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), 0 ms)
↳2 CpxTRS
    ↳3 CpxTrsMatchBoundsTAProof (⇔, 115 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 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳10 CdtProblem
        ↳11 CdtUsableRulesProof (⇔, 0 ms)
        ↳12 CdtProblem
        ↳13 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 131 ms)
        ↳14 CdtProblem
        ↳15 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 44 ms)
        ↳16 CdtProblem
        ↳17 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 0 ms)
        ↳18 CdtProblem
        ↳19 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 0 ms)
        ↳20 CdtProblem
        ↳21 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 15 ms)
        ↳22 CdtProblem
        ↳23 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 0 ms)
        ↳24 CdtProblem
        ↳25 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 0 ms)
        ↳26 CdtProblem
        ↳27 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 18 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(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(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:

proper(true) → ok(true)
add(ok(X1), ok(X2)) → ok(add(X1, X2))
top(ok(X)) → top(active(X))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
s(ok(X)) → ok(s(X))
proper(false) → ok(false)
proper(0) → ok(0)
first(X1, mark(X2)) → mark(first(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
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
add0(0, 0) → 2
top0(0) → 3
from0(0) → 4
cons0(0, 0) → 5
and0(0, 0) → 6
first0(0, 0) → 7
s0(0) → 8
if0(0, 0, 0) → 9
true1() → 10
ok1(10) → 1
add1(0, 0) → 11
ok1(11) → 2
active1(0) → 12
top1(12) → 3
nil1() → 13
ok1(13) → 1
from1(0) → 14
ok1(14) → 4
cons1(0, 0) → 15
ok1(15) → 5
and1(0, 0) → 16
mark1(16) → 6
first1(0, 0) → 17
ok1(17) → 7
first1(0, 0) → 18
mark1(18) → 7
and1(0, 0) → 19
ok1(19) → 6
s1(0) → 20
ok1(20) → 8
false1() → 21
ok1(21) → 1
01() → 22
ok1(22) → 1
add1(0, 0) → 23
mark1(23) → 2
if1(0, 0, 0) → 24
mark1(24) → 9
if1(0, 0, 0) → 25
ok1(25) → 9
proper1(0) → 26
top1(26) → 3
ok1(10) → 26
ok1(11) → 11
ok1(11) → 23
ok1(13) → 26
ok1(14) → 14
ok1(15) → 15
mark1(16) → 16
mark1(16) → 19
ok1(17) → 17
ok1(17) → 18
mark1(18) → 17
mark1(18) → 18
ok1(19) → 16
ok1(19) → 19
ok1(20) → 20
ok1(21) → 26
ok1(22) → 26
mark1(23) → 11
mark1(23) → 23
mark1(24) → 24
mark1(24) → 25
ok1(25) → 24
ok1(25) → 25
active2(10) → 27
top2(27) → 3
active2(13) → 27
active2(21) → 27
active2(22) → 27

(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)
add(ok(z0), ok(z1)) → ok(add(z0, z1))
add(mark(z0), z1) → mark(add(z0, z1))
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
from(ok(z0)) → ok(from(z0))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
and(mark(z0), z1) → mark(and(z0, z1))
and(ok(z0), ok(z1)) → ok(and(z0, z1))
first(ok(z0), ok(z1)) → ok(first(z0, z1))
first(mark(z0), z1) → mark(first(z0, z1))
first(z0, mark(z1)) → mark(first(z0, z1))
s(ok(z0)) → ok(s(z0))
if(mark(z0), z1, z2) → mark(if(z0, z1, z2))
if(ok(z0), ok(z1), ok(z2)) → ok(if(z0, z1, z2))
Tuples:

PROPER(true) → c
PROPER(nil) → c1
PROPER(false) → c2
PROPER(0) → c3
ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
TOP(ok(z0)) → c6(TOP(active(z0)))
TOP(mark(z0)) → c7(TOP(proper(z0)), PROPER(z0))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
S tuples:

PROPER(true) → c
PROPER(nil) → c1
PROPER(false) → c2
PROPER(0) → c3
ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
TOP(ok(z0)) → c6(TOP(active(z0)))
TOP(mark(z0)) → c7(TOP(proper(z0)), PROPER(z0))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
K tuples:none
Defined Rule Symbols:

proper, add, top, from, cons, and, first, s, if

Defined Pair Symbols:

PROPER, ADD, TOP, FROM, CONS, AND, FIRST, S, IF

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12, c13, c14, c15, c16, c17

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

Removed 5 trailing nodes:

PROPER(0) → c3
PROPER(nil) → c1
PROPER(false) → c2
TOP(ok(z0)) → c6(TOP(active(z0)))
PROPER(true) → c

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(true) → ok(true)
proper(nil) → ok(nil)
proper(false) → ok(false)
proper(0) → ok(0)
add(ok(z0), ok(z1)) → ok(add(z0, z1))
add(mark(z0), z1) → mark(add(z0, z1))
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
from(ok(z0)) → ok(from(z0))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
and(mark(z0), z1) → mark(and(z0, z1))
and(ok(z0), ok(z1)) → ok(and(z0, z1))
first(ok(z0), ok(z1)) → ok(first(z0, z1))
first(mark(z0), z1) → mark(first(z0, z1))
first(z0, mark(z1)) → mark(first(z0, z1))
s(ok(z0)) → ok(s(z0))
if(mark(z0), z1, z2) → mark(if(z0, z1, z2))
if(ok(z0), ok(z1), ok(z2)) → ok(if(z0, z1, z2))
Tuples:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
TOP(mark(z0)) → c7(TOP(proper(z0)), PROPER(z0))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
S tuples:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
TOP(mark(z0)) → c7(TOP(proper(z0)), PROPER(z0))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
K tuples:none
Defined Rule Symbols:

proper, add, top, from, cons, and, first, s, if

Defined Pair Symbols:

ADD, TOP, FROM, CONS, AND, FIRST, S, IF

Compound Symbols:

c4, c5, c7, c8, c9, c10, c11, c12, c13, c14, c15, c16, c17

(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)
add(ok(z0), ok(z1)) → ok(add(z0, z1))
add(mark(z0), z1) → mark(add(z0, z1))
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
from(ok(z0)) → ok(from(z0))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
and(mark(z0), z1) → mark(and(z0, z1))
and(ok(z0), ok(z1)) → ok(and(z0, z1))
first(ok(z0), ok(z1)) → ok(first(z0, z1))
first(mark(z0), z1) → mark(first(z0, z1))
first(z0, mark(z1)) → mark(first(z0, z1))
s(ok(z0)) → ok(s(z0))
if(mark(z0), z1, z2) → mark(if(z0, z1, z2))
if(ok(z0), ok(z1), ok(z2)) → ok(if(z0, z1, z2))
Tuples:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
S tuples:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
K tuples:none
Defined Rule Symbols:

proper, add, top, from, cons, and, first, s, if

Defined Pair Symbols:

ADD, FROM, CONS, AND, FIRST, S, IF, TOP

Compound Symbols:

c4, c5, c8, c9, c10, c11, c12, c13, c14, c15, c16, c17, c7

(11) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

add(ok(z0), ok(z1)) → ok(add(z0, z1))
add(mark(z0), z1) → mark(add(z0, z1))
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
from(ok(z0)) → ok(from(z0))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
and(mark(z0), z1) → mark(and(z0, z1))
and(ok(z0), ok(z1)) → ok(and(z0, z1))
first(ok(z0), ok(z1)) → ok(first(z0, z1))
first(mark(z0), z1) → mark(first(z0, z1))
first(z0, mark(z1)) → mark(first(z0, z1))
s(ok(z0)) → ok(s(z0))
if(mark(z0), z1, z2) → mark(if(z0, z1, z2))
if(ok(z0), ok(z1), ok(z2)) → ok(if(z0, z1, z2))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
S tuples:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
K tuples:none
Defined Rule Symbols:

proper

Defined Pair Symbols:

ADD, FROM, CONS, AND, FIRST, S, IF, TOP

Compound Symbols:

c4, c5, c8, c9, c10, c11, c12, c13, c14, c15, c16, c17, c7

(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.

