2ND(ok(z0)) → c10(2ND(z0))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
S tuples:

FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
2ND(mark(z0)) → c9(2ND(z0))
2ND(ok(z0)) → c10(2ND(z0))
K tuples:

CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

FROM, CONS1, CONS, 2ND, S

Compound Symbols:

c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12

(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)) → c7(CONS(z0, z1))
2ND(ok(z0)) → c10(2ND(z0))
We considered the (Usable) Rules:none
And the Tuples:

FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
2ND(mark(z0)) → c9(2ND(z0))
2ND(ok(z0)) → c10(2ND(z0))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(2ND(x1)) = x1   
POL(CONS(x1, x2)) = x2   
POL(CONS1(x1, x2)) = 0   
POL(FROM(x1)) = 0   
POL(S(x1)) = x1   
POL(c10(x1)) = x1   
POL(c11(x1)) = x1   
POL(c12(x1)) = x1   
POL(c2(x1)) = x1   
POL(c3(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(ok(x1)) = [1] + x1   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
2ND(mark(z0)) → c9(2ND(z0))
2ND(ok(z0)) → c10(2ND(z0))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
S tuples:

FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
2ND(mark(z0)) → c9(2ND(z0))
K tuples:

CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
2ND(ok(z0)) → c10(2ND(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

FROM, CONS1, CONS, 2ND, S

Compound Symbols:

c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12

(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.

2ND(mark(z0)) → c9(2ND(z0))
We considered the (Usable) Rules:none
And the Tuples:

FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
2ND(mark(z0)) → c9(2ND(z0))
2ND(ok(z0)) → c10(2ND(z0))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(2ND(x1)) = x1   
POL(CONS(x1, x2)) = 0   
POL(CONS1(x1, x2)) = 0   
POL(FROM(x1)) = 0   
POL(S(x1)) = 0   
POL(c10(x1)) = x1   
POL(c11(x1)) = x1   
POL(c12(x1)) = x1   
POL(c2(x1)) = x1   
POL(c3(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(ok(x1)) = x1   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
2ND(mark(z0)) → c9(2ND(z0))
2ND(ok(z0)) → c10(2ND(z0))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
S tuples:

FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
K tuples:

CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
2ND(ok(z0)) → c10(2ND(z0))
2ND(mark(z0)) → c9(2ND(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

FROM, CONS1, CONS, 2ND, S

Compound Symbols:

c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12

(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.

FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
2ND(mark(z0)) → c9(2ND(z0))
2ND(ok(z0)) → c10(2ND(z0))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(2ND(x1)) = 0   
POL(CONS(x1, x2)) = x1   
POL(CONS1(x1, x2)) = 0   
POL(FROM(x1)) = x1   
POL(S(x1)) = x1   
POL(c10(x1)) = x1   
POL(c11(x1)) = x1   
POL(c12(x1)) = x1   
POL(c2(x1)) = x1   
POL(c3(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(ok(x1)) = [1] + x1   

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
2ND(mark(z0)) → c9(2ND(z0))
2ND(ok(z0)) → c10(2ND(z0))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
S tuples:

CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
K tuples:

CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
2ND(ok(z0)) → c10(2ND(z0))
2ND(mark(z0)) → c9(2ND(z0))
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
Defined Rule Symbols:none

Defined Pair Symbols:

FROM, CONS1, CONS, 2ND, S

Compound Symbols:

c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12

(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.

CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
2ND(mark(z0)) → c9(2ND(z0))
2ND(ok(z0)) → c10(2ND(z0))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(2ND(x1)) = 0   
POL(CONS(x1, x2)) = 0   
POL(CONS1(x1, x2)) = x2   
POL(FROM(x1)) = 0   
POL(S(x1)) = 0   
POL(c10(x1)) = x1   
POL(c11(x1)) = x1   
POL(c12(x1)) = x1   
POL(c2(x1)) = x1   
POL(c3(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(ok(x1)) = x1   

(20) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
2ND(mark(z0)) → c9(2ND(z0))
2ND(ok(z0)) → c10(2ND(z0))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
S tuples:none
K tuples:

CONS1(ok(z0), ok(z1)) → c4(CONS1(z0, z1))
CONS1(mark(z0), z1) → c5(CONS1(z0, z1))
S(ok(z0)) → c11(S(z0))
S(mark(z0)) → c12(S(z0))
CONS(ok(z0), ok(z1)) → c7(CONS(z0, z1))
2ND(ok(z0)) → c10(2ND(z0))
2ND(mark(z0)) → c9(2ND(z0))
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
CONS(mark(z0), z1) → c8(CONS(z0, z1))
CONS1(z0, mark(z1)) → c6(CONS1(z0, z1))
Defined Rule Symbols:none

Defined Pair Symbols:

FROM, CONS1, CONS, 2ND, S

Compound Symbols:

c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12

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

The set S is empty

(22) BOUNDS(1, 1)