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:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

app(nil, z0) → z0
app(z0, nil) → z0
app(cons(z0, z1), z2) → cons(z0, app(z1, z2))
sum(cons(z0, nil)) → cons(z0, nil)
sum(cons(z0, cons(z1, z2))) → sum(cons(plus(z0, z1), z2))
sum(app(z0, cons(z1, cons(z2, z3)))) → sum(app(z0, sum(cons(z1, cons(z2, z3)))))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))

Tuples:

APP(cons(z0, z1), z2) → c2(APP(z1, z2))
SUM(cons(z0, cons(z1, z2))) → c4(SUM(cons(plus(z0, z1), z2)), PLUS(z0, z1))
SUM(app(z0, cons(z1, cons(z2, z3)))) → c5(SUM(app(z0, sum(cons(z1, cons(z2, z3))))), APP(z0, sum(cons(z1, cons(z2, z3)))), SUM(cons(z1, cons(z2, z3))))
PLUS(s(z0), z1) → c7(PLUS(z0, z1))

S tuples:

APP(cons(z0, z1), z2) → c2(APP(z1, z2))
SUM(cons(z0, cons(z1, z2))) → c4(SUM(cons(plus(z0, z1), z2)), PLUS(z0, z1))
SUM(app(z0, cons(z1, cons(z2, z3)))) → c5(SUM(app(z0, sum(cons(z1, cons(z2, z3))))), APP(z0, sum(cons(z1, cons(z2, z3)))), SUM(cons(z1, cons(z2, z3))))
PLUS(s(z0), z1) → c7(PLUS(z0, z1))

K tuples:none
Defined Rule Symbols:

app, sum, plus

Defined Pair Symbols:

APP, SUM, PLUS

Compound Symbols:

c2, c4, c5, c7

(3) CdtUnreachableProof (EQUIVALENT transformation)

The following tuples could be removed as they are not reachable from basic start terms:

SUM(app(z0, cons(z1, cons(z2, z3)))) → c5(SUM(app(z0, sum(cons(z1, cons(z2, z3))))), APP(z0, sum(cons(z1, cons(z2, z3)))), SUM(cons(z1, cons(z2, z3))))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

app(nil, z0) → z0
app(z0, nil) → z0
app(cons(z0, z1), z2) → cons(z0, app(z1, z2))
sum(cons(z0, nil)) → cons(z0, nil)
sum(cons(z0, cons(z1, z2))) → sum(cons(plus(z0, z1), z2))
sum(app(z0, cons(z1, cons(z2, z3)))) → sum(app(z0, sum(cons(z1, cons(z2, z3)))))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))

Tuples:

APP(cons(z0, z1), z2) → c2(APP(z1, z2))
SUM(cons(z0, cons(z1, z2))) → c4(SUM(cons(plus(z0, z1), z2)), PLUS(z0, z1))
PLUS(s(z0), z1) → c7(PLUS(z0, z1))

S tuples:

APP(cons(z0, z1), z2) → c2(APP(z1, z2))
SUM(cons(z0, cons(z1, z2))) → c4(SUM(cons(plus(z0, z1), z2)), PLUS(z0, z1))
PLUS(s(z0), z1) → c7(PLUS(z0, z1))

K tuples:none
Defined Rule Symbols:

app, sum, plus

Defined Pair Symbols:

APP, SUM, PLUS

Compound Symbols:

c2, c4, c7

(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.

APP(cons(z0, z1), z2) → c2(APP(z1, z2))
SUM(cons(z0, cons(z1, z2))) → c4(SUM(cons(plus(z0, z1), z2)), PLUS(z0, z1))

We considered the (Usable) Rules:

plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))

And the Tuples:

APP(cons(z0, z1), z2) → c2(APP(z1, z2))
SUM(cons(z0, cons(z1, z2))) → c4(SUM(cons(plus(z0, z1), z2)), PLUS(z0, z1))
PLUS(s(z0), z1) → c7(PLUS(z0, z1))

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

POL(0) = 0   
POL(APP(x₁, x₂)) = [4]x₁   
POL(PLUS(x₁, x₂)) = [2]   
POL(SUM(x₁)) = [4]x₁   
POL(c2(x₁)) = x₁   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c7(x₁)) = x₁   
POL(cons(x₁, x₂)) = [4] + x₂   
POL(plus(x₁, x₂)) = [4] + [4]x₁ + x₂   
POL(s(x₁)) = [3] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

app(nil, z0) → z0
app(z0, nil) → z0
app(cons(z0, z1), z2) → cons(z0, app(z1, z2))
sum(cons(z0, nil)) → cons(z0, nil)
sum(cons(z0, cons(z1, z2))) → sum(cons(plus(z0, z1), z2))
sum(app(z0, cons(z1, cons(z2, z3)))) → sum(app(z0, sum(cons(z1, cons(z2, z3)))))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))

Tuples:

APP(cons(z0, z1), z2) → c2(APP(z1, z2))
SUM(cons(z0, cons(z1, z2))) → c4(SUM(cons(plus(z0, z1), z2)), PLUS(z0, z1))
PLUS(s(z0), z1) → c7(PLUS(z0, z1))

S tuples:

PLUS(s(z0), z1) → c7(PLUS(z0, z1))

K tuples:

APP(cons(z0, z1), z2) → c2(APP(z1, z2))
SUM(cons(z0, cons(z1, z2))) → c4(SUM(cons(plus(z0, z1), z2)), PLUS(z0, z1))

Defined Rule Symbols:

app, sum, plus

Defined Pair Symbols:

APP, SUM, PLUS

Compound Symbols:

c2, c4, c7

(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.

PLUS(s(z0), z1) → c7(PLUS(z0, z1))

We considered the (Usable) Rules:

plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))

And the Tuples:

APP(cons(z0, z1), z2) → c2(APP(z1, z2))
SUM(cons(z0, cons(z1, z2))) → c4(SUM(cons(plus(z0, z1), z2)), PLUS(z0, z1))
PLUS(s(z0), z1) → c7(PLUS(z0, z1))

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

POL(0) = 0   
POL(APP(x₁, x₂)) = [2]x₁·x₂   
POL(PLUS(x₁, x₂)) = x₁ + [3]x₂   
POL(SUM(x₁)) = [2]x₁²   
POL(c2(x₁)) = x₁   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c7(x₁)) = x₁   
POL(cons(x₁, x₂)) = [2] + x₁ + x₂   
POL(plus(x₁, x₂)) = x₁ + x₂   
POL(s(x₁)) = [1] + x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

app(nil, z0) → z0
app(z0, nil) → z0
app(cons(z0, z1), z2) → cons(z0, app(z1, z2))
sum(cons(z0, nil)) → cons(z0, nil)
sum(cons(z0, cons(z1, z2))) → sum(cons(plus(z0, z1), z2))
sum(app(z0, cons(z1, cons(z2, z3)))) → sum(app(z0, sum(cons(z1, cons(z2, z3)))))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))

Tuples:

APP(cons(z0, z1), z2) → c2(APP(z1, z2))
SUM(cons(z0, cons(z1, z2))) → c4(SUM(cons(plus(z0, z1), z2)), PLUS(z0, z1))
PLUS(s(z0), z1) → c7(PLUS(z0, z1))

S tuples:none
K tuples:

APP(cons(z0, z1), z2) → c2(APP(z1, z2))
SUM(cons(z0, cons(z1, z2))) → c4(SUM(cons(plus(z0, z1), z2)), PLUS(z0, z1))
PLUS(s(z0), z1) → c7(PLUS(z0, z1))

Defined Rule Symbols:

app, sum, plus

Defined Pair Symbols:

APP, SUM, PLUS

Compound Symbols:

c2, c4, c7

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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