We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ t(N) -> cs(r(q(N)), nt(ns(N)))
, t(X) -> nt(X)
, q(0()) -> 0()
, q(s(X)) -> s(p(q(X), d(X)))
, s(X) -> ns(X)
, p(X, 0()) -> X
, p(0(), X) -> X
, p(s(X), s(Y)) -> s(s(p(X, Y)))
, d(0()) -> 0()
, d(s(X)) -> s(s(d(X)))
, f(X1, X2) -> nf(X1, X2)
, f(0(), X) -> nil()
, f(s(X), cs(Y, Z)) -> cs(Y, nf(X, a(Z)))
, a(X) -> X
, a(nt(X)) -> t(a(X))
, a(ns(X)) -> s(a(X))
, a(nf(X1, X2)) -> f(a(X1), a(X2)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
Arguments of following rules are not normal-forms:
{ q(s(X)) -> s(p(q(X), d(X)))
, p(s(X), s(Y)) -> s(s(p(X, Y)))
, d(s(X)) -> s(s(d(X)))
, f(s(X), cs(Y, Z)) -> cs(Y, nf(X, a(Z))) }
All above mentioned rules can be savely removed.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ t(N) -> cs(r(q(N)), nt(ns(N)))
, t(X) -> nt(X)
, q(0()) -> 0()
, s(X) -> ns(X)
, p(X, 0()) -> X
, p(0(), X) -> X
, d(0()) -> 0()
, f(X1, X2) -> nf(X1, X2)
, f(0(), X) -> nil()
, a(X) -> X
, a(nt(X)) -> t(a(X))
, a(ns(X)) -> s(a(X))
, a(nf(X1, X2)) -> f(a(X1), a(X2)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs: { d(0()) -> 0() }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(t) = {1}, Uargs(cs) = {1}, Uargs(r) = {1}, Uargs(s) = {1},
Uargs(f) = {1, 2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[t](x1) = [1] x1 + [0]
[cs](x1, x2) = [1] x1 + [0]
[r](x1) = [1] x1 + [0]
[q](x1) = [0]
[nt](x1) = [1] x1 + [0]
[ns](x1) = [1] x1 + [0]
[0] = [0]
[s](x1) = [1] x1 + [0]
[p](x1, x2) = [3] x1 + [1] x2 + [0]
[d](x1) = [4]
[f](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
[nf](x1, x2) = [1] x1 + [1] x2 + [0]
[a](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[t(N)] = [1] N + [0]
>= [0]
= [cs(r(q(N)), nt(ns(N)))]
[t(X)] = [1] X + [0]
>= [1] X + [0]
= [nt(X)]
[q(0())] = [0]
>= [0]
= [0()]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [ns(X)]
[p(X, 0())] = [3] X + [0]
>= [1] X + [0]
= [X]
[p(0(), X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[d(0())] = [4]
> [0]
= [0()]
[f(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [nf(X1, X2)]
[f(0(), X)] = [1] X + [0]
>= [0]
= [nil()]
[a(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[a(nt(X))] = [1] X + [0]
>= [1] X + [0]
= [t(a(X))]
[a(ns(X))] = [1] X + [0]
>= [1] X + [0]
= [s(a(X))]
[a(nf(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [f(a(X1), a(X2))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ t(N) -> cs(r(q(N)), nt(ns(N)))
, t(X) -> nt(X)
, q(0()) -> 0()
, s(X) -> ns(X)
, p(X, 0()) -> X
, p(0(), X) -> X
, f(X1, X2) -> nf(X1, X2)
, f(0(), X) -> nil()
, a(X) -> X
, a(nt(X)) -> t(a(X))
, a(ns(X)) -> s(a(X))
, a(nf(X1, X2)) -> f(a(X1), a(X2)) }
Weak Trs: { d(0()) -> 0() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs:
{ f(X1, X2) -> nf(X1, X2)
, f(0(), X) -> nil()
, a(X) -> X
, a(nf(X1, X2)) -> f(a(X1), a(X2)) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(t) = {1}, Uargs(cs) = {1}, Uargs(r) = {1}, Uargs(s) = {1},
Uargs(f) = {1, 2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[t](x1) = [1] x1 + [0]
[cs](x1, x2) = [1] x1 + [0]
[r](x1) = [1] x1 + [0]
[q](x1) = [0]
[nt](x1) = [1] x1 + [0]
[ns](x1) = [1] x1 + [0]
[0] = [0]
[s](x1) = [1] x1 + [0]
[p](x1, x2) = [2] x1 + [1] x2 + [0]
[d](x1) = [5]
[f](x1, x2) = [1] x1 + [1] x2 + [4]
[nil] = [0]
[nf](x1, x2) = [1] x1 + [1] x2 + [2]
[a](x1) = [4] x1 + [1]
The order satisfies the following ordering constraints:
[t(N)] = [1] N + [0]
>= [0]
= [cs(r(q(N)), nt(ns(N)))]
[t(X)] = [1] X + [0]
>= [1] X + [0]
= [nt(X)]
[q(0())] = [0]
>= [0]
= [0()]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [ns(X)]
[p(X, 0())] = [2] X + [0]
>= [1] X + [0]
= [X]
[p(0(), X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[d(0())] = [5]
> [0]
= [0()]
[f(X1, X2)] = [1] X1 + [1] X2 + [4]
> [1] X1 + [1] X2 + [2]
= [nf(X1, X2)]
[f(0(), X)] = [1] X + [4]
> [0]
= [nil()]
[a(X)] = [4] X + [1]
> [1] X + [0]
= [X]
[a(nt(X))] = [4] X + [1]
>= [4] X + [1]
= [t(a(X))]
[a(ns(X))] = [4] X + [1]
>= [4] X + [1]
= [s(a(X))]
[a(nf(X1, X2))] = [4] X1 + [4] X2 + [9]
> [4] X1 + [4] X2 + [6]
= [f(a(X1), a(X2))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ t(N) -> cs(r(q(N)), nt(ns(N)))
, t(X) -> nt(X)
, q(0()) -> 0()
, s(X) -> ns(X)
, p(X, 0()) -> X
, p(0(), X) -> X
, a(nt(X)) -> t(a(X))
, a(ns(X)) -> s(a(X)) }
Weak Trs:
{ d(0()) -> 0()
, f(X1, X2) -> nf(X1, X2)
, f(0(), X) -> nil()
, a(X) -> X
, a(nf(X1, X2)) -> f(a(X1), a(X2)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(t) = {1}, Uargs(cs) = {1}, Uargs(r) = {1}, Uargs(s) = {1},
Uargs(f) = {1, 2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[t](x1) = [1] x1 + [0]
[cs](x1, x2) = [1] x1 + [1]
[r](x1) = [1] x1 + [1]
[q](x1) = [0]
[nt](x1) = [1] x1 + [0]
[ns](x1) = [1] x1 + [0]
[0] = [1]
[s](x1) = [1] x1 + [0]
[p](x1, x2) = [1] x1 + [1] x2 + [0]
[d](x1) = [1] x1 + [4]
[f](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
[nf](x1, x2) = [1] x1 + [1] x2 + [0]
[a](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[t(N)] = [1] N + [0]
? [2]
= [cs(r(q(N)), nt(ns(N)))]
[t(X)] = [1] X + [0]
>= [1] X + [0]
= [nt(X)]
[q(0())] = [0]
? [1]
= [0()]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [ns(X)]
[p(X, 0())] = [1] X + [1]
> [1] X + [0]
= [X]
[p(0(), X)] = [1] X + [1]
> [1] X + [0]
= [X]
[d(0())] = [5]
> [1]
= [0()]
[f(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [nf(X1, X2)]
[f(0(), X)] = [1] X + [1]
> [0]
= [nil()]
[a(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[a(nt(X))] = [1] X + [0]
>= [1] X + [0]
= [t(a(X))]
[a(ns(X))] = [1] X + [0]
>= [1] X + [0]
= [s(a(X))]
[a(nf(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [f(a(X1), a(X2))]
Further, it can be verified that all rules not oriented are covered by the weightgap condition.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ t(N) -> cs(r(q(N)), nt(ns(N)))
, t(X) -> nt(X)
, q(0()) -> 0()
, s(X) -> ns(X)
, a(nt(X)) -> t(a(X))
, a(ns(X)) -> s(a(X)) }
Weak Trs:
{ p(X, 0()) -> X
, p(0(), X) -> X
, d(0()) -> 0()
, f(X1, X2) -> nf(X1, X2)
, f(0(), X) -> nil()
, a(X) -> X
, a(nf(X1, X2)) -> f(a(X1), a(X2)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(t) = {1}, Uargs(cs) = {1}, Uargs(r) = {1}, Uargs(s) = {1},
Uargs(f) = {1, 2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[t](x1) = [1] x1 + [0]
[cs](x1, x2) = [1] x1 + [0]
[r](x1) = [1] x1 + [1]
[q](x1) = [1] x1 + [2]
[nt](x1) = [1] x1 + [0]
[ns](x1) = [1] x1 + [0]
[0] = [0]
[s](x1) = [1] x1 + [0]
[p](x1, x2) = [1] x1 + [1] x2 + [0]
[d](x1) = [4]
[f](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
[nf](x1, x2) = [1] x1 + [1] x2 + [0]
[a](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[t(N)] = [1] N + [0]
? [1] N + [3]
= [cs(r(q(N)), nt(ns(N)))]
[t(X)] = [1] X + [0]
>= [1] X + [0]
= [nt(X)]
[q(0())] = [2]
> [0]
= [0()]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [ns(X)]
[p(X, 0())] = [1] X + [0]
>= [1] X + [0]
= [X]
[p(0(), X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[d(0())] = [4]
> [0]
= [0()]
[f(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [nf(X1, X2)]
[f(0(), X)] = [1] X + [0]
>= [0]
= [nil()]
[a(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[a(nt(X))] = [1] X + [0]
>= [1] X + [0]
= [t(a(X))]
[a(ns(X))] = [1] X + [0]
>= [1] X + [0]
= [s(a(X))]
[a(nf(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [f(a(X1), a(X2))]
Further, it can be verified that all rules not oriented are covered by the weightgap condition.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ t(N) -> cs(r(q(N)), nt(ns(N)))
, t(X) -> nt(X)
, s(X) -> ns(X)
, a(nt(X)) -> t(a(X))
, a(ns(X)) -> s(a(X)) }
Weak Trs:
{ q(0()) -> 0()
, p(X, 0()) -> X
, p(0(), X) -> X
, d(0()) -> 0()
, f(X1, X2) -> nf(X1, X2)
, f(0(), X) -> nil()
, a(X) -> X
, a(nf(X1, X2)) -> f(a(X1), a(X2)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(t) = {1}, Uargs(cs) = {1}, Uargs(r) = {1}, Uargs(s) = {1},
Uargs(f) = {1, 2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[t](x1) = [1] x1 + [4]
[cs](x1, x2) = [1] x1 + [1]
[r](x1) = [1] x1 + [1]
[q](x1) = [1] x1 + [0]
[nt](x1) = [1] x1 + [0]
[ns](x1) = [1] x1 + [0]
[0] = [0]
[s](x1) = [1] x1 + [0]
[p](x1, x2) = [1] x1 + [1] x2 + [2]
[d](x1) = [1] x1 + [4]
[f](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
[nf](x1, x2) = [1] x1 + [1] x2 + [0]
[a](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[t(N)] = [1] N + [4]
> [1] N + [2]
= [cs(r(q(N)), nt(ns(N)))]
[t(X)] = [1] X + [4]
> [1] X + [0]
= [nt(X)]
[q(0())] = [0]
>= [0]
= [0()]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [ns(X)]
[p(X, 0())] = [1] X + [2]
> [1] X + [0]
= [X]
[p(0(), X)] = [1] X + [2]
> [1] X + [0]
= [X]
[d(0())] = [4]
> [0]
= [0()]
[f(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [nf(X1, X2)]
[f(0(), X)] = [1] X + [0]
>= [0]
= [nil()]
[a(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[a(nt(X))] = [1] X + [0]
? [1] X + [4]
= [t(a(X))]
[a(ns(X))] = [1] X + [0]
>= [1] X + [0]
= [s(a(X))]
[a(nf(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [f(a(X1), a(X2))]
Further, it can be verified that all rules not oriented are covered by the weightgap condition.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ s(X) -> ns(X)
, a(nt(X)) -> t(a(X))
, a(ns(X)) -> s(a(X)) }
Weak Trs:
{ t(N) -> cs(r(q(N)), nt(ns(N)))
, t(X) -> nt(X)
, q(0()) -> 0()
, p(X, 0()) -> X
, p(0(), X) -> X
, d(0()) -> 0()
, f(X1, X2) -> nf(X1, X2)
, f(0(), X) -> nil()
, a(X) -> X
, a(nf(X1, X2)) -> f(a(X1), a(X2)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs: { a(ns(X)) -> s(a(X)) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(t) = {1}, Uargs(cs) = {1}, Uargs(r) = {1}, Uargs(s) = {1},
Uargs(f) = {1, 2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[t](x1) = [1] x1 + [0]
[cs](x1, x2) = [1] x1 + [0]
[r](x1) = [1] x1 + [0]
[q](x1) = [0]
[nt](x1) = [1] x1 + [0]
[ns](x1) = [1] x1 + [4]
[0] = [0]
[s](x1) = [1] x1 + [4]
[p](x1, x2) = [4] x1 + [4] x2 + [0]
[d](x1) = [5]
[f](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
[nf](x1, x2) = [1] x1 + [1] x2 + [0]
[a](x1) = [2] x1 + [0]
The order satisfies the following ordering constraints:
[t(N)] = [1] N + [0]
>= [0]
= [cs(r(q(N)), nt(ns(N)))]
[t(X)] = [1] X + [0]
>= [1] X + [0]
= [nt(X)]
[q(0())] = [0]
>= [0]
= [0()]
[s(X)] = [1] X + [4]
>= [1] X + [4]
= [ns(X)]
[p(X, 0())] = [4] X + [0]
>= [1] X + [0]
= [X]
[p(0(), X)] = [4] X + [0]
>= [1] X + [0]
= [X]
[d(0())] = [5]
> [0]
= [0()]
[f(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [nf(X1, X2)]
[f(0(), X)] = [1] X + [0]
>= [0]
= [nil()]
[a(X)] = [2] X + [0]
>= [1] X + [0]
= [X]
[a(nt(X))] = [2] X + [0]
>= [2] X + [0]
= [t(a(X))]
[a(ns(X))] = [2] X + [8]
> [2] X + [4]
= [s(a(X))]
[a(nf(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [f(a(X1), a(X2))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ s(X) -> ns(X)
, a(nt(X)) -> t(a(X)) }
Weak Trs:
{ t(N) -> cs(r(q(N)), nt(ns(N)))
, t(X) -> nt(X)
, q(0()) -> 0()
, p(X, 0()) -> X
, p(0(), X) -> X
, d(0()) -> 0()
, f(X1, X2) -> nf(X1, X2)
, f(0(), X) -> nil()
, a(X) -> X
, a(ns(X)) -> s(a(X))
, a(nf(X1, X2)) -> f(a(X1), a(X2)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs: { s(X) -> ns(X) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(t) = {1}, Uargs(cs) = {1}, Uargs(r) = {1}, Uargs(s) = {1},
Uargs(f) = {1, 2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[t](x1) = [1] x1 + [0]
[cs](x1, x2) = [1] x1 + [0]
[r](x1) = [1] x1 + [0]
[q](x1) = [0]
[nt](x1) = [1] x1 + [0]
[ns](x1) = [1] x1 + [4]
[0] = [0]
[s](x1) = [1] x1 + [7]
[p](x1, x2) = [4] x1 + [4] x2 + [0]
[d](x1) = [5]
[f](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
[nf](x1, x2) = [1] x1 + [1] x2 + [0]
[a](x1) = [2] x1 + [0]
The order satisfies the following ordering constraints:
[t(N)] = [1] N + [0]
>= [0]
= [cs(r(q(N)), nt(ns(N)))]
[t(X)] = [1] X + [0]
>= [1] X + [0]
= [nt(X)]
[q(0())] = [0]
>= [0]
= [0()]
[s(X)] = [1] X + [7]
> [1] X + [4]
= [ns(X)]
[p(X, 0())] = [4] X + [0]
>= [1] X + [0]
= [X]
[p(0(), X)] = [4] X + [0]
>= [1] X + [0]
= [X]
[d(0())] = [5]
> [0]
= [0()]
[f(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [nf(X1, X2)]
[f(0(), X)] = [1] X + [0]
>= [0]
= [nil()]
[a(X)] = [2] X + [0]
>= [1] X + [0]
= [X]
[a(nt(X))] = [2] X + [0]
>= [2] X + [0]
= [t(a(X))]
[a(ns(X))] = [2] X + [8]
> [2] X + [7]
= [s(a(X))]
[a(nf(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [f(a(X1), a(X2))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs: { a(nt(X)) -> t(a(X)) }
Weak Trs:
{ t(N) -> cs(r(q(N)), nt(ns(N)))
, t(X) -> nt(X)
, q(0()) -> 0()
, s(X) -> ns(X)
, p(X, 0()) -> X
, p(0(), X) -> X
, d(0()) -> 0()
, f(X1, X2) -> nf(X1, X2)
, f(0(), X) -> nil()
, a(X) -> X
, a(ns(X)) -> s(a(X))
, a(nf(X1, X2)) -> f(a(X1), a(X2)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs: { a(nt(X)) -> t(a(X)) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(t) = {1}, Uargs(cs) = {1}, Uargs(r) = {1}, Uargs(s) = {1},
Uargs(f) = {1, 2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[t](x1) = [1] x1 + [1]
[cs](x1, x2) = [1] x1 + [0]
[r](x1) = [1] x1 + [1]
[q](x1) = [0]
[nt](x1) = [1] x1 + [1]
[ns](x1) = [1] x1 + [0]
[0] = [0]
[s](x1) = [1] x1 + [0]
[p](x1, x2) = [4] x1 + [4] x2 + [0]
[d](x1) = [4]
[f](x1, x2) = [1] x1 + [1] x2 + [1]
[nil] = [0]
[nf](x1, x2) = [1] x1 + [1] x2 + [1]
[a](x1) = [5] x1 + [3]
The order satisfies the following ordering constraints:
[t(N)] = [1] N + [1]
>= [1]
= [cs(r(q(N)), nt(ns(N)))]
[t(X)] = [1] X + [1]
>= [1] X + [1]
= [nt(X)]
[q(0())] = [0]
>= [0]
= [0()]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [ns(X)]
[p(X, 0())] = [4] X + [0]
>= [1] X + [0]
= [X]
[p(0(), X)] = [4] X + [0]
>= [1] X + [0]
= [X]
[d(0())] = [4]
> [0]
= [0()]
[f(X1, X2)] = [1] X1 + [1] X2 + [1]
>= [1] X1 + [1] X2 + [1]
= [nf(X1, X2)]
[f(0(), X)] = [1] X + [1]
> [0]
= [nil()]
[a(X)] = [5] X + [3]
> [1] X + [0]
= [X]
[a(nt(X))] = [5] X + [8]
> [5] X + [4]
= [t(a(X))]
[a(ns(X))] = [5] X + [3]
>= [5] X + [3]
= [s(a(X))]
[a(nf(X1, X2))] = [5] X1 + [5] X2 + [8]
> [5] X1 + [5] X2 + [7]
= [f(a(X1), a(X2))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ t(N) -> cs(r(q(N)), nt(ns(N)))
, t(X) -> nt(X)
, q(0()) -> 0()
, s(X) -> ns(X)
, p(X, 0()) -> X
, p(0(), X) -> X
, d(0()) -> 0()
, f(X1, X2) -> nf(X1, X2)
, f(0(), X) -> nil()
, a(X) -> X
, a(nt(X)) -> t(a(X))
, a(ns(X)) -> s(a(X))
, a(nf(X1, X2)) -> f(a(X1), a(X2)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^1))