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), 7 ms)
↳2 CpxTRS
    ↳3 CpxTrsMatchBoundsTAProof (⇔, 169 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)), 137 ms)
        ↳14 CdtProblem
        ↳15 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 50 ms)
        ↳16 CdtProblem
        ↳17 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 29 ms)
        ↳18 CdtProblem
        ↳19 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 0 ms)
        ↳20 CdtProblem
        ↳21 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 5 ms)
        ↳22 CdtProblem
        ↳23 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 28 ms)
        ↳24 CdtProblem
        ↳25 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 28 ms)
        ↳26 CdtProblem
        ↳27 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 15 ms)
        ↳28 CdtProblem
        ↳29 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 21 ms)
        ↳30 CdtProblem
        ↳31 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 29 ms)
        ↳32 CdtProblem
        ↳33 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳34 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(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
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(terms(X)) → terms(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(recip(X)) → recip(active(X))
active(sqr(X)) → sqr(active(X))
active(s(X)) → s(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(dbl(X)) → dbl(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
terms(mark(X)) → mark(terms(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
recip(mark(X)) → mark(recip(X))
sqr(mark(X)) → mark(sqr(X))
s(mark(X)) → mark(s(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
dbl(mark(X)) → mark(dbl(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(terms(X)) → terms(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(recip(X)) → recip(proper(X))
proper(sqr(X)) → sqr(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(dbl(X)) → dbl(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
terms(ok(X)) → ok(terms(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
recip(ok(X)) → ok(recip(X))
sqr(ok(X)) → ok(sqr(X))
s(ok(X)) → ok(s(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
dbl(ok(X)) → ok(dbl(X))
first(ok(X1), ok(X2)) → ok(first(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(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
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(terms(X)) → terms(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(recip(X)) → recip(active(X))
active(sqr(X)) → sqr(active(X))
active(s(X)) → s(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(dbl(X)) → dbl(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
proper(terms(X)) → terms(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(recip(X)) → recip(proper(X))
proper(sqr(X)) → sqr(proper(X))
proper(s(X)) → s(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(dbl(X)) → dbl(proper(X))
proper(first(X1, X2)) → first(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:

sqr(ok(X)) → ok(sqr(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
top(ok(X)) → top(active(X))
proper(nil) → ok(nil)
terms(ok(X)) → ok(terms(X))
dbl(mark(X)) → mark(dbl(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
terms(mark(X)) → mark(terms(X))
first(mark(X1), X2) → mark(first(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
s(ok(X)) → ok(s(X))
s(mark(X)) → mark(s(X))
recip(ok(X)) → ok(recip(X))
recip(mark(X)) → mark(recip(X))
proper(0) → ok(0)
dbl(ok(X)) → ok(dbl(X))
first(X1, mark(X2)) → mark(first(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
sqr(mark(X)) → mark(sqr(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, 8, 9, 10]
transitions:
ok0(0) → 0
active0(0) → 0
nil0() → 0
mark0(0) → 0
00() → 0
sqr0(0) → 1
add0(0, 0) → 2
top0(0) → 3
proper0(0) → 4
terms0(0) → 5
dbl0(0) → 6
cons0(0, 0) → 7
first0(0, 0) → 8
s0(0) → 9
recip0(0) → 10
sqr1(0) → 11
ok1(11) → 1
add1(0, 0) → 12
ok1(12) → 2
active1(0) → 13
top1(13) → 3
nil1() → 14
ok1(14) → 4
terms1(0) → 15
ok1(15) → 5
dbl1(0) → 16
mark1(16) → 6
cons1(0, 0) → 17
ok1(17) → 7
first1(0, 0) → 18
ok1(18) → 8
terms1(0) → 19
mark1(19) → 5
first1(0, 0) → 20
mark1(20) → 8
add1(0, 0) → 21
mark1(21) → 2
s1(0) → 22
ok1(22) → 9
s1(0) → 23
mark1(23) → 9
recip1(0) → 24
ok1(24) → 10
recip1(0) → 25
mark1(25) → 10
01() → 26
ok1(26) → 4
dbl1(0) → 27
ok1(27) → 6
sqr1(0) → 28
mark1(28) → 1
cons1(0, 0) → 29
mark1(29) → 7
proper1(0) → 30
top1(30) → 3
ok1(11) → 11
ok1(11) → 28
ok1(12) → 12
ok1(12) → 21
ok1(14) → 30
ok1(15) → 15
ok1(15) → 19
mark1(16) → 16
mark1(16) → 27
ok1(17) → 17
ok1(17) → 29
ok1(18) → 18
ok1(18) → 20
mark1(19) → 15
mark1(19) → 19
mark1(20) → 18
mark1(20) → 20
mark1(21) → 12
mark1(21) → 21
ok1(22) → 22
ok1(22) → 23
mark1(23) → 22
mark1(23) → 23
ok1(24) → 24
ok1(24) → 25
mark1(25) → 24
mark1(25) → 25
ok1(26) → 30
ok1(27) → 16
ok1(27) → 27
mark1(28) → 11
mark1(28) → 28
mark1(29) → 17
mark1(29) → 29
active2(14) → 31
top2(31) → 3
active2(26) → 31

(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:

sqr(ok(z0)) → ok(sqr(z0))
sqr(mark(z0)) → mark(sqr(z0))
add(ok(z0), ok(z1)) → ok(add(z0, z1))
add(z0, mark(z1)) → mark(add(z0, z1))
add(mark(z0), z1) → mark(add(z0, z1))
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
proper(nil) → ok(nil)
proper(0) → ok(0)
terms(ok(z0)) → ok(terms(z0))
terms(mark(z0)) → mark(terms(z0))
dbl(mark(z0)) → mark(dbl(z0))
dbl(ok(z0)) → ok(dbl(z0))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
cons(mark(z0), z1) → mark(cons(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))
s(mark(z0)) → mark(s(z0))
recip(ok(z0)) → ok(recip(z0))
recip(mark(z0)) → mark(recip(z0))
Tuples:

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TOP(ok(z0)) → c5(TOP(active(z0)))
TOP(mark(z0)) → c6(TOP(proper(z0)), PROPER(z0))
PROPER(nil) → c7
PROPER(0) → c8
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
S tuples:

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TOP(ok(z0)) → c5(TOP(active(z0)))
TOP(mark(z0)) → c6(TOP(proper(z0)), PROPER(z0))
PROPER(nil) → c7
PROPER(0) → c8
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
K tuples:none
Defined Rule Symbols:

sqr, add, top, proper, terms, dbl, cons, first, s, recip

Defined Pair Symbols:

SQR, ADD, TOP, PROPER, TERMS, DBL, CONS, FIRST, S, RECIP

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12, c13, c14, c15, c16, c17, c18, c19, c20, c21

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

Removed 3 trailing nodes:

PROPER(0) → c8
PROPER(nil) → c7
TOP(ok(z0)) → c5(TOP(active(z0)))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

sqr(ok(z0)) → ok(sqr(z0))
sqr(mark(z0)) → mark(sqr(z0))
add(ok(z0), ok(z1)) → ok(add(z0, z1))
add(z0, mark(z1)) → mark(add(z0, z1))
add(mark(z0), z1) → mark(add(z0, z1))
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
proper(nil) → ok(nil)
proper(0) → ok(0)
terms(ok(z0)) → ok(terms(z0))
terms(mark(z0)) → mark(terms(z0))
dbl(mark(z0)) → mark(dbl(z0))
dbl(ok(z0)) → ok(dbl(z0))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
cons(mark(z0), z1) → mark(cons(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))
s(mark(z0)) → mark(s(z0))
recip(ok(z0)) → ok(recip(z0))
recip(mark(z0)) → mark(recip(z0))
Tuples:

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TOP(mark(z0)) → c6(TOP(proper(z0)), PROPER(z0))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
S tuples:

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TOP(mark(z0)) → c6(TOP(proper(z0)), PROPER(z0))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
K tuples:none
Defined Rule Symbols:

sqr, add, top, proper, terms, dbl, cons, first, s, recip

Defined Pair Symbols:

SQR, ADD, TOP, TERMS, DBL, CONS, FIRST, S, RECIP

Compound Symbols:

c, c1, c2, c3, c4, c6, c9, c10, c11, c12, c13, c14, c15, c16, c17, c18, c19, c20, c21

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

Removed 1 trailing tuple parts

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

sqr(ok(z0)) → ok(sqr(z0))
sqr(mark(z0)) → mark(sqr(z0))
add(ok(z0), ok(z1)) → ok(add(z0, z1))
add(z0, mark(z1)) → mark(add(z0, z1))
add(mark(z0), z1) → mark(add(z0, z1))
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
proper(nil) → ok(nil)
proper(0) → ok(0)
terms(ok(z0)) → ok(terms(z0))
terms(mark(z0)) → mark(terms(z0))
dbl(mark(z0)) → mark(dbl(z0))
dbl(ok(z0)) → ok(dbl(z0))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
cons(mark(z0), z1) → mark(cons(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))
s(mark(z0)) → mark(s(z0))
recip(ok(z0)) → ok(recip(z0))
recip(mark(z0)) → mark(recip(z0))
Tuples:

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
K tuples:none
Defined Rule Symbols:

sqr, add, top, proper, terms, dbl, cons, first, s, recip

Defined Pair Symbols:

SQR, ADD, TERMS, DBL, CONS, FIRST, S, RECIP, TOP

Compound Symbols:

c, c1, c2, c3, c4, c9, c10, c11, c12, c13, c14, c15, c16, c17, c18, c19, c20, c21, c6

(11) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

sqr(ok(z0)) → ok(sqr(z0))
sqr(mark(z0)) → mark(sqr(z0))
add(ok(z0), ok(z1)) → ok(add(z0, z1))
add(z0, mark(z1)) → mark(add(z0, z1))
add(mark(z0), z1) → mark(add(z0, z1))
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
terms(ok(z0)) → ok(terms(z0))
terms(mark(z0)) → mark(terms(z0))
dbl(mark(z0)) → mark(dbl(z0))
dbl(ok(z0)) → ok(dbl(z0))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
cons(mark(z0), z1) → mark(cons(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))
s(mark(z0)) → mark(s(z0))
recip(ok(z0)) → ok(recip(z0))
recip(mark(z0)) → mark(recip(z0))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
K tuples:none
Defined Rule Symbols:

proper

Defined Pair Symbols:

SQR, ADD, TERMS, DBL, CONS, FIRST, S, RECIP, TOP

Compound Symbols:

c, c1, c2, c3, c4, c9, c10, c11, c12, c13, c14, c15, c16, c17, c18, c19, c20, c21, c6

(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)) → c6(TOP(proper(z0)))
We considered the (Usable) Rules:

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

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [1]   
POL(ADD(x1, x2)) = 0   
POL(CONS(x1, x2)) = 0   
POL(DBL(x1)) = 0   
POL(FIRST(x1, x2)) = 0   
POL(RECIP(x1)) = 0   
POL(S(x1)) = 0   
POL(SQR(x1)) = 0   
POL(TERMS(x1)) = 0   
POL(TOP(x1)) = x1   
POL(c(x1)) = x1   
POL(c1(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(c18(x1)) = x1   
POL(c19(x1)) = x1   
POL(c2(x1)) = x1   
POL(c20(x1)) = x1   
POL(c21(x1)) = x1   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(c6(x1)) = x1   
POL(c9(x1)) = x1   
POL(mark(x1)) = [1] + x1   
POL(nil) = [1]   
POL(ok(x1)) = 0   
POL(proper(x1)) = x1   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
K tuples:

TOP(mark(z0)) → c6(TOP(proper(z0)))
Defined Rule Symbols:

proper

Defined Pair Symbols:

SQR, ADD, TERMS, DBL, CONS, FIRST, S, RECIP, TOP

Compound Symbols:

c, c1, c2, c3, c4, c9, c10, c11, c12, c13, c14, c15, c16, c17, c18, c19, c20, c21, c6

(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.

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

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ADD(x1, x2)) = 0   
POL(CONS(x1, x2)) = x1 + x2   
POL(DBL(x1)) = 0   
POL(FIRST(x1, x2)) = 0   
POL(RECIP(x1)) = 0   
POL(S(x1)) = 0   
POL(SQR(x1)) = 0   
POL(TERMS(x1)) = 0   
POL(TOP(x1)) = 0   
POL(c(x1)) = x1   
POL(c1(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(c18(x1)) = x1   
POL(c19(x1)) = x1   
POL(c2(x1)) = x1   
POL(c20(x1)) = x1   
POL(c21(x1)) = x1   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(c6(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:

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
K tuples:

TOP(mark(z0)) → c6(TOP(proper(z0)))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
Defined Rule Symbols:

proper

Defined Pair Symbols:

SQR, ADD, TERMS, DBL, CONS, FIRST, S, RECIP, TOP

Compound Symbols:

c, c1, c2, c3, c4, c9, c10, c11, c12, c13, c14, c15, c16, c17, c18, c19, c20, c21, c6

(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.

SQR(mark(z0)) → c1(SQR(z0))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
RECIP(mark(z0)) → c21(RECIP(z0))
We considered the (Usable) Rules:none
And the Tuples:

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(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(CONS(x1, x2)) = 0   
POL(DBL(x1)) = 0   
POL(FIRST(x1, x2)) = 0   
POL(RECIP(x1)) = x1   
POL(S(x1)) = 0   
POL(SQR(x1)) = x1   
POL(TERMS(x1)) = 0   
POL(TOP(x1)) = 0   
POL(c(x1)) = x1   
POL(c1(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(c18(x1)) = x1   
POL(c19(x1)) = x1   
POL(c2(x1)) = x1   
POL(c20(x1)) = x1   
POL(c21(x1)) = x1   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(c6(x1)) = x1   
POL(c9(x1)) = x1   
POL(mark(x1)) = [1] + x1   
POL(nil) = 0   
POL(ok(x1)) = x1   
POL(proper(x1)) = 0   

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:

SQR(ok(z0)) → c(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
K tuples:

TOP(mark(z0)) → c6(TOP(proper(z0)))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
SQR(mark(z0)) → c1(SQR(z0))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
RECIP(mark(z0)) → c21(RECIP(z0))
Defined Rule Symbols:

proper

Defined Pair Symbols:

SQR, ADD, TERMS, DBL, CONS, FIRST, S, RECIP, TOP

Compound Symbols:

c, c1, c2, c3, c4, c9, c10, c11, c12, c13, c14, c15, c16, c17, c18, c19, c20, c21, c6

(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.

ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
DBL(ok(z0)) → c12(DBL(z0))
S(ok(z0)) → c18(S(z0))
We considered the (Usable) Rules:none
And the Tuples:

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ADD(x1, x2)) = x1   
POL(CONS(x1, x2)) = [2]x1 + x2   
POL(DBL(x1)) = x1   
POL(FIRST(x1, x2)) = 0   
POL(RECIP(x1)) = 0   
POL(S(x1)) = [2]x1   
POL(SQR(x1)) = 0   
POL(TERMS(x1)) = 0   
POL(TOP(x1)) = 0   
POL(c(x1)) = x1   
POL(c1(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(c18(x1)) = x1   
POL(c19(x1)) = x1   
POL(c2(x1)) = x1   
POL(c20(x1)) = x1   
POL(c21(x1)) = x1   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(c6(x1)) = x1   
POL(c9(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(ok(x1)) = [1] + x1   
POL(proper(x1)) = 0   

(20) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:

SQR(ok(z0)) → c(SQR(z0))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
K tuples:

TOP(mark(z0)) → c6(TOP(proper(z0)))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
SQR(mark(z0)) → c1(SQR(z0))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
RECIP(mark(z0)) → c21(RECIP(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
DBL(ok(z0)) → c12(DBL(z0))
S(ok(z0)) → c18(S(z0))
Defined Rule Symbols:

proper

Defined Pair Symbols:

SQR, ADD, TERMS, DBL, CONS, FIRST, S, RECIP, TOP

Compound Symbols:

c, c1, c2, c3, c4, c9, c10, c11, c12, c13, c14, c15, c16, c17, c18, c19, c20, c21, c6

(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.

SQR(ok(z0)) → c(SQR(z0))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ADD(x1, x2)) = 0   
POL(CONS(x1, x2)) = 0   
POL(DBL(x1)) = 0   
POL(FIRST(x1, x2)) = x2   
POL(RECIP(x1)) = 0   
POL(S(x1)) = 0   
POL(SQR(x1)) = x1   
POL(TERMS(x1)) = 0   
POL(TOP(x1)) = 0   
POL(c(x1)) = x1   
POL(c1(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(c18(x1)) = x1   
POL(c19(x1)) = x1   
POL(c2(x1)) = x1   
POL(c20(x1)) = x1   
POL(c21(x1)) = x1   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(c6(x1)) = x1   
POL(c9(x1)) = x1   
POL(mark(x1)) = [1] + x1   
POL(nil) = 0   
POL(ok(x1)) = [1] + x1   
POL(proper(x1)) = 0   

(22) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:

TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
K tuples:

TOP(mark(z0)) → c6(TOP(proper(z0)))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
SQR(mark(z0)) → c1(SQR(z0))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
RECIP(mark(z0)) → c21(RECIP(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
DBL(ok(z0)) → c12(DBL(z0))
S(ok(z0)) → c18(S(z0))
SQR(ok(z0)) → c(SQR(z0))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
Defined Rule Symbols:

proper

Defined Pair Symbols:

SQR, ADD, TERMS, DBL, CONS, FIRST, S, RECIP, TOP

Compound Symbols:

c, c1, c2, c3, c4, c9, c10, c11, c12, c13, c14, c15, c16, c17, c18, c19, c20, c21, c6

(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.

DBL(mark(z0)) → c11(DBL(z0))
S(mark(z0)) → c19(S(z0))
We considered the (Usable) Rules:

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

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ADD(x1, x2)) = 0   
POL(CONS(x1, x2)) = x2   
POL(DBL(x1)) = x1   
POL(FIRST(x1, x2)) = [2]x2   
POL(RECIP(x1)) = 0   
POL(S(x1)) = x1   
POL(SQR(x1)) = x1   
POL(TERMS(x1)) = 0   
POL(TOP(x1)) = [2]x1   
POL(c(x1)) = x1   
POL(c1(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(c18(x1)) = x1   
POL(c19(x1)) = x1   
POL(c2(x1)) = x1   
POL(c20(x1)) = x1   
POL(c21(x1)) = x1   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(c6(x1)) = x1   
POL(c9(x1)) = x1   
POL(mark(x1)) = [1] + x1   
POL(nil) = 0   
POL(ok(x1)) = [1] + x1   
POL(proper(x1)) = [1]   

(24) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:

TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
RECIP(ok(z0)) → c20(RECIP(z0))
K tuples:

TOP(mark(z0)) → c6(TOP(proper(z0)))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
SQR(mark(z0)) → c1(SQR(z0))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
RECIP(mark(z0)) → c21(RECIP(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
DBL(ok(z0)) → c12(DBL(z0))
S(ok(z0)) → c18(S(z0))
SQR(ok(z0)) → c(SQR(z0))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
DBL(mark(z0)) → c11(DBL(z0))
S(mark(z0)) → c19(S(z0))
Defined Rule Symbols:

proper

Defined Pair Symbols:

SQR, ADD, TERMS, DBL, CONS, FIRST, S, RECIP, TOP

Compound Symbols:

c, c1, c2, c3, c4, c9, c10, c11, c12, c13, c14, c15, c16, c17, c18, c19, c20, c21, c6

(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.

TERMS(ok(z0)) → c9(TERMS(z0))
We considered the (Usable) Rules:none
And the Tuples:

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ADD(x1, x2)) = 0   
POL(CONS(x1, x2)) = 0   
POL(DBL(x1)) = 0   
POL(FIRST(x1, x2)) = 0   
POL(RECIP(x1)) = 0   
POL(S(x1)) = 0   
POL(SQR(x1)) = 0   
POL(TERMS(x1)) = [2]x1   
POL(TOP(x1)) = 0   
POL(c(x1)) = x1   
POL(c1(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(c18(x1)) = x1   
POL(c19(x1)) = x1   
POL(c2(x1)) = x1   
POL(c20(x1)) = x1   
POL(c21(x1)) = x1   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(c6(x1)) = x1   
POL(c9(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = [1]   
POL(ok(x1)) = [2] + x1   
POL(proper(x1)) = [3] + x1   

(26) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:

TERMS(mark(z0)) → c10(TERMS(z0))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
RECIP(ok(z0)) → c20(RECIP(z0))
K tuples:

TOP(mark(z0)) → c6(TOP(proper(z0)))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
SQR(mark(z0)) → c1(SQR(z0))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
RECIP(mark(z0)) → c21(RECIP(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
DBL(ok(z0)) → c12(DBL(z0))
S(ok(z0)) → c18(S(z0))
SQR(ok(z0)) → c(SQR(z0))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
DBL(mark(z0)) → c11(DBL(z0))
S(mark(z0)) → c19(S(z0))
TERMS(ok(z0)) → c9(TERMS(z0))
Defined Rule Symbols:

proper

Defined Pair Symbols:

SQR, ADD, TERMS, DBL, CONS, FIRST, S, RECIP, TOP

Compound Symbols:

c, c1, c2, c3, c4, c9, c10, c11, c12, c13, c14, c15, c16, c17, c18, c19, c20, c21, c6

(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.

RECIP(ok(z0)) → c20(RECIP(z0))
We considered the (Usable) Rules:none
And the Tuples:

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ADD(x1, x2)) = 0   
POL(CONS(x1, x2)) = [2]x2   
POL(DBL(x1)) = [2]x1   
POL(FIRST(x1, x2)) = [2]x2   
POL(RECIP(x1)) = [2]x1   
POL(S(x1)) = [2]x1   
POL(SQR(x1)) = 0   
POL(TERMS(x1)) = 0   
POL(TOP(x1)) = 0   
POL(c(x1)) = x1   
POL(c1(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(c18(x1)) = x1   
POL(c19(x1)) = x1   
POL(c2(x1)) = x1   
POL(c20(x1)) = x1   
POL(c21(x1)) = x1   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(c6(x1)) = x1   
POL(c9(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(ok(x1)) = [1] + x1   
POL(proper(x1)) = 0   

(28) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:

TERMS(mark(z0)) → c10(TERMS(z0))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
K tuples:

TOP(mark(z0)) → c6(TOP(proper(z0)))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
SQR(mark(z0)) → c1(SQR(z0))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
RECIP(mark(z0)) → c21(RECIP(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
DBL(ok(z0)) → c12(DBL(z0))
S(ok(z0)) → c18(S(z0))
SQR(ok(z0)) → c(SQR(z0))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
DBL(mark(z0)) → c11(DBL(z0))
S(mark(z0)) → c19(S(z0))
TERMS(ok(z0)) → c9(TERMS(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
Defined Rule Symbols:

proper

Defined Pair Symbols:

SQR, ADD, TERMS, DBL, CONS, FIRST, S, RECIP, TOP

Compound Symbols:

c, c1, c2, c3, c4, c9, c10, c11, c12, c13, c14, c15, c16, c17, c18, c19, c20, c21, c6

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

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

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

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [1]   
POL(ADD(x1, x2)) = x1   
POL(CONS(x1, x2)) = 0   
POL(DBL(x1)) = 0   
POL(FIRST(x1, x2)) = x1   
POL(RECIP(x1)) = 0   
POL(S(x1)) = 0   
POL(SQR(x1)) = 0   
POL(TERMS(x1)) = 0   
POL(TOP(x1)) = 0   
POL(c(x1)) = x1   
POL(c1(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(c18(x1)) = x1   
POL(c19(x1)) = x1   
POL(c2(x1)) = x1   
POL(c20(x1)) = x1   
POL(c21(x1)) = x1   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(c6(x1)) = x1   
POL(c9(x1)) = x1   
POL(mark(x1)) = [1] + x1   
POL(nil) = 0   
POL(ok(x1)) = x1   
POL(proper(x1)) = [1]   

(30) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:

TERMS(mark(z0)) → c10(TERMS(z0))
K tuples:

TOP(mark(z0)) → c6(TOP(proper(z0)))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
SQR(mark(z0)) → c1(SQR(z0))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
RECIP(mark(z0)) → c21(RECIP(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
DBL(ok(z0)) → c12(DBL(z0))
S(ok(z0)) → c18(S(z0))
SQR(ok(z0)) → c(SQR(z0))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
DBL(mark(z0)) → c11(DBL(z0))
S(mark(z0)) → c19(S(z0))
TERMS(ok(z0)) → c9(TERMS(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
Defined Rule Symbols:

proper

Defined Pair Symbols:

SQR, ADD, TERMS, DBL, CONS, FIRST, S, RECIP, TOP

Compound Symbols:

c, c1, c2, c3, c4, c9, c10, c11, c12, c13, c14, c15, c16, c17, c18, c19, c20, c21, c6

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

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

TERMS(mark(z0)) → c10(TERMS(z0))
We considered the (Usable) Rules:none
And the Tuples:

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ADD(x1, x2)) = 0   
POL(CONS(x1, x2)) = 0   
POL(DBL(x1)) = 0   
POL(FIRST(x1, x2)) = 0   
POL(RECIP(x1)) = 0   
POL(S(x1)) = 0   
POL(SQR(x1)) = 0   
POL(TERMS(x1)) = [2]x1   
POL(TOP(x1)) = 0   
POL(c(x1)) = x1   
POL(c1(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(c18(x1)) = x1   
POL(c19(x1)) = x1   
POL(c2(x1)) = x1   
POL(c20(x1)) = x1   
POL(c21(x1)) = x1   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(c6(x1)) = x1   
POL(c9(x1)) = x1   
POL(mark(x1)) = [2] + x1   
POL(nil) = 0   
POL(ok(x1)) = x1   
POL(proper(x1)) = 0   

(32) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

SQR(ok(z0)) → c(SQR(z0))
SQR(mark(z0)) → c1(SQR(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
TERMS(ok(z0)) → c9(TERMS(z0))
TERMS(mark(z0)) → c10(TERMS(z0))
DBL(mark(z0)) → c11(DBL(z0))
DBL(ok(z0)) → c12(DBL(z0))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
S(ok(z0)) → c18(S(z0))
S(mark(z0)) → c19(S(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
RECIP(mark(z0)) → c21(RECIP(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:none
K tuples:

TOP(mark(z0)) → c6(TOP(proper(z0)))
CONS(ok(z0), ok(z1)) → c13(CONS(z0, z1))
CONS(mark(z0), z1) → c14(CONS(z0, z1))
SQR(mark(z0)) → c1(SQR(z0))
ADD(z0, mark(z1)) → c3(ADD(z0, z1))
ADD(mark(z0), z1) → c4(ADD(z0, z1))
RECIP(mark(z0)) → c21(RECIP(z0))
ADD(ok(z0), ok(z1)) → c2(ADD(z0, z1))
DBL(ok(z0)) → c12(DBL(z0))
S(ok(z0)) → c18(S(z0))
SQR(ok(z0)) → c(SQR(z0))
FIRST(ok(z0), ok(z1)) → c15(FIRST(z0, z1))
FIRST(z0, mark(z1)) → c17(FIRST(z0, z1))
DBL(mark(z0)) → c11(DBL(z0))
S(mark(z0)) → c19(S(z0))
TERMS(ok(z0)) → c9(TERMS(z0))
RECIP(ok(z0)) → c20(RECIP(z0))
FIRST(mark(z0), z1) → c16(FIRST(z0, z1))
TERMS(mark(z0)) → c10(TERMS(z0))
Defined Rule Symbols:

proper

Defined Pair Symbols:

SQR, ADD, TERMS, DBL, CONS, FIRST, S, RECIP, TOP

Compound Symbols:

c, c1, c2, c3, c4, c9, c10, c11, c12, c13, c14, c15, c16, c17, c18, c19, c20, c21, c6

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

The set S is empty

(34) BOUNDS(1, 1)