### (0) Obligation:

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

foldl(x, Cons(S(0), xs)) → foldl(S(x), xs)
foldl(S(0), Cons(x, xs)) → foldl(S(x), xs)
foldr(a, Cons(x, xs)) → op(x, foldr(a, xs))
foldr(a, Nil) → a
foldl(a, Nil) → a
notEmpty(Cons(x, xs)) → True
notEmpty(Nil) → False
op(x, S(0)) → S(x)
op(S(0), y) → S(y)
fold(a, xs) → Cons(foldl(a, xs), Cons(foldr(a, xs), Nil))

Rewrite Strategy: INNERMOST

### (1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

foldl(z0, Cons(S(0), z1)) → foldl(S(z0), z1)
foldl(S(0), Cons(z0, z1)) → foldl(S(z0), z1)
foldl(z0, Nil) → z0
foldr(z0, Cons(z1, z2)) → op(z1, foldr(z0, z2))
foldr(z0, Nil) → z0
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
op(z0, S(0)) → S(z0)
op(S(0), z0) → S(z0)
fold(z0, z1) → Cons(foldl(z0, z1), Cons(foldr(z0, z1), Nil))
Tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1))
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1))
FOLDL(z0, Nil) → c2
FOLDR(z0, Cons(z1, z2)) → c3(OP(z1, foldr(z0, z2)), FOLDR(z0, z2))
FOLDR(z0, Nil) → c4
NOTEMPTY(Cons(z0, z1)) → c5
NOTEMPTY(Nil) → c6
OP(z0, S(0)) → c7
OP(S(0), z0) → c8
FOLD(z0, z1) → c9(FOLDL(z0, z1), FOLDR(z0, z1))
S tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1))
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1))
FOLDL(z0, Nil) → c2
FOLDR(z0, Cons(z1, z2)) → c3(OP(z1, foldr(z0, z2)), FOLDR(z0, z2))
FOLDR(z0, Nil) → c4
NOTEMPTY(Cons(z0, z1)) → c5
NOTEMPTY(Nil) → c6
OP(z0, S(0)) → c7
OP(S(0), z0) → c8
FOLD(z0, z1) → c9(FOLDL(z0, z1), FOLDR(z0, z1))
K tuples:none
Defined Rule Symbols:

foldl, foldr, notEmpty, op, fold

Defined Pair Symbols:

FOLDL, FOLDR, NOTEMPTY, OP, FOLD

Compound Symbols:

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

### (3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

FOLD(z0, z1) → c9(FOLDL(z0, z1), FOLDR(z0, z1))
Removed 6 trailing nodes:

OP(S(0), z0) → c8
NOTEMPTY(Cons(z0, z1)) → c5
FOLDL(z0, Nil) → c2
NOTEMPTY(Nil) → c6
OP(z0, S(0)) → c7
FOLDR(z0, Nil) → c4

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

foldl(z0, Cons(S(0), z1)) → foldl(S(z0), z1)
foldl(S(0), Cons(z0, z1)) → foldl(S(z0), z1)
foldl(z0, Nil) → z0
foldr(z0, Cons(z1, z2)) → op(z1, foldr(z0, z2))
foldr(z0, Nil) → z0
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
op(z0, S(0)) → S(z0)
op(S(0), z0) → S(z0)
fold(z0, z1) → Cons(foldl(z0, z1), Cons(foldr(z0, z1), Nil))
Tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1))
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1))
FOLDR(z0, Cons(z1, z2)) → c3(OP(z1, foldr(z0, z2)), FOLDR(z0, z2))
S tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1))
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1))
FOLDR(z0, Cons(z1, z2)) → c3(OP(z1, foldr(z0, z2)), FOLDR(z0, z2))
K tuples:none
Defined Rule Symbols:

foldl, foldr, notEmpty, op, fold

Defined Pair Symbols:

FOLDL, FOLDR

Compound Symbols:

c, c1, c3

### (5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

foldl(z0, Cons(S(0), z1)) → foldl(S(z0), z1)
foldl(S(0), Cons(z0, z1)) → foldl(S(z0), z1)
foldl(z0, Nil) → z0
foldr(z0, Cons(z1, z2)) → op(z1, foldr(z0, z2))
foldr(z0, Nil) → z0
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
op(z0, S(0)) → S(z0)
op(S(0), z0) → S(z0)
fold(z0, z1) → Cons(foldl(z0, z1), Cons(foldr(z0, z1), Nil))
Tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1))
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1))
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
S tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1))
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1))
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
K tuples:none
Defined Rule Symbols:

foldl, foldr, notEmpty, op, fold

Defined Pair Symbols:

FOLDL, FOLDR

Compound Symbols:

c, c1, c3

### (7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

foldl(z0, Cons(S(0), z1)) → foldl(S(z0), z1)
foldl(S(0), Cons(z0, z1)) → foldl(S(z0), z1)
foldl(z0, Nil) → z0
foldr(z0, Cons(z1, z2)) → op(z1, foldr(z0, z2))
foldr(z0, Nil) → z0
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
op(z0, S(0)) → S(z0)
op(S(0), z0) → S(z0)
fold(z0, z1) → Cons(foldl(z0, z1), Cons(foldr(z0, z1), Nil))

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1))
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1))
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
S tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1))
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1))
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

FOLDL, FOLDR

Compound Symbols:

c, c1, c3

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

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

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1))
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1))
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
We considered the (Usable) Rules:none
And the Tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1))
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1))
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [3]
POL(Cons(x1, x2)) = [5] + x1 + x2
POL(FOLDL(x1, x2)) = [4]x1 + [4]x2
POL(FOLDR(x1, x2)) = [5]x2
POL(S(x1)) = [3] + x1
POL(c(x1)) = x1
POL(c1(x1)) = x1
POL(c3(x1)) = x1

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1))
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1))
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
S tuples:none
K tuples:

FOLDL(z0, Cons(S(0), z1)) → c(FOLDL(S(z0), z1))
FOLDL(S(0), Cons(z0, z1)) → c1(FOLDL(S(z0), z1))
FOLDR(z0, Cons(z1, z2)) → c3(FOLDR(z0, z2))
Defined Rule Symbols:none

Defined Pair Symbols:

FOLDL, FOLDR

Compound Symbols:

c, c1, c3

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

The set S is empty