We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ fact(X) ->
if(zero(X), n__s(n__0()), n__prod(X, n__fact(n__p(X))))
, fact(X) -> n__fact(X)
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y)
, zero(0()) -> true()
, zero(s(X)) -> false()
, add(0(), X) -> X
, add(s(X), Y) -> s(add(X, Y))
, 0() -> n__0()
, s(X) -> n__s(X)
, prod(X1, X2) -> n__prod(X1, X2)
, prod(0(), X) -> 0()
, prod(s(X), Y) -> add(Y, prod(X, Y))
, activate(X) -> X
, activate(n__s(X)) -> s(activate(X))
, activate(n__0()) -> 0()
, activate(n__prod(X1, X2)) -> prod(activate(X1), activate(X2))
, activate(n__fact(X)) -> fact(activate(X))
, activate(n__p(X)) -> p(activate(X))
, p(X) -> n__p(X)
, p(s(X)) -> X }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
Arguments of following rules are not normal-forms:
{ zero(0()) -> true()
, zero(s(X)) -> false()
, add(0(), X) -> X
, add(s(X), Y) -> s(add(X, Y))
, prod(0(), X) -> 0()
, prod(s(X), Y) -> add(Y, prod(X, Y))
, p(s(X)) -> X }
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:
{ fact(X) ->
if(zero(X), n__s(n__0()), n__prod(X, n__fact(n__p(X))))
, fact(X) -> n__fact(X)
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y)
, 0() -> n__0()
, s(X) -> n__s(X)
, prod(X1, X2) -> n__prod(X1, X2)
, activate(X) -> X
, activate(n__s(X)) -> s(activate(X))
, activate(n__0()) -> 0()
, activate(n__prod(X1, X2)) -> prod(activate(X1), activate(X2))
, activate(n__fact(X)) -> fact(activate(X))
, activate(n__p(X)) -> p(activate(X))
, p(X) -> n__p(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We add the following dependency tuples:
Strict DPs:
{ fact^#(X) ->
c_1(if^#(zero(X), n__s(n__0()), n__prod(X, n__fact(n__p(X)))))
, fact^#(X) -> c_2()
, if^#(true(), X, Y) -> c_3(activate^#(X))
, if^#(false(), X, Y) -> c_4(activate^#(Y))
, activate^#(X) -> c_8()
, activate^#(n__s(X)) -> c_9(s^#(activate(X)), activate^#(X))
, activate^#(n__0()) -> c_10(0^#())
, activate^#(n__prod(X1, X2)) ->
c_11(prod^#(activate(X1), activate(X2)),
activate^#(X1),
activate^#(X2))
, activate^#(n__fact(X)) ->
c_12(fact^#(activate(X)), activate^#(X))
, activate^#(n__p(X)) -> c_13(p^#(activate(X)), activate^#(X))
, 0^#() -> c_5()
, s^#(X) -> c_6()
, prod^#(X1, X2) -> c_7()
, p^#(X) -> c_14() }
and mark the set of starting terms.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ fact^#(X) ->
c_1(if^#(zero(X), n__s(n__0()), n__prod(X, n__fact(n__p(X)))))
, fact^#(X) -> c_2()
, if^#(true(), X, Y) -> c_3(activate^#(X))
, if^#(false(), X, Y) -> c_4(activate^#(Y))
, activate^#(X) -> c_8()
, activate^#(n__s(X)) -> c_9(s^#(activate(X)), activate^#(X))
, activate^#(n__0()) -> c_10(0^#())
, activate^#(n__prod(X1, X2)) ->
c_11(prod^#(activate(X1), activate(X2)),
activate^#(X1),
activate^#(X2))
, activate^#(n__fact(X)) ->
c_12(fact^#(activate(X)), activate^#(X))
, activate^#(n__p(X)) -> c_13(p^#(activate(X)), activate^#(X))
, 0^#() -> c_5()
, s^#(X) -> c_6()
, prod^#(X1, X2) -> c_7()
, p^#(X) -> c_14() }
Weak Trs:
{ fact(X) ->
if(zero(X), n__s(n__0()), n__prod(X, n__fact(n__p(X))))
, fact(X) -> n__fact(X)
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y)
, 0() -> n__0()
, s(X) -> n__s(X)
, prod(X1, X2) -> n__prod(X1, X2)
, activate(X) -> X
, activate(n__s(X)) -> s(activate(X))
, activate(n__0()) -> 0()
, activate(n__prod(X1, X2)) -> prod(activate(X1), activate(X2))
, activate(n__fact(X)) -> fact(activate(X))
