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

(0) Obligation:

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

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(x, xs) → member(x, xs)

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
member[Ite][True][Ite](False, x', Cons(x, xs)) → member(x', xs)
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
member[Ite][True][Ite](False, z0, Cons(z1, z2)) → member(z0, z2)
member[Ite][True][Ite](True, z0, z1) → True
member(z0, Cons(z1, z2)) → member[Ite][True][Ite](!EQ(z0, z1), z0, Cons(z1, z2))
member(z0, Nil) → False
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
goal(z0, z1) → member(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
MEMBER[ITE][TRUE][ITE](False, z0, Cons(z1, z2)) → c4(MEMBER(z0, z2))
MEMBER[ITE][TRUE][ITE](True, z0, z1) → c5
MEMBER(z0, Cons(z1, z2)) → c6(MEMBER[ITE][TRUE][ITE](!EQ(z0, z1), z0, Cons(z1, z2)), !EQ'(z0, z1))
MEMBER(z0, Nil) → c7
NOTEMPTY(Cons(z0, z1)) → c8
NOTEMPTY(Nil) → c9
GOAL(z0, z1) → c10(MEMBER(z0, z1))

S tuples:

MEMBER(z0, Cons(z1, z2)) → c6(MEMBER[ITE][TRUE][ITE](!EQ(z0, z1), z0, Cons(z1, z2)), !EQ'(z0, z1))
MEMBER(z0, Nil) → c7
NOTEMPTY(Cons(z0, z1)) → c8
NOTEMPTY(Nil) → c9
GOAL(z0, z1) → c10(MEMBER(z0, z1))

K tuples:none
Defined Rule Symbols:

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

Defined Pair Symbols:

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

Compound Symbols:

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

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

GOAL(z0, z1) → c10(MEMBER(z0, z1))

Removed 6 trailing nodes:

!EQ'(S(z0), 0) → c2
!EQ'(0, S(z0)) → c1
!EQ'(0, 0) → c3
NOTEMPTY(Nil) → c9
NOTEMPTY(Cons(z0, z1)) → c8
MEMBER[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
member[Ite][True][Ite](False, z0, Cons(z1, z2)) → member(z0, z2)
member[Ite][True][Ite](True, z0, z1) → True
member(z0, Cons(z1, z2)) → member[Ite][True][Ite](!EQ(z0, z1), z0, Cons(z1, z2))
member(z0, Nil) → False
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
goal(z0, z1) → member(z0, z1)

Tuples:

!EQ'(S(z0), S(z1)) → c(!EQ'(z0, z1))
MEMBER[ITE][TRUE][ITE](False, z0, Cons(z1, z2)) → c4(MEMBER(z0, z2))
MEMBER(z0, Cons(z1, z2)) → c6(MEMBER[ITE][TRUE][ITE](!EQ(z0, z1), z0, Cons(z1, z2)), !EQ'(z0, z1))
MEMBER(z0, Nil) → c7

S tuples:

MEMBER(z0, Cons(z1, z2)) → c6(MEMBER[ITE][TRUE][ITE](!EQ(z0, z1), z0, Cons(z1, z2)), !EQ'(z0, z1))
MEMBER(z0, Nil) → c7

K tuples:none
Defined Rule Symbols:

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

Defined Pair Symbols:

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

Compound Symbols:

c, c4, c6, c7

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(6) 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

Tuples:

!EQ'(S(z0), S(z1)) → c(!EQ'(z0, z1))
MEMBER[ITE][TRUE][ITE](False, z0, Cons(z1, z2)) → c4(MEMBER(z0, z2))
MEMBER(z0, Cons(z1, z2)) → c6(MEMBER[ITE][TRUE][ITE](!EQ(z0, z1), z0, Cons(z1, z2)), !EQ'(z0, z1))
MEMBER(z0, Nil) → c7

S tuples:

MEMBER(z0, Cons(z1, z2)) → c6(MEMBER[ITE][TRUE][ITE](!EQ(z0, z1), z0, Cons(z1, z2)), !EQ'(z0, z1))
MEMBER(z0, Nil) → c7

K tuples:none
Defined Rule Symbols:

!EQ

Defined Pair Symbols:

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

Compound Symbols:

c, c4, c6, c7

(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.