S(ok(z0)) → c15(S(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:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ADD(x1, x2)) = 0   
POL(AND(x1, x2)) = 0   
POL(CONS(x1, x2)) = 0   
POL(FIRST(x1, x2)) = 0   
POL(FROM(x1)) = 0   
POL(IF(x1, x2, x3)) = 0   
POL(S(x1)) = [2]x1   
POL(TOP(x1)) = [2]x1   
POL(c10(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(c17(x1)) = x1   
POL(c4(x1)) = x1   
POL(c5(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(false) = [1]   
POL(mark(x1)) = [2] + x1   
POL(nil) = 0   
POL(ok(x1)) = [1] + x1   
POL(proper(x1)) = [2]   
POL(true) = 0   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
S tuples:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
K tuples:

S(ok(z0)) → c15(S(z0))
Defined Rule Symbols:

proper

Defined Pair Symbols:

ADD, FROM, CONS, AND, FIRST, S, IF, TOP

Compound Symbols:

c4, c5, c8, c9, c10, c11, c12, c13, c14, c15, c16, c17, c7

(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.

TOP(mark(z0)) → c7(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:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ADD(x1, x2)) = 0   
POL(AND(x1, x2)) = 0   
POL(CONS(x1, x2)) = 0   
POL(FIRST(x1, x2)) = 0   
POL(FROM(x1)) = 0   
POL(IF(x1, x2, x3)) = 0   
POL(S(x1)) = 0   
POL(TOP(x1)) = x1   
POL(c10(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(c17(x1)) = x1   
POL(c4(x1)) = x1   
POL(c5(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)) = 0   
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:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
S tuples:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
K tuples:

S(ok(z0)) → c15(S(z0))
TOP(mark(z0)) → c7(TOP(proper(z0)))
Defined Rule Symbols:

proper

Defined Pair Symbols:

ADD, FROM, CONS, AND, FIRST, S, IF, TOP

Compound Symbols:

c4, c5, c8, c9, c10, c11, c12, c13, c14, c15, c16, c17, c7

(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)) → c9(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [1]   
POL(ADD(x1, x2)) = 0   
POL(AND(x1, x2)) = 0   
POL(CONS(x1, x2)) = x2   
POL(FIRST(x1, x2)) = [2]x1 + [2]x2   
POL(FROM(x1)) = 0   
POL(IF(x1, x2, x3)) = 0   
POL(S(x1)) = x1   
POL(TOP(x1)) = 0   
POL(c10(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(c17(x1)) = x1   
POL(c4(x1)) = x1   
POL(c5(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(false) = [3]   
POL(mark(x1)) = [1] + x1   
POL(nil) = [2]   
POL(ok(x1)) = [2] + x1   
POL(proper(x1)) = [3]x1   
POL(true) = [2]   

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
S tuples:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
K tuples:

S(ok(z0)) → c15(S(z0))
TOP(mark(z0)) → c7(TOP(proper(z0)))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
Defined Rule Symbols:

proper

Defined Pair Symbols:

ADD, FROM, CONS, AND, FIRST, S, IF, TOP

Compound Symbols:

c4, c5, c8, c9, c10, c11, c12, c13, c14, c15, c16, c17, c7

(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)) → c8(FROM(z0))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ADD(x1, x2)) = 0   
POL(AND(x1, x2)) = x1   
POL(CONS(x1, x2)) = x2   
POL(FIRST(x1, x2)) = 0   
POL(FROM(x1)) = x1   
POL(IF(x1, x2, x3)) = 0   
POL(S(x1)) = 0   
POL(TOP(x1)) = 0   
POL(c10(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(c17(x1)) = x1   
POL(c4(x1)) = x1   
POL(c5(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   

(20) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
S tuples:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
K tuples:

S(ok(z0)) → c15(S(z0))
TOP(mark(z0)) → c7(TOP(proper(z0)))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
Defined Rule Symbols:

proper

Defined Pair Symbols:

ADD, FROM, CONS, AND, FIRST, S, IF, TOP

Compound Symbols:

c4, c5, c8, c9, c10, c11, c12, c13, c14, c15, c16, c17, c7

(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.

IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
We considered the (Usable) Rules:none
And the Tuples:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ADD(x1, x2)) = 0   
POL(AND(x1, x2)) = x2   
POL(CONS(x1, x2)) = 0   
POL(FIRST(x1, x2)) = 0   
POL(FROM(x1)) = x1   
POL(IF(x1, x2, x3)) = x3   
POL(S(x1)) = 0   
POL(TOP(x1)) = 0   
POL(c10(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(c17(x1)) = x1   
POL(c4(x1)) = x1   
POL(c5(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(false) = 0   
POL(mark(x1)) = 0   
POL(nil) = 0   
POL(ok(x1)) = [1] + 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:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
S tuples:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
K tuples:

S(ok(z0)) → c15(S(z0))
TOP(mark(z0)) → c7(TOP(proper(z0)))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
Defined Rule Symbols:

proper

Defined Pair Symbols:

ADD, FROM, CONS, AND, FIRST, S, IF, TOP

Compound Symbols:

c4, c5, c8, c9, c10, c11, c12, c13, c14, c15, c16, c17, c7

(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.

IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
We considered the (Usable) Rules:none
And the Tuples:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ADD(x1, x2)) = 0   
POL(AND(x1, x2)) = 0   
POL(CONS(x1, x2)) = 0   
POL(FIRST(x1, x2)) = 0   
POL(FROM(x1)) = 0   
POL(IF(x1, x2, x3)) = [2]x1 + [2]x3   
POL(S(x1)) = [3]x1   
POL(TOP(x1)) = 0   
POL(c10(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(c17(x1)) = x1   
POL(c4(x1)) = x1   
POL(c5(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   

(24) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
S tuples:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
K tuples:

S(ok(z0)) → c15(S(z0))
TOP(mark(z0)) → c7(TOP(proper(z0)))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
Defined Rule Symbols:

proper

Defined Pair Symbols:

ADD, FROM, CONS, AND, FIRST, S, IF, TOP

Compound Symbols:

c4, c5, c8, c9, c10, c11, c12, c13, c14, c15, c16, c17, c7

(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.

ADD(mark(z0), z1) → c5(ADD(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ADD(x1, x2)) = x1   
POL(AND(x1, x2)) = 0   
POL(CONS(x1, x2)) = 0   
POL(FIRST(x1, x2)) = 0   
POL(FROM(x1)) = x1   
POL(IF(x1, x2, x3)) = 0   
POL(S(x1)) = x1   
POL(TOP(x1)) = 0   
POL(c10(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(c17(x1)) = x1   
POL(c4(x1)) = x1   
POL(c5(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   

(26) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
S tuples:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
K tuples:

S(ok(z0)) → c15(S(z0))
TOP(mark(z0)) → c7(TOP(proper(z0)))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
Defined Rule Symbols:

proper

Defined Pair Symbols:

ADD, FROM, CONS, AND, FIRST, S, IF, TOP

Compound Symbols:

c4, c5, c8, c9, c10, c11, c12, c13, c14, c15, c16, c17, c7

(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.

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ADD(x1, x2)) = x1 + x2   
POL(AND(x1, x2)) = 0   
POL(CONS(x1, x2)) = 0   
POL(FIRST(x1, x2)) = 0   
POL(FROM(x1)) = 0   
POL(IF(x1, x2, x3)) = 0   
POL(S(x1)) = 0   
POL(TOP(x1)) = 0   
POL(c10(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(c17(x1)) = x1   
POL(c4(x1)) = x1   
POL(c5(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:

ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
S(ok(z0)) → c15(S(z0))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
TOP(mark(z0)) → c7(TOP(proper(z0)))
S tuples:none
K tuples:

S(ok(z0)) → c15(S(z0))
TOP(mark(z0)) → c7(TOP(proper(z0)))
CONS(ok(z0), ok(z1)) → c9(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c12(FIRST(z0, z1))
FIRST(mark(z0), z1) → c13(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c14(FIRST(z0, z1))
FROM(ok(z0)) → c8(FROM(z0))
AND(mark(z0), z1) → c10(AND(z0, z1))
AND(ok(z0), ok(z1)) → c11(AND(z0, z1))
IF(ok(z0), ok(z1), ok(z2)) → c17(IF(z0, z1, z2))
IF(mark(z0), z1, z2) → c16(IF(z0, z1, z2))
ADD(mark(z0), z1) → c5(ADD(z0, z1))
ADD(ok(z0), ok(z1)) → c4(ADD(z0, z1))
Defined Rule Symbols:

proper

Defined Pair Symbols:

ADD, FROM, CONS, AND, FIRST, S, IF, TOP

Compound Symbols:

c4, c5, c8, c9, c10, c11, c12, c13, c14, c15, c16, c17, c7

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

The set S is empty

(30) BOUNDS(1, 1)