, activate(n__p(X)) -> p(activate(X))
, p(X) -> n__p(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We estimate the number of application of {1,2,5,11,12,13,14} by
applications of Pre({1,2,5,11,12,13,14}) = {3,4,6,7,8,9,10}. Here
rules are labeled as follows:
DPs:
{ 1: fact^#(X) ->
c_1(if^#(zero(X), n__s(n__0()), n__prod(X, n__fact(n__p(X)))))
, 2: fact^#(X) -> c_2()
, 3: if^#(true(), X, Y) -> c_3(activate^#(X))
, 4: if^#(false(), X, Y) -> c_4(activate^#(Y))
, 5: activate^#(X) -> c_8()
, 6: activate^#(n__s(X)) -> c_9(s^#(activate(X)), activate^#(X))
, 7: activate^#(n__0()) -> c_10(0^#())
, 8: activate^#(n__prod(X1, X2)) ->
c_11(prod^#(activate(X1), activate(X2)),
activate^#(X1),
activate^#(X2))
, 9: activate^#(n__fact(X)) ->
c_12(fact^#(activate(X)), activate^#(X))
, 10: activate^#(n__p(X)) -> c_13(p^#(activate(X)), activate^#(X))
, 11: 0^#() -> c_5()
, 12: s^#(X) -> c_6()
, 13: prod^#(X1, X2) -> c_7()
, 14: p^#(X) -> c_14() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ if^#(true(), X, Y) -> c_3(activate^#(X))
, if^#(false(), X, Y) -> c_4(activate^#(Y))
, activate^#(n__s(X)) -> c_9(s^#(activate(X)), activate^#(X))
, activate^#(n__0()) -> c_10(0^#())
, activate^#(n__prod(X1, X2)) ->
c_11(prod^#(activate(X1), activate(X2)),
activate^#(X1),
activate^#(X2))
, activate^#(n__fact(X)) ->
c_12(fact^#(activate(X)), activate^#(X))
, activate^#(n__p(X)) -> c_13(p^#(activate(X)), activate^#(X)) }
Weak DPs:
{ fact^#(X) ->
c_1(if^#(zero(X), n__s(n__0()), n__prod(X, n__fact(n__p(X)))))
, fact^#(X) -> c_2()
, activate^#(X) -> c_8()
, 0^#() -> c_5()
, s^#(X) -> c_6()
, prod^#(X1, X2) -> c_7()
, p^#(X) -> c_14() }
Weak Trs:
{ fact(X) ->
if(zero(X), n__s(n__0()), n__prod(X, n__fact(n__p(X))))
, fact(X) -> n__fact(X)
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y)
, 0() -> n__0()
, s(X) -> n__s(X)
, prod(X1, X2) -> n__prod(X1, X2)
, activate(X) -> X
, activate(n__s(X)) -> s(activate(X))
, activate(n__0()) -> 0()
, activate(n__prod(X1, X2)) -> prod(activate(X1), activate(X2))
, activate(n__fact(X)) -> fact(activate(X))
, activate(n__p(X)) -> p(activate(X))
, p(X) -> n__p(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We estimate the number of application of {4} by applications of
Pre({4}) = {1,2,3,5,6,7}. Here rules are labeled as follows:
DPs:
{ 1: if^#(true(), X, Y) -> c_3(activate^#(X))
, 2: if^#(false(), X, Y) -> c_4(activate^#(Y))
, 3: activate^#(n__s(X)) -> c_9(s^#(activate(X)), activate^#(X))
, 4: activate^#(n__0()) -> c_10(0^#())
, 5: activate^#(n__prod(X1, X2)) ->
c_11(prod^#(activate(X1), activate(X2)),
activate^#(X1),
activate^#(X2))
, 6: activate^#(n__fact(X)) ->
c_12(fact^#(activate(X)), activate^#(X))
, 7: activate^#(n__p(X)) -> c_13(p^#(activate(X)), activate^#(X))
, 8: fact^#(X) ->
c_1(if^#(zero(X), n__s(n__0()), n__prod(X, n__fact(n__p(X)))))
, 9: fact^#(X) -> c_2()
, 10: activate^#(X) -> c_8()
, 11: 0^#() -> c_5()
, 12: s^#(X) -> c_6()
, 13: prod^#(X1, X2) -> c_7()
, 14: p^#(X) -> c_14() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ if^#(true(), X, Y) -> c_3(activate^#(X))
, if^#(false(), X, Y) -> c_4(activate^#(Y))
, activate^#(n__s(X)) -> c_9(s^#(activate(X)), activate^#(X))
, activate^#(n__prod(X1, X2)) ->
c_11(prod^#(activate(X1), activate(X2)),
activate^#(X1),
activate^#(X2))
, activate^#(n__fact(X)) ->
c_12(fact^#(activate(X)), activate^#(X))
, activate^#(n__p(X)) -> c_13(p^#(activate(X)), activate^#(X)) }
Weak DPs:
{ fact^#(X) ->
c_1(if^#(zero(X), n__s(n__0()), n__prod(X, n__fact(n__p(X)))))
, fact^#(X) -> c_2()
, activate^#(X) -> c_8()
, activate^#(n__0()) -> c_10(0^#())
, 0^#() -> c_5()
, s^#(X) -> c_6()
, prod^#(X1, X2) -> c_7()
, p^#(X) -> c_14() }
Weak Trs:
{ fact(X) ->
if(zero(X), n__s(n__0()), n__prod(X, n__fact(n__p(X))))
, fact(X) -> n__fact(X)
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y)
, 0() -> n__0()
, s(X) -> n__s(X)
, prod(X1, X2) -> n__prod(X1, X2)
, activate(X) -> X
, activate(n__s(X)) -> s(activate(X))
, activate(n__0()) -> 0()
, activate(n__prod(X1, X2)) -> prod(activate(X1), activate(X2))
, activate(n__fact(X)) -> fact(activate(X))
, activate(n__p(X)) -> p(activate(X))
, p(X) -> n__p(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ fact^#(X) ->
c_1(if^#(zero(X), n__s(n__0()), n__prod(X, n__fact(n__p(X)))))
, fact^#(X) -> c_2()
, activate^#(X) -> c_8()
, activate^#(n__0()) -> c_10(0^#())
, 0^#() -> c_5()
, s^#(X) -> c_6()
, prod^#(X1, X2) -> c_7()
, p^#(X) -> c_14() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ if^#(true(), X, Y) -> c_3(activate^#(X))
, if^#(false(), X, Y) -> c_4(activate^#(Y))
, activate^#(n__s(X)) -> c_9(s^#(activate(X)), activate^#(X))
, activate^#(n__prod(X1, X2)) ->
c_11(prod^#(activate(X1), activate(X2)),
activate^#(X1),
activate^#(X2))
, activate^#(n__fact(X)) ->
c_12(fact^#(activate(X)), activate^#(X))
, activate^#(n__p(X)) -> c_13(p^#(activate(X)), activate^#(X)) }
Weak Trs:
{ fact(X) ->
if(zero(X), n__s(n__0()), n__prod(X, n__fact(n__p(X))))
, fact(X) -> n__fact(X)
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y)
, 0() -> n__0()
, s(X) -> n__s(X)
, prod(X1, X2) -> n__prod(X1, X2)
, activate(X) -> X
, activate(n__s(X)) -> s(activate(X))
, activate(n__0()) -> 0()
, activate(n__prod(X1, X2)) -> prod(activate(X1), activate(X2))
, activate(n__fact(X)) -> fact(activate(X))
, activate(n__p(X)) -> p(activate(X))
, p(X) -> n__p(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:
{ activate^#(n__s(X)) -> c_9(s^#(activate(X)), activate^#(X))
, activate^#(n__prod(X1, X2)) ->
c_11(prod^#(activate(X1), activate(X2)),
activate^#(X1),
activate^#(X2))
, activate^#(n__fact(X)) ->
c_12(fact^#(activate(X)), activate^#(X))
, activate^#(n__p(X)) -> c_13(p^#(activate(X)), activate^#(X)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ if^#(true(), X, Y) -> c_1(activate^#(X))
, if^#(false(), X, Y) -> c_2(activate^#(Y))
, activate^#(n__s(X)) -> c_3(activate^#(X))
, activate^#(n__prod(X1, X2)) ->
c_4(activate^#(X1), activate^#(X2))
, activate^#(n__fact(X)) -> c_5(activate^#(X))
, activate^#(n__p(X)) -> c_6(activate^#(X)) }
Weak Trs:
{ fact(X) ->
if(zero(X), n__s(n__0()), n__prod(X, n__fact(n__p(X))))
, fact(X) -> n__fact(X)
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y)
, 0() -> n__0()
, s(X) -> n__s(X)
, prod(X1, X2) -> n__prod(X1, X2)
, activate(X) -> X
, activate(n__s(X)) -> s(activate(X))
, activate(n__0()) -> 0()
, activate(n__prod(X1, X2)) -> prod(activate(X1), activate(X2))
, activate(n__fact(X)) -> fact(activate(X))
, activate(n__p(X)) -> p(activate(X))
, p(X) -> n__p(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
No rule is usable, rules are removed from the input problem.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ if^#(true(), X, Y) -> c_1(activate^#(X))
, if^#(false(), X, Y) -> c_2(activate^#(Y))
, activate^#(n__s(X)) -> c_3(activate^#(X))
, activate^#(n__prod(X1, X2)) ->
c_4(activate^#(X1), activate^#(X2))
, activate^#(n__fact(X)) -> c_5(activate^#(X))
, activate^#(n__p(X)) -> c_6(activate^#(X)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
Consider the dependency graph
1: if^#(true(), X, Y) -> c_1(activate^#(X))
-->_1 activate^#(n__p(X)) -> c_6(activate^#(X)) :6
-->_1 activate^#(n__fact(X)) -> c_5(activate^#(X)) :5
-->_1 activate^#(n__prod(X1, X2)) ->
c_4(activate^#(X1), activate^#(X2)) :4
-->_1 activate^#(n__s(X)) -> c_3(activate^#(X)) :3
2: if^#(false(), X, Y) -> c_2(activate^#(Y))
-->_1 activate^#(n__p(X)) -> c_6(activate^#(X)) :6
-->_1 activate^#(n__fact(X)) -> c_5(activate^#(X)) :5
-->_1 activate^#(n__prod(X1, X2)) ->
c_4(activate^#(X1), activate^#(X2)) :4
-->_1 activate^#(n__s(X)) -> c_3(activate^#(X)) :3
3: activate^#(n__s(X)) -> c_3(activate^#(X))
-->_1 activate^#(n__p(X)) -> c_6(activate^#(X)) :6
-->_1 activate^#(n__fact(X)) -> c_5(activate^#(X)) :5
-->_1 activate^#(n__prod(X1, X2)) ->
c_4(activate^#(X1), activate^#(X2)) :4
-->_1 activate^#(n__s(X)) -> c_3(activate^#(X)) :3
4: activate^#(n__prod(X1, X2)) ->
c_4(activate^#(X1), activate^#(X2))
-->_2 activate^#(n__p(X)) -> c_6(activate^#(X)) :6
-->_1 activate^#(n__p(X)) -> c_6(activate^#(X)) :6
-->_2 activate^#(n__fact(X)) -> c_5(activate^#(X)) :5
-->_1 activate^#(n__fact(X)) -> c_5(activate^#(X)) :5
-->_2 activate^#(n__prod(X1, X2)) ->
c_4(activate^#(X1), activate^#(X2)) :4
-->_1 activate^#(n__prod(X1, X2)) ->
c_4(activate^#(X1), activate^#(X2)) :4
-->_2 activate^#(n__s(X)) -> c_3(activate^#(X)) :3
-->_1 activate^#(n__s(X)) -> c_3(activate^#(X)) :3
5: activate^#(n__fact(X)) -> c_5(activate^#(X))
-->_1 activate^#(n__p(X)) -> c_6(activate^#(X)) :6
-->_1 activate^#(n__fact(X)) -> c_5(activate^#(X)) :5
-->_1 activate^#(n__prod(X1, X2)) ->
c_4(activate^#(X1), activate^#(X2)) :4
-->_1 activate^#(n__s(X)) -> c_3(activate^#(X)) :3
6: activate^#(n__p(X)) -> c_6(activate^#(X))
-->_1 activate^#(n__p(X)) -> c_6(activate^#(X)) :6
-->_1 activate^#(n__fact(X)) -> c_5(activate^#(X)) :5
-->_1 activate^#(n__prod(X1, X2)) ->
c_4(activate^#(X1), activate^#(X2)) :4
-->_1 activate^#(n__s(X)) -> c_3(activate^#(X)) :3
Following roots of the dependency graph are removed, as the
considered set of starting terms is closed under reduction with
respect to these rules (modulo compound contexts).
{ if^#(true(), X, Y) -> c_1(activate^#(X))
, if^#(false(), X, Y) -> c_2(activate^#(Y)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ activate^#(n__s(X)) -> c_3(activate^#(X))
, activate^#(n__prod(X1, X2)) ->
c_4(activate^#(X1), activate^#(X2))
, activate^#(n__fact(X)) -> c_5(activate^#(X))
, activate^#(n__p(X)) -> c_6(activate^#(X)) }
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.
DPs:
{ 3: activate^#(n__fact(X)) -> c_5(activate^#(X))
, 4: activate^#(n__p(X)) -> c_6(activate^#(X)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_3) = {1}, Uargs(c_4) = {1, 2}, Uargs(c_5) = {1},
Uargs(c_6) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[n__s](x1) = [1] x1 + [0]
[n__prod](x1, x2) = [1] x1 + [1] x2 + [0]
[n__fact](x1) = [1] x1 + [2]
[n__p](x1) = [1] x1 + [2]
[activate^#](x1) = [4] x1 + [0]
[c_3](x1) = [1] x1 + