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

(0) Obligation:

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

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(X1, X2)
activate(n__from(X)) → from(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__len(X)) → len(X)
activate(X) → X

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

fst(0, z0) → nil
fst(s(z0), cons(z1, z2)) → cons(z1, n__fst(activate(z0), activate(z2)))
fst(z0, z1) → n__fst(z0, z1)
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
add(0, z0) → z0
add(s(z0), z1) → s(n__add(activate(z0), z1))
add(z0, z1) → n__add(z0, z1)
len(nil) → 0
len(cons(z0, z1)) → s(n__len(activate(z1)))
len(z0) → n__len(z0)
activate(n__fst(z0, z1)) → fst(z0, z1)
activate(n__from(z0)) → from(z0)
activate(n__add(z0, z1)) → add(z0, z1)
activate(n__len(z0)) → len(z0)
activate(z0) → z0

Tuples:

FST(s(z0), cons(z1, z2)) → c1(ACTIVATE(z0), ACTIVATE(z2))
ADD(s(z0), z1) → c6(ACTIVATE(z0))
LEN(cons(z0, z1)) → c9(ACTIVATE(z1))
ACTIVATE(n__fst(z0, z1)) → c11(FST(z0, z1))
ACTIVATE(n__from(z0)) → c12(FROM(z0))
ACTIVATE(n__add(z0, z1)) → c13(ADD(z0, z1))
ACTIVATE(n__len(z0)) → c14(LEN(z0))

S tuples:

FST(s(z0), cons(z1, z2)) → c1(ACTIVATE(z0), ACTIVATE(z2))
ADD(s(z0), z1) → c6(ACTIVATE(z0))
LEN(cons(z0, z1)) → c9(ACTIVATE(z1))
ACTIVATE(n__fst(z0, z1)) → c11(FST(z0, z1))
ACTIVATE(n__from(z0)) → c12(FROM(z0))
ACTIVATE(n__add(z0, z1)) → c13(ADD(z0, z1))
ACTIVATE(n__len(z0)) → c14(LEN(z0))

K tuples:none
Defined Rule Symbols:

fst, from, add, len, activate

Defined Pair Symbols:

FST, ADD, LEN, ACTIVATE

Compound Symbols:

c1, c6, c9, c11, c12, c13, c14

(3) CdtGraphRemoveDanglingProof (ComplexityIfPolyImplication transformation)

Removed 1 of 7 dangling nodes:

ACTIVATE(n__from(z0)) → c12(FROM(z0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

fst(0, z0) → nil
fst(s(z0), cons(z1, z2)) → cons(z1, n__fst(activate(z0), activate(z2)))
fst(z0, z1) → n__fst(z0, z1)
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
add(0, z0) → z0
add(s(z0), z1) → s(n__add(activate(z0), z1))
add(z0, z1) → n__add(z0, z1)
len(nil) → 0
len(cons(z0, z1)) → s(n__len(activate(z1)))
len(z0) → n__len(z0)
activate(n__fst(z0, z1)) → fst(z0, z1)
activate(n__from(z0)) → from(z0)
activate(n__add(z0, z1)) → add(z0, z1)
activate(n__len(z0)) → len(z0)
activate(z0) → z0

Tuples:

FST(s(z0), cons(z1, z2)) → c1(ACTIVATE(z0), ACTIVATE(z2))
ADD(s(z0), z1) → c6(ACTIVATE(z0))
LEN(cons(z0, z1)) → c9(ACTIVATE(z1))
ACTIVATE(n__fst(z0, z1)) → c11(FST(z0, z1))
ACTIVATE(n__add(z0, z1)) → c13(ADD(z0, z1))
ACTIVATE(n__len(z0)) → c14(LEN(z0))

S tuples:

FST(s(z0), cons(z1, z2)) → c1(ACTIVATE(z0), ACTIVATE(z2))
ADD(s(z0), z1) → c6(ACTIVATE(z0))
LEN(cons(z0, z1)) → c9(ACTIVATE(z1))
ACTIVATE(n__fst(z0, z1)) → c11(FST(z0, z1))
ACTIVATE(n__add(z0, z1)) → c13(ADD(z0, z1))
ACTIVATE(n__len(z0)) → c14(LEN(z0))