MEMBER(z0, Nil) → c7

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

!EQ'(S(z0), S(z1)) → c(!EQ'(z0, z1))
MEMBER[ITE][TRUE][ITE](False, z0, Cons(z1, z2)) → c4(MEMBER(z0, z2))
MEMBER(z0, Cons(z1, z2)) → c6(MEMBER[ITE][TRUE][ITE](!EQ(z0, z1), z0, Cons(z1, z2)), !EQ'(z0, z1))
MEMBER(z0, Nil) → c7

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

POL(!EQ(x₁, x₂)) = 0   
POL(!EQ'(x₁, x₂)) = 0   
POL(0) = [5]   
POL(Cons(x₁, x₂)) = x₂   
POL(False) = [5]   
POL(MEMBER(x₁, x₂)) = x₂   
POL(MEMBER[ITE][TRUE][ITE](x₁, x₂, x₃)) = x₃   
POL(Nil) = [2]   
POL(S(x₁)) = [5]   
POL(True) = [1]   
POL(c(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(c7) = 0   

(8) 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

Tuples:

!EQ'(S(z0), S(z1)) → c(!EQ'(z0, z1))
MEMBER[ITE][TRUE][ITE](False, z0, Cons(z1, z2)) → c4(MEMBER(z0, z2))
MEMBER(z0, Cons(z1, z2)) → c6(MEMBER[ITE][TRUE][ITE](!EQ(z0, z1), z0, Cons(z1, z2)), !EQ'(z0, z1))
MEMBER(z0, Nil) → c7

S tuples:

MEMBER(z0, Cons(z1, z2)) → c6(MEMBER[ITE][TRUE][ITE](!EQ(z0, z1), z0, Cons(z1, z2)), !EQ'(z0, z1))

K tuples:

MEMBER(z0, Nil) → c7

Defined Rule Symbols:

!EQ

Defined Pair Symbols:

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

Compound Symbols:

c, c4, c6, c7

(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, Cons(z1, z2)) → c6(MEMBER[ITE][TRUE][ITE](!EQ(z0, z1), z0, Cons(z1, z2)), !EQ'(z0, z1))

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

!EQ'(S(z0), S(z1)) → c(!EQ'(z0, z1))
MEMBER[ITE][TRUE][ITE](False, z0, Cons(z1, z2)) → c4(MEMBER(z0, z2))
MEMBER(z0, Cons(z1, z2)) → c6(MEMBER[ITE][TRUE][ITE](!EQ(z0, z1), z0, Cons(z1, z2)), !EQ'(z0, z1))
MEMBER(z0, Nil) → c7

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

POL(!EQ(x₁, x₂)) = 0   
POL(!EQ'(x₁, x₂)) = 0   
POL(0) = [2]   
POL(Cons(x₁, x₂)) = [4] + x₂   
POL(False) = 0   
POL(MEMBER(x₁, x₂)) = [2] + [4]x₂   
POL(MEMBER[ITE][TRUE][ITE](x₁, x₂, x₃)) = [4]x₃   
POL(Nil) = 0   
POL(S(x₁)) = [2]   
POL(True) = [5]   
POL(c(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(c7) = 0   

(10) 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

Tuples:

!EQ'(S(z0), S(z1)) → c(!EQ'(z0, z1))
MEMBER[ITE][TRUE][ITE](False, z0, Cons(z1, z2)) → c4(MEMBER(z0, z2))
MEMBER(z0, Cons(z1, z2)) → c6(MEMBER[ITE][TRUE][ITE](!EQ(z0, z1), z0, Cons(z1, z2)), !EQ'(z0, z1))
MEMBER(z0, Nil) → c7

S tuples:none
K tuples:

MEMBER(z0, Nil) → c7
MEMBER(z0, Cons(z1, z2)) → c6(MEMBER[ITE][TRUE][ITE](!EQ(z0, z1), z0, Cons(z1, z2)), !EQ'(z0, z1))

Defined Rule Symbols:

!EQ

Defined Pair Symbols:

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

Compound Symbols:

c, c4, c6, c7

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

The set S is empty

(12) BOUNDS(1, 1)