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

(0) Obligation:

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

add0(x', Cons(x, xs)) → add0(Cons(Cons(Nil, Nil), x'), xs)
notEmpty(Cons(x, xs)) → True
notEmpty(Nil) → False
add0(x, Nil) → x
goal(x, y) → add0(x, y)

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

add0(z0, Cons(z1, z2)) → add0(Cons(Cons(Nil, Nil), z0), z2)
add0(z0, Nil) → z0
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
goal(z0, z1) → add0(z0, z1)

Tuples:

ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))
ADD0(z0, Nil) → c1
NOTEMPTY(Cons(z0, z1)) → c2
NOTEMPTY(Nil) → c3
GOAL(z0, z1) → c4(ADD0(z0, z1))

S tuples:

ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))
ADD0(z0, Nil) → c1
NOTEMPTY(Cons(z0, z1)) → c2
NOTEMPTY(Nil) → c3
GOAL(z0, z1) → c4(ADD0(z0, z1))

K tuples:none
Defined Rule Symbols:

add0, notEmpty, goal

Defined Pair Symbols:

ADD0, NOTEMPTY, GOAL

Compound Symbols:

c, c1, c2, c3, c4

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

GOAL(z0, z1) → c4(ADD0(z0, z1))

Removed 3 trailing nodes:

NOTEMPTY(Nil) → c3
NOTEMPTY(Cons(z0, z1)) → c2
ADD0(z0, Nil) → c1

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

add0(z0, Cons(z1, z2)) → add0(Cons(Cons(Nil, Nil), z0), z2)
add0(z0, Nil) → z0
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
goal(z0, z1) → add0(z0, z1)

Tuples:

ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))

S tuples:

ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))

K tuples:none
Defined Rule Symbols:

add0, notEmpty, goal

Defined Pair Symbols:

ADD0

Compound Symbols:

c

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

add0(z0, Cons(z1, z2)) → add0(Cons(Cons(Nil, Nil), z0), z2)
add0(z0, Nil) → z0
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
goal(z0, z1) → add0(z0, z1)

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))

S tuples:

ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

ADD0

Compound Symbols:

c

(7) 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.

ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))

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

ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))

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

POL(ADD0(x₁, x₂)) = [4]x₂   
POL(Cons(x₁, x₂)) = [4] + x₂   
POL(Nil) = [3]   
POL(c(x₁)) = x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))

S tuples:none
K tuples:

ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))

Defined Rule Symbols:none

Defined Pair Symbols:

ADD0

Compound Symbols:

c

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

The set S is empty

(10) BOUNDS(O(1), O(1))