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:

t(N) → cs(r(q(N)), nt(ns(N)))
q(0) → 0
q(s(X)) → s(p(q(X), d(X)))
d(0) → 0
d(s(X)) → s(s(d(X)))
p(0, X) → X
p(X, 0) → X
p(s(X), s(Y)) → s(s(p(X, Y)))
f(0, X) → nil
f(s(X), cs(Y, Z)) → cs(Y, nf(X, a(Z)))
t(X) → nt(X)
s(X) → ns(X)
f(X1, X2) → nf(X1, X2)
a(nt(X)) → t(a(X))
a(ns(X)) → s(a(X))
a(nf(X1, X2)) → f(a(X1), a(X2))
a(X) → X

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

t(z0) → cs(r(q(z0)), nt(ns(z0)))
t(z0) → nt(z0)
q(0) → 0
q(s(z0)) → s(p(q(z0), d(z0)))
d(0) → 0
d(s(z0)) → s(s(d(z0)))
p(0, z0) → z0
p(z0, 0) → z0
p(s(z0), s(z1)) → s(s(p(z0, z1)))
f(0, z0) → nil
f(s(z0), cs(z1, z2)) → cs(z1, nf(z0, a(z2)))
f(z0, z1) → nf(z0, z1)
s(z0) → ns(z0)
a(nt(z0)) → t(a(z0))
a(ns(z0)) → s(a(z0))
a(nf(z0, z1)) → f(a(z0), a(z1))
a(z0) → z0

Tuples:

T(z0) → c(Q(z0))
Q(s(z0)) → c3(S(p(q(z0), d(z0))), P(q(z0), d(z0)), Q(z0), D(z0))
D(s(z0)) → c5(S(s(d(z0))), S(d(z0)), D(z0))
P(s(z0), s(z1)) → c8(S(s(p(z0, z1))), S(p(z0, z1)), P(z0, z1))
F(s(z0), cs(z1, z2)) → c10(A(z2))
A(nt(z0)) → c13(T(a(z0)), A(z0))
A(ns(z0)) → c14(S(a(z0)), A(z0))
A(nf(z0, z1)) → c15(F(a(z0), a(z1)), A(z0), A(z1))

S tuples:

T(z0) → c(Q(z0))
Q(s(z0)) → c3(S(p(q(z0), d(z0))), P(q(z0), d(z0)), Q(z0), D(z0))
D(s(z0)) → c5(S(s(d(z0))), S(d(z0)), D(z0))
P(s(z0), s(z1)) → c8(S(s(p(z0, z1))), S(p(z0, z1)), P(z0, z1))
F(s(z0), cs(z1, z2)) → c10(A(z2))
A(nt(z0)) → c13(T(a(z0)), A(z0))
A(ns(z0)) → c14(S(a(z0)), A(z0))
A(nf(z0, z1)) → c15(F(a(z0), a(z1)), A(z0), A(z1))

K tuples:none
Defined Rule Symbols:

t, q, d, p, f, s, a

Defined Pair Symbols:

T, Q, D, P, F, A

Compound Symbols:

c, c3, c5, c8, c10, c13, c14, c15

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

F(s(z0), cs(z1, z2)) → c10(A(z2))

Removed 4 trailing nodes:

T(z0) → c(Q(z0))
P(s(z0), s(z1)) → c8(S(s(p(z0, z1))), S(p(z0, z1)), P(z0, z1))
D(s(z0)) → c5(S(s(d(z0))), S(d(z0)), D(z0))
Q(s(z0)) → c3(S(p(q(z0), d(z0))), P(q(z0), d(z0)), Q(z0), D(z0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

t(z0) → cs(r(q(z0)), nt(ns(z0)))
t(z0) → nt(z0)
q(0) → 0
q(s(z0)) → s(p(q(z0), d(z0)))
d(0) → 0
d(s(z0)) → s(s(d(z0)))
p(0, z0) → z0
p(z0, 0) → z0
p(s(z0), s(z1)) → s(s(p(z0, z1)))
f(0, z0) → nil
f(s(z0), cs(z1, z2)) → cs(z1, nf(z0, a(z2)))
f(z0, z1) → nf(z0, z1)
s(z0) → ns(z0)
a(nt(z0)) → t(a(z0))
a(ns(z0)) → s(a(z0))
a(nf(z0, z1)) → f(a(z0), a(z1))
a(z0) → z0

Tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0))
A(ns(z0)) → c14(S(a(z0)), A(z0))
A(nf(z0, z1)) → c15(F(a(z0), a(z1)), A(z0), A(z1))

S tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0))
A(ns(z0)) → c14(S(a(z0)), A(z0))
A(nf(z0, z1)) → c15(F(a(z0), a(z1)), A(z0), A(z1))

K tuples:none
Defined Rule Symbols:

t, q, d, p, f, s, a

Defined Pair Symbols:

A

Compound Symbols:

c13, c14, c15

(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.

A(nt(z0)) → c13(T(a(z0)), A(z0))
A(ns(z0)) → c14(S(a(z0)), A(z0))

We considered the (Usable) Rules:

a(nt(z0)) → t(a(z0))
a(ns(z0)) → s(a(z0))
a(nf(z0, z1)) → f(a(z0), a(z1))
a(z0) → z0
f(0, z0) → nil
f(z0, z1) → nf(z0, z1)
s(z0) → ns(z0)
t(z0) → cs(r(q(z0)), nt(ns(z0)))
t(z0) → nt(z0)
q(0) → 0

And the Tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0))
A(ns(z0)) → c14(S(a(z0)), A(z0))
A(nf(z0, z1)) → c15(F(a(z0), a(z1)), A(z0), A(z1))

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

POL(0) = [1]   
POL(A(x₁)) = [2]x₁   
POL(F(x₁, x₂)) = 0   
POL(S(x₁)) = 0   
POL(T(x₁)) = [1]   
POL(a(x₁)) = [4] + [2]x₁   
POL(c13(x₁, x₂)) = x₁ + x₂   
POL(c14(x₁, x₂)) = x₁ + x₂   
POL(c15(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(cs(x₁, x₂)) = [2]   
POL(f(x₁, x₂)) = [1] + [3]x₁ + [5]x₂   
POL(nf(x₁, x₂)) = x₁ + x₂   
POL(nil) = [2]   
POL(ns(x₁)) = [1] + x₁   
POL(nt(x₁)) = [1] + x₁   
POL(q(x₁)) = [4] + [2]x₁   
POL(r(x₁)) = [5]   
POL(s(x₁)) = [2] + [5]x₁   
POL(t(x₁)) = [5]   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

t(z0) → cs(r(q(z0)), nt(ns(z0)))
t(z0) → nt(z0)
q(0) → 0
q(s(z0)) → s(p(q(z0), d(z0)))
d(0) → 0
d(s(z0)) → s(s(d(z0)))
p(0, z0) → z0
p(z0, 0) → z0
p(s(z0), s(z1)) → s(s(p(z0, z1)))
f(0, z0) → nil
f(s(z0), cs(z1, z2)) → cs(z1, nf(z0, a(z2)))
f(z0, z1) → nf(z0, z1)
s(z0) → ns(z0)
a(nt(z0)) → t(a(z0))
a(ns(z0)) → s(a(z0))
a(nf(z0, z1)) → f(a(z0), a(z1))
a(z0) → z0

Tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0))
A(ns(z0)) → c14(S(a(z0)), A(z0))
A(nf(z0, z1)) → c15(F(a(z0), a(z1)), A(z0), A(z1))

S tuples:

A(nf(z0, z1)) → c15(F(a(z0), a(z1)), A(z0), A(z1))

K tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0))
A(ns(z0)) → c14(S(a(z0)), A(z0))

Defined Rule Symbols:

t, q, d, p, f, s, a

Defined Pair Symbols:

A

Compound Symbols:

c13, c14, c15

(7) 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.

A(nf(z0, z1)) → c15(F(a(z0), a(z1)), A(z0), A(z1))

We considered the (Usable) Rules:

a(nt(z0)) → t(a(z0))
a(ns(z0)) → s(a(z0))
a(nf(z0, z1)) → f(a(z0), a(z1))
a(z0) → z0
f(0, z0) → nil
f(z0, z1) → nf(z0, z1)
s(z0) → ns(z0)
t(z0) → cs(r(q(z0)), nt(ns(z0)))
t(z0) → nt(z0)
q(0) → 0

And the Tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0))
A(ns(z0)) → c14(S(a(z0)), A(z0))
A(nf(z0, z1)) → c15(F(a(z0), a(z1)), A(z0), A(z1))

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

POL(0) = [5]   
POL(A(x₁)) = [2]x₁   
POL(F(x₁, x₂)) = 0   
POL(S(x₁)) = [2]   
POL(T(x₁)) = [3]   
POL(a(x₁)) = [4]   
POL(c13(x₁, x₂)) = x₁ + x₂   
POL(c14(x₁, x₂)) = x₁ + x₂   
POL(c15(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(cs(x₁, x₂)) = [3]   
POL(f(x₁, x₂)) = [4] + [3]x₁ + [2]x₂   
POL(nf(x₁, x₂)) = [1] + x₁ + x₂   
POL(nil) = [1]   
POL(ns(x₁)) = [2] + x₁   
POL(nt(x₁)) = [3] + x₁   
POL(q(x₁)) = [5]x₁   
POL(r(x₁)) = [3]   
POL(s(x₁)) = [3] + [5]x₁   
POL(t(x₁)) = [4] + [4]x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

t(z0) → cs(r(q(z0)), nt(ns(z0)))
t(z0) → nt(z0)
q(0) → 0
q(s(z0)) → s(p(q(z0), d(z0)))
d(0) → 0
d(s(z0)) → s(s(d(z0)))
p(0, z0) → z0
p(z0, 0) → z0
p(s(z0), s(z1)) → s(s(p(z0, z1)))
f(0, z0) → nil
f(s(z0), cs(z1, z2)) → cs(z1, nf(z0, a(z2)))
f(z0, z1) → nf(z0, z1)
s(z0) → ns(z0)
a(nt(z0)) → t(a(z0))
a(ns(z0)) → s(a(z0))
a(nf(z0, z1)) → f(a(z0), a(z1))
a(z0) → z0

Tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0))
A(ns(z0)) → c14(S(a(z0)), A(z0))
A(nf(z0, z1)) → c15(F(a(z0), a(z1)), A(z0), A(z1))

S tuples:none
K tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0))
A(ns(z0)) → c14(S(a(z0)), A(z0))
A(nf(z0, z1)) → c15(F(a(z0), a(z1)), A(z0), A(z1))

Defined Rule Symbols:

t, q, d, p, f, s, a

Defined Pair Symbols:

A

Compound Symbols:

c13, c14, c15

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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