K tuples:none
Defined Rule Symbols:

fst, from, add, len, activate

Defined Pair Symbols:

FST, ADD, LEN, ACTIVATE

Compound Symbols:

c1, c6, c9, c11, c13, c14

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

ADD(s(z0), z1) → c6(ACTIVATE(z0))

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

FST(s(z0), cons(z1, z2)) → c1(ACTIVATE(z0), ACTIVATE(z2))
ADD(s(z0), z1) → c6(ACTIVATE(z0))
LEN(cons(z0, z1)) → c9(ACTIVATE(z1))
ACTIVATE(n__fst(z0, z1)) → c11(FST(z0, z1))
ACTIVATE(n__add(z0, z1)) → c13(ADD(z0, z1))
ACTIVATE(n__len(z0)) → c14(LEN(z0))

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

POL(ACTIVATE(x₁)) = [2]x₁   
POL(ADD(x₁, x₂)) = [1] + [2]x₁ + x₂   
POL(FST(x₁, x₂)) = [2]x₁ + [2]x₂   
POL(LEN(x₁)) = [2]x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c11(x₁)) = x₁   
POL(c13(x₁)) = x₁   
POL(c14(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(cons(x₁, x₂)) = x₂   
POL(n__add(x₁, x₂)) = [2] + x₁ + x₂   
POL(n__fst(x₁, x₂)) = x₁ + x₂   
POL(n__len(x₁)) = x₁   
POL(s(x₁)) = x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

fst(0, z0) → nil
fst(s(z0), cons(z1, z2)) → cons(z1, n__fst(activate(z0), activate(z2)))
fst(z0, z1) → n__fst(z0, z1)
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
add(0, z0) → z0
add(s(z0), z1) → s(n__add(activate(z0), z1))
add(z0, z1) → n__add(z0, z1)
len(nil) → 0
len(cons(z0, z1)) → s(n__len(activate(z1)))
len(z0) → n__len(z0)
activate(n__fst(z0, z1)) → fst(z0, z1)
activate(n__from(z0)) → from(z0)
activate(n__add(z0, z1)) → add(z0, z1)
activate(n__len(z0)) → len(z0)
activate(z0) → z0

Tuples:

FST(s(z0), cons(z1, z2)) → c1(ACTIVATE(z0), ACTIVATE(z2))
ADD(s(z0), z1) → c6(ACTIVATE(z0))
LEN(cons(z0, z1)) → c9(ACTIVATE(z1))
ACTIVATE(n__fst(z0, z1)) → c11(FST(z0, z1))
ACTIVATE(n__add(z0, z1)) → c13(ADD(z0, z1))
ACTIVATE(n__len(z0)) → c14(LEN(z0))

S tuples:

FST(s(z0), cons(z1, z2)) → c1(ACTIVATE(z0), ACTIVATE(z2))
LEN(cons(z0, z1)) → c9(ACTIVATE(z1))
ACTIVATE(n__fst(z0, z1)) → c11(FST(z0, z1))
ACTIVATE(n__add(z0, z1)) → c13(ADD(z0, z1))
ACTIVATE(n__len(z0)) → c14(LEN(z0))

K tuples:

ADD(s(z0), z1) → c6(ACTIVATE(z0))

Defined Rule Symbols:

fst, from, add, len, activate

Defined Pair Symbols:

FST, ADD, LEN, ACTIVATE

Compound Symbols:

c1, c6, c9, c11, c13, c14

(7) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

ACTIVATE(n__fst(z0, z1)) → c11(FST(z0, z1))

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

FST(s(z0), cons(z1, z2)) → c1(ACTIVATE(z0), ACTIVATE(z2))
ADD(s(z0), z1) → c6(ACTIVATE(z0))
LEN(cons(z0, z1)) → c9(ACTIVATE(z1))
ACTIVATE(n__fst(z0, z1)) → c11(FST(z0, z1))
ACTIVATE(n__add(z0, z1)) → c13(ADD(z0, z1))
ACTIVATE(n__len(z0)) → c14(LEN(z0))

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

POL(ACTIVATE(x₁)) = x₁   
POL(ADD(x₁, x₂)) = x₁   
POL(FST(x₁, x₂)) = x₁ + x₂   
POL(LEN(x₁)) = [1] + x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c11(x₁)) = x₁   
POL(c13(x₁)) = x₁   
POL(c14(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(cons(x₁, x₂)) = x₂   
POL(n__add(x₁, x₂)) = x₁   
POL(n__fst(x₁, x₂)) = [2] + x₁ + x₂   
POL(n__len(x₁)) = [2] + x₁   
POL(s(x₁)) = x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

fst(0, z0) → nil
fst(s(z0), cons(z1, z2)) → cons(z1, n__fst(activate(z0), activate(z2)))
fst(z0, z1) → n__fst(z0, z1)
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
add(0, z0) → z0
add(s(z0), z1) → s(n__add(activate(z0), z1))
add(z0, z1) → n__add(z0, z1)
len(nil) → 0
len(cons(z0, z1)) → s(n__len(activate(z1)))
len(z0) → n__len(z0)
activate(n__fst(z0, z1)) → fst(z0, z1)
activate(n__from(z0)) → from(z0)
activate(n__add(z0, z1)) → add(z0, z1)
activate(n__len(z0)) → len(z0)
activate(z0) → z0

Tuples:

FST(s(z0), cons(z1, z2)) → c1(ACTIVATE(z0), ACTIVATE(z2))
ADD(s(z0), z1) → c6(ACTIVATE(z0))
LEN(cons(z0, z1)) → c9(ACTIVATE(z1))
ACTIVATE(n__fst(z0, z1)) → c11(FST(z0, z1))
ACTIVATE(n__add(z0, z1)) → c13(ADD(z0, z1))
ACTIVATE(n__len(z0)) → c14(LEN(z0))

S tuples:

FST(s(z0), cons(z1, z2)) → c1(ACTIVATE(z0), ACTIVATE(z2))
LEN(cons(z0, z1)) → c9(ACTIVATE(z1))
ACTIVATE(n__add(z0, z1)) → c13(ADD(z0, z1))
ACTIVATE(n__len(z0)) → c14(LEN(z0))

K tuples:

ADD(s(z0), z1) → c6(ACTIVATE(z0))
ACTIVATE(n__fst(z0, z1)) → c11(FST(z0, z1))

Defined Rule Symbols:

fst, from, add, len, activate

Defined Pair Symbols:

FST, ADD, LEN, ACTIVATE

Compound Symbols:

c1, c6, c9, c11, c13, c14

(9) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

FST(s(z0), cons(z1, z2)) → c1(ACTIVATE(z0), ACTIVATE(z2))
ACTIVATE(n__fst(z0, z1)) → c11(FST(z0, z1))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

fst(0, z0) → nil
fst(s(z0), cons(z1, z2)) → cons(z1, n__fst(activate(z0), activate(z2)))
fst(z0, z1) → n__fst(z0, z1)
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
add(0, z0) → z0
add(s(z0), z1) → s(n__add(activate(z0), z1))
add(z0, z1) → n__add(z0, z1)
len(nil) → 0
len(cons(z0, z1)) → s(n__len(activate(z1)))
len(z0) → n__len(z0)
activate(n__fst(z0, z1)) → fst(z0, z1)
activate(n__from(z0)) → from(z0)
activate(n__add(z0, z1)) → add(z0, z1)
activate(n__len(z0)) → len(z0)
activate(z0) → z0

Tuples:

FST(s(z0), cons(z1, z2)) → c1(ACTIVATE(z0), ACTIVATE(z2))
ADD(s(z0), z1) → c6(ACTIVATE(z0))
LEN(cons(z0, z1)) → c9(ACTIVATE(z1))
ACTIVATE(n__fst(z0, z1)) → c11(FST(z0, z1))
ACTIVATE(n__add(z0, z1)) → c13(ADD(z0, z1))
ACTIVATE(n__len(z0)) → c14(LEN(z0))

S tuples:

LEN(cons(z0, z1)) → c9(ACTIVATE(z1))
ACTIVATE(n__add(z0, z1)) → c13(ADD(z0, z1))
ACTIVATE(n__len(z0)) → c14(LEN(z0))

K tuples:

ADD(s(z0), z1) → c6(ACTIVATE(z0))
ACTIVATE(n__fst(z0, z1)) → c11(FST(z0, z1))
FST(s(z0), cons(z1, z2)) → c1(ACTIVATE(z0), ACTIVATE(z2))

Defined Rule Symbols:

fst, from, add, len, activate

Defined Pair Symbols:

FST, ADD, LEN, ACTIVATE

Compound Symbols:

c1, c6, c9, c11, c13, c14

(11) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

LEN(cons(z0, z1)) → c9(ACTIVATE(z1))
ACTIVATE(n__add(z0, z1)) → c13(ADD(z0, z1))

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

FST(s(z0), cons(z1, z2)) → c1(ACTIVATE(z0), ACTIVATE(z2))
ADD(s(z0), z1) → c6(ACTIVATE(z0))
LEN(cons(z0, z1)) → c9(ACTIVATE(z1))
ACTIVATE(n__fst(z0, z1)) → c11(FST(z0, z1))
ACTIVATE(n__add(z0, z1)) → c13(ADD(z0, z1))
ACTIVATE(n__len(z0)) → c14(LEN(z0))

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

POL(ACTIVATE(x₁)) = [4] + [4]x₁   
POL(ADD(x₁, x₂)) = [3] + [4]x₁ + [3]x₂   
POL(FST(x₁, x₂)) = [3] + [4]x₁ + [4]x₂   
POL(LEN(x₁)) = [3] + [4]x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c11(x₁)) = x₁   
POL(c13(x₁)) = x₁   
POL(c14(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(cons(x₁, x₂)) = [5] + x₂   
POL(n__add(x₁, x₂)) = x₁ + x₂   
POL(n__fst(x₁, x₂)) = [5] + x₁ + x₂   
POL(n__len(x₁)) = [5] + x₁   
POL(s(x₁)) = [2] + x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

fst(0, z0) → nil
fst(s(z0), cons(z1, z2)) → cons(z1, n__fst(activate(z0), activate(z2)))
fst(z0, z1) → n__fst(z0, z1)
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
add(0, z0) → z0
add(s(z0), z1) → s(n__add(activate(z0), z1))
add(z0, z1) → n__add(z0, z1)
len(nil) → 0
len(cons(z0, z1)) → s(n__len(activate(z1)))
len(z0) → n__len(z0)
activate(n__fst(z0, z1)) → fst(z0, z1)
activate(n__from(z0)) → from(z0)
activate(n__add(z0, z1)) → add(z0, z1)
activate(n__len(z0)) → len(z0)
activate(z0) → z0

Tuples:

FST(s(z0), cons(z1, z2)) → c1(ACTIVATE(z0), ACTIVATE(z2))
ADD(s(z0), z1) → c6(ACTIVATE(z0))
LEN(cons(z0, z1)) → c9(ACTIVATE(z1))
ACTIVATE(n__fst(z0, z1)) → c11(FST(z0, z1))
ACTIVATE(n__add(z0, z1)) → c13(ADD(z0, z1))
ACTIVATE(n__len(z0)) → c14(LEN(z0))

S tuples:

ACTIVATE(n__len(z0)) → c14(LEN(z0))

K tuples:

ADD(s(z0), z1) → c6(ACTIVATE(z0))
ACTIVATE(n__fst(z0, z1)) → c11(FST(z0, z1))
FST(s(z0), cons(z1, z2)) → c1(ACTIVATE(z0), ACTIVATE(z2))
LEN(cons(z0, z1)) → c9(ACTIVATE(z1))
ACTIVATE(n__add(z0, z1)) → c13(ADD(z0, z1))

Defined Rule Symbols:

fst, from, add, len, activate

Defined Pair Symbols:

FST, ADD, LEN, ACTIVATE

Compound Symbols:

c1, c6, c9, c11, c13, c14

(13) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

ACTIVATE(n__len(z0)) → c14(LEN(z0))
LEN(cons(z0, z1)) → c9(ACTIVATE(z1))

Now S is empty

(14) BOUNDS(O(1), O(1))