Runtime Complexity (innermost) proof of /tmp/tmptIdFEE/overlap.xml

The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(1, n^2).

0 CpxRelTRS

↳1 CpxTrsToCdtProof (UPPER BOUND(ID), 0 ms)

↳2 CdtProblem

↳3 CdtLeafRemovalProof (ComplexityIfPolyImplication, 0 ms)

↳4 CdtProblem

↳5 CdtUsableRulesProof (⇔, 0 ms)

↳6 CdtProblem

↳7 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 235 ms)

↳8 CdtProblem

↳9 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 53 ms)

↳10 CdtProblem

↳11 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 65 ms)

↳12 CdtProblem

↳13 CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)), 180 ms)

↳14 CdtProblem

↳15 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)

↳16 BOUNDS(1, 1)

(0) Obligation:

Runtime Complexity Relative TRS:
The TRS R consists of the following rules:

overlap(Cons(x, xs), ys) → overlap[Ite][True][Ite](member(x, ys), Cons(x, xs), ys)
overlap(Nil, ys) → False
member(x', Cons(x, xs)) → member[Ite][True][Ite](!EQ(x, x'), x', Cons(x, xs))
member(x, Nil) → False
notEmpty(Cons(x, xs)) → True
notEmpty(Nil) → False
goal(xs, ys) → overlap(xs, ys)

The (relative) TRS S consists of the following rules:

!EQ(S(x), S(y)) → !EQ(x, y)
!EQ(0, S(y)) → False
!EQ(S(x), 0) → False
!EQ(0, 0) → True
overlap[Ite][True][Ite](False, Cons(x, xs), ys) → overlap(xs, ys)
member[Ite][True][Ite](False, x', Cons(x, xs)) → member(x', xs)
overlap[Ite][True][Ite](True, xs, ys) → True
member[Ite][True][Ite](True, x, xs) → True

Rewrite Strategy: INNERMOST

(1) CpxTrsToCdtProof (UPPER BOUND(ID) transformation)

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

!EQ(S(z0), S(z1)) → !EQ(z0, z1)
!EQ(0, S(z0)) → False
!EQ(S(z0), 0) → False
!EQ(0, 0) → True
overlap[Ite][True][Ite](False, Cons(z0, z1), z2) → overlap(z1, z2)
overlap[Ite][True][Ite](True, z0, z1) → True
member[Ite][True][Ite](False, z0, Cons(z1, z2)) → member(z0, z2)
member[Ite][True][Ite](True, z0, z1) → True
overlap(Cons(z0, z1), z2) → overlap[Ite][True][Ite](member(z0, z2), Cons(z0, z1), z2)
overlap(Nil, z0) → False
member(z0, Cons(z1, z2)) → member[Ite][True][Ite](!EQ(z1, z0), z0, Cons(z1, z2))
member(z0, Nil) → False
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
goal(z0, z1) → overlap(z0, z1)

Tuples:

!EQ'(S(z0), S(z1)) → c(!EQ'(z0, z1))
!EQ'(0, S(z0)) → c1
!EQ'(S(z0), 0) → c2
!EQ'(0, 0) → c3
OVERLAP[ITE][TRUE][ITE](False, Cons(z0, z1), z2) → c4(OVERLAP(z1, z2))
OVERLAP[ITE][TRUE][ITE](True, z0, z1) → c5
MEMBER[ITE][TRUE][ITE](False, z0, Cons(z1, z2)) → c6(MEMBER(z0, z2))
MEMBER[ITE][TRUE][ITE](True, z0, z1) → c7
OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))
OVERLAP(Nil, z0) → c9
MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))
MEMBER(z0, Nil) → c11
NOTEMPTY(Cons(z0, z1)) → c12
NOTEMPTY(Nil) → c13
GOAL(z0, z1) → c14(OVERLAP(z0, z1))

S tuples:

OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))
OVERLAP(Nil, z0) → c9
MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))
MEMBER(z0, Nil) → c11
NOTEMPTY(Cons(z0, z1)) → c12
NOTEMPTY(Nil) → c13
GOAL(z0, z1) → c14(OVERLAP(z0, z1))

K tuples:none
Defined Rule Symbols:

overlap, member, notEmpty, goal, !EQ, overlap[Ite][True][Ite], member[Ite][True][Ite]

Defined Pair Symbols:

!EQ', OVERLAP[ITE][TRUE][ITE], MEMBER[ITE][TRUE][ITE], OVERLAP, MEMBER, NOTEMPTY, GOAL

Compound Symbols:

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

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

GOAL(z0, z1) → c14(OVERLAP(z0, z1))

Removed 7 trailing nodes:

!EQ'(S(z0), 0) → c2
MEMBER[ITE][TRUE][ITE](True, z0, z1) → c7
!EQ'(0, S(z0)) → c1
NOTEMPTY(Nil) → c13
NOTEMPTY(Cons(z0, z1)) → c12
!EQ'(0, 0) → c3
OVERLAP[ITE][TRUE][ITE](True, z0, z1) → c5

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

!EQ(S(z0), S(z1)) → !EQ(z0, z1)
!EQ(0, S(z0)) → False
!EQ(S(z0), 0) → False
!EQ(0, 0) → True
overlap[Ite][True][Ite](False, Cons(z0, z1), z2) → overlap(z1, z2)
overlap[Ite][True][Ite](True, z0, z1) → True
member[Ite][True][Ite](False, z0, Cons(z1, z2)) → member(z0, z2)
member[Ite][True][Ite](True, z0, z1) → True
overlap(Cons(z0, z1), z2) → overlap[Ite][True][Ite](member(z0, z2), Cons(z0, z1), z2)
overlap(Nil, z0) → False
member(z0, Cons(z1, z2)) → member[Ite][True][Ite](!EQ(z1, z0), z0, Cons(z1, z2))
member(z0, Nil) → False
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
goal(z0, z1) → overlap(z0, z1)

Tuples:

!EQ'(S(z0), S(z1)) → c(!EQ'(z0, z1))
OVERLAP[ITE][TRUE][ITE](False, Cons(z0, z1), z2) → c4(OVERLAP(z1, z2))
MEMBER[ITE][TRUE][ITE](False, z0, Cons(z1, z2)) → c6(MEMBER(z0, z2))
OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))
OVERLAP(Nil, z0) → c9
MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))
MEMBER(z0, Nil) → c11

S tuples:

OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))
OVERLAP(Nil, z0) → c9
MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))
MEMBER(z0, Nil) → c11

K tuples:none
Defined Rule Symbols:

overlap, member, notEmpty, goal, !EQ, overlap[Ite][True][Ite], member[Ite][True][Ite]

Defined Pair Symbols:

!EQ', OVERLAP[ITE][TRUE][ITE], MEMBER[ITE][TRUE][ITE], OVERLAP, MEMBER

Compound Symbols:

c, c4, c6, c8, c9, c10, c11

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

overlap[Ite][True][Ite](False, Cons(z0, z1), z2) → overlap(z1, z2)
overlap[Ite][True][Ite](True, z0, z1) → True
overlap(Cons(z0, z1), z2) → overlap[Ite][True][Ite](member(z0, z2), Cons(z0, z1), z2)
overlap(Nil, z0) → False
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
goal(z0, z1) → overlap(z0, z1)

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

member(z0, Cons(z1, z2)) → member[Ite][True][Ite](!EQ(z1, z0), z0, Cons(z1, z2))
member(z0, Nil) → False
member[Ite][True][Ite](False, z0, Cons(z1, z2)) → member(z0, z2)
member[Ite][True][Ite](True, z0, z1) → True
!EQ(S(z0), S(z1)) → !EQ(z0, z1)
!EQ(0, S(z0)) → False
!EQ(S(z0), 0) → False
!EQ(0, 0) → True

Tuples:

!EQ'(S(z0), S(z1)) → c(!EQ'(z0, z1))
OVERLAP[ITE][TRUE][ITE](False, Cons(z0, z1), z2) → c4(OVERLAP(z1, z2))
MEMBER[ITE][TRUE][ITE](False, z0, Cons(z1, z2)) → c6(MEMBER(z0, z2))
OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))
OVERLAP(Nil, z0) → c9
MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))
MEMBER(z0, Nil) → c11

S tuples:

OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))
OVERLAP(Nil, z0) → c9
MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))
MEMBER(z0, Nil) → c11

K tuples:none
Defined Rule Symbols:

member, member[Ite][True][Ite], !EQ

Defined Pair Symbols:

!EQ', OVERLAP[ITE][TRUE][ITE], MEMBER[ITE][TRUE][ITE], OVERLAP, MEMBER

Compound Symbols:

c, c4, c6, c8, c9, c10, c11

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

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

OVERLAP(Nil, z0) → c9

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

!EQ'(S(z0), S(z1)) → c(!EQ'(z0, z1))
OVERLAP[ITE][TRUE][ITE](False, Cons(z0, z1), z2) → c4(OVERLAP(z1, z2))
MEMBER[ITE][TRUE][ITE](False, z0, Cons(z1, z2)) → c6(MEMBER(z0, z2))
OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))
OVERLAP(Nil, z0) → c9
MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))
MEMBER(z0, Nil) → c11

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

POL(!EQ(x₁, x₂)) = [2]
POL(!EQ'(x₁, x₂)) = 0
POL(0) = [2]
POL(Cons(x₁, x₂)) = 0
POL(False) = [4]
POL(MEMBER(x₁, x₂)) = 0
POL(MEMBER[ITE][TRUE][ITE](x₁, x₂, x₃)) = 0
POL(Nil) = [2]
POL(OVERLAP(x₁, x₂)) = [4]
POL(OVERLAP[ITE][TRUE][ITE](x₁, x₂, x₃)) = [4]
POL(S(x₁)) = 0
POL(True) = [1]
POL(c(x₁)) = x₁
POL(c10(x₁, x₂)) = x₁ + x₂
POL(c11) = 0
POL(c4(x₁)) = x₁
POL(c6(x₁)) = x₁
POL(c8(x₁, x₂)) = x₁ + x₂
POL(c9) = 0
POL(member(x₁, x₂)) = [5] + [5]x₁ + [2]x₂
POL(member[Ite][True][Ite](x₁, x₂, x₃)) = [3] + [2]x₁ + [3]x₂ + [3]x₃

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

member(z0, Cons(z1, z2)) → member[Ite][True][Ite](!EQ(z1, z0), z0, Cons(z1, z2))
member(z0, Nil) → False
member[Ite][True][Ite](False, z0, Cons(z1, z2)) → member(z0, z2)
member[Ite][True][Ite](True, z0, z1) → True
!EQ(S(z0), S(z1)) → !EQ(z0, z1)
!EQ(0, S(z0)) → False
!EQ(S(z0), 0) → False
!EQ(0, 0) → True

Tuples:

!EQ'(S(z0), S(z1)) → c(!EQ'(z0, z1))
OVERLAP[ITE][TRUE][ITE](False, Cons(z0, z1), z2) → c4(OVERLAP(z1, z2))
MEMBER[ITE][TRUE][ITE](False, z0, Cons(z1, z2)) → c6(MEMBER(z0, z2))
OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))
OVERLAP(Nil, z0) → c9
MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))
MEMBER(z0, Nil) → c11

S tuples:

OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))
MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))
MEMBER(z0, Nil) → c11

K tuples:

OVERLAP(Nil, z0) → c9

Defined Rule Symbols:

member, member[Ite][True][Ite], !EQ

Defined Pair Symbols:

!EQ', OVERLAP[ITE][TRUE][ITE], MEMBER[ITE][TRUE][ITE], OVERLAP, MEMBER

Compound Symbols:

c, c4, c6, c8, c9, c10, c11

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

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

MEMBER(z0, Nil) → c11

We considered the (Usable) Rules:

member(z0, Cons(z1, z2)) → member[Ite][True][Ite](!EQ(z1, z0), z0, Cons(z1, z2))
member[Ite][True][Ite](False, z0, Cons(z1, z2)) → member(z0, z2)
member[Ite][True][Ite](True, z0, z1) → True
member(z0, Nil) → False

And the Tuples:

!EQ'(S(z0), S(z1)) → c(!EQ'(z0, z1))
OVERLAP[ITE][TRUE][ITE](False, Cons(z0, z1), z2) → c4(OVERLAP(z1, z2))
MEMBER[ITE][TRUE][ITE](False, z0, Cons(z1, z2)) → c6(MEMBER(z0, z2))
OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))
OVERLAP(Nil, z0) → c9
MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))
MEMBER(z0, Nil) → c11

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

POL(!EQ(x₁, x₂)) = [4] + [2]x₁
POL(!EQ'(x₁, x₂)) = 0
POL(0) = [2]
POL(Cons(x₁, x₂)) = [4] + x₂
POL(False) = 0
POL(MEMBER(x₁, x₂)) = [1]
POL(MEMBER[ITE][TRUE][ITE](x₁, x₂, x₃)) = [1]
POL(Nil) = 0
POL(OVERLAP(x₁, x₂)) = [2] + [4]x₁
POL(OVERLAP[ITE][TRUE][ITE](x₁, x₂, x₃)) = x₁ + [4]x₂
POL(S(x₁)) = [2]
POL(True) = 0
POL(c(x₁)) = x₁
POL(c10(x₁, x₂)) = x₁ + x₂
POL(c11) = 0
POL(c4(x₁)) = x₁
POL(c6(x₁)) = x₁
POL(c8(x₁, x₂)) = x₁ + x₂
POL(c9) = 0
POL(member(x₁, x₂)) = [1]
POL(member[Ite][True][Ite](x₁, x₂, x₃)) = [1]

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

member(z0, Cons(z1, z2)) → member[Ite][True][Ite](!EQ(z1, z0), z0, Cons(z1, z2))
member(z0, Nil) → False
member[Ite][True][Ite](False, z0, Cons(z1, z2)) → member(z0, z2)
member[Ite][True][Ite](True, z0, z1) → True
!EQ(S(z0), S(z1)) → !EQ(z0, z1)
!EQ(0, S(z0)) → False
!EQ(S(z0), 0) → False
!EQ(0, 0) → True

Tuples:

!EQ'(S(z0), S(z1)) → c(!EQ'(z0, z1))
OVERLAP[ITE][TRUE][ITE](False, Cons(z0, z1), z2) → c4(OVERLAP(z1, z2))
MEMBER[ITE][TRUE][ITE](False, z0, Cons(z1, z2)) → c6(MEMBER(z0, z2))
OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))
OVERLAP(Nil, z0) → c9
MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))
MEMBER(z0, Nil) → c11

S tuples:

OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))
MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))

K tuples:

OVERLAP(Nil, z0) → c9
MEMBER(z0, Nil) → c11

Defined Rule Symbols:

member, member[Ite][True][Ite], !EQ

Defined Pair Symbols:

!EQ', OVERLAP[ITE][TRUE][ITE], MEMBER[ITE][TRUE][ITE], OVERLAP, MEMBER

Compound Symbols:

c, c4, c6, c8, c9, c10, c11

(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.

OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))

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

!EQ'(S(z0), S(z1)) → c(!EQ'(z0, z1))
OVERLAP[ITE][TRUE][ITE](False, Cons(z0, z1), z2) → c4(OVERLAP(z1, z2))
MEMBER[ITE][TRUE][ITE](False, z0, Cons(z1, z2)) → c6(MEMBER(z0, z2))
OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))
OVERLAP(Nil, z0) → c9
MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))
MEMBER(z0, Nil) → c11

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

POL(!EQ(x₁, x₂)) = [3]x₁ + [2]x₂
POL(!EQ'(x₁, x₂)) = 0
POL(0) = 0
POL(Cons(x₁, x₂)) = [2] + x₂
POL(False) = 0
POL(MEMBER(x₁, x₂)) = 0
POL(MEMBER[ITE][TRUE][ITE](x₁, x₂, x₃)) = 0
POL(Nil) = [3]
POL(OVERLAP(x₁, x₂)) = [1] + [4]x₁
POL(OVERLAP[ITE][TRUE][ITE](x₁, x₂, x₃)) = [4]x₂
POL(S(x₁)) = [4]
POL(True) = 0
POL(c(x₁)) = x₁
POL(c10(x₁, x₂)) = x₁ + x₂
POL(c11) = 0
POL(c4(x₁)) = x₁
POL(c6(x₁)) = x₁
POL(c8(x₁, x₂)) = x₁ + x₂
POL(c9) = 0
POL(member(x₁, x₂)) = [3]x₁
POL(member[Ite][True][Ite](x₁, x₂, x₃)) = [1] + [2]x₁ + x₂ + [5]x₃

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

member(z0, Cons(z1, z2)) → member[Ite][True][Ite](!EQ(z1, z0), z0, Cons(z1, z2))
member(z0, Nil) → False
member[Ite][True][Ite](False, z0, Cons(z1, z2)) → member(z0, z2)
member[Ite][True][Ite](True, z0, z1) → True
!EQ(S(z0), S(z1)) → !EQ(z0, z1)
!EQ(0, S(z0)) → False
!EQ(S(z0), 0) → False
!EQ(0, 0) → True

Tuples:

!EQ'(S(z0), S(z1)) → c(!EQ'(z0, z1))
OVERLAP[ITE][TRUE][ITE](False, Cons(z0, z1), z2) → c4(OVERLAP(z1, z2))
MEMBER[ITE][TRUE][ITE](False, z0, Cons(z1, z2)) → c6(MEMBER(z0, z2))
OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))
OVERLAP(Nil, z0) → c9
MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))
MEMBER(z0, Nil) → c11

S tuples:

MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))

K tuples:

OVERLAP(Nil, z0) → c9
MEMBER(z0, Nil) → c11
OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))

Defined Rule Symbols:

member, member[Ite][True][Ite], !EQ

Defined Pair Symbols:

!EQ', OVERLAP[ITE][TRUE][ITE], MEMBER[ITE][TRUE][ITE], OVERLAP, MEMBER

Compound Symbols:

c, c4, c6, c8, c9, c10, c11

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

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

MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))

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

!EQ'(S(z0), S(z1)) → c(!EQ'(z0, z1))
OVERLAP[ITE][TRUE][ITE](False, Cons(z0, z1), z2) → c4(OVERLAP(z1, z2))
MEMBER[ITE][TRUE][ITE](False, z0, Cons(z1, z2)) → c6(MEMBER(z0, z2))
OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))
OVERLAP(Nil, z0) → c9
MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))
MEMBER(z0, Nil) → c11

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

POL(!EQ(x₁, x₂)) = 0
POL(!EQ'(x₁, x₂)) = 0
POL(0) = 0
POL(Cons(x₁, x₂)) = [1] + x₂
POL(False) = 0
POL(MEMBER(x₁, x₂)) = [3] + x₂
POL(MEMBER[ITE][TRUE][ITE](x₁, x₂, x₃)) = [2] + x₃
POL(Nil) = 0
POL(OVERLAP(x₁, x₂)) = [2] + [3]x₁ + x₂ + x₁·x₂ + x₁²
POL(OVERLAP[ITE][TRUE][ITE](x₁, x₂, x₃)) = x₂ + x₂·x₃ + x₂²
POL(S(x₁)) = 0
POL(True) = 0
POL(c(x₁)) = x₁
POL(c10(x₁, x₂)) = x₁ + x₂
POL(c11) = 0
POL(c4(x₁)) = x₁
POL(c6(x₁)) = x₁
POL(c8(x₁, x₂)) = x₁ + x₂
POL(c9) = 0
POL(member(x₁, x₂)) = 0
POL(member[Ite][True][Ite](x₁, x₂, x₃)) = 0

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

member(z0, Cons(z1, z2)) → member[Ite][True][Ite](!EQ(z1, z0), z0, Cons(z1, z2))
member(z0, Nil) → False
member[Ite][True][Ite](False, z0, Cons(z1, z2)) → member(z0, z2)
member[Ite][True][Ite](True, z0, z1) → True
!EQ(S(z0), S(z1)) → !EQ(z0, z1)
!EQ(0, S(z0)) → False
!EQ(S(z0), 0) → False
!EQ(0, 0) → True

Tuples:

!EQ'(S(z0), S(z1)) → c(!EQ'(z0, z1))
OVERLAP[ITE][TRUE][ITE](False, Cons(z0, z1), z2) → c4(OVERLAP(z1, z2))
MEMBER[ITE][TRUE][ITE](False, z0, Cons(z1, z2)) → c6(MEMBER(z0, z2))
OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))
OVERLAP(Nil, z0) → c9
MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))
MEMBER(z0, Nil) → c11

S tuples:none
K tuples:

OVERLAP(Nil, z0) → c9
MEMBER(z0, Nil) → c11
OVERLAP(Cons(z0, z1), z2) → c8(OVERLAP[ITE][TRUE][ITE](member(z0, z2), Cons(z0, z1), z2), MEMBER(z0, z2))
MEMBER(z0, Cons(z1, z2)) → c10(MEMBER[ITE][TRUE][ITE](!EQ(z1, z0), z0, Cons(z1, z2)), !EQ'(z1, z0))

Defined Rule Symbols:

member, member[Ite][True][Ite], !EQ

Defined Pair Symbols:

!EQ', OVERLAP[ITE][TRUE][ITE], MEMBER[ITE][TRUE][ITE], OVERLAP, MEMBER

Compound Symbols:

c, c4, c6, c8, c9, c10, c11

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

The set S is empty

(0) Obligation:

(1) CpxTrsToCdtProof (UPPER BOUND(ID) transformation)

(2) Obligation:

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

(4) Obligation:

(5) CdtUsableRulesProof (EQUIVALENT transformation)

(6) Obligation:

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

(8) Obligation:

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

(10) Obligation:

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

(12) Obligation:

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

(14) Obligation:

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

(16) BOUNDS(1, 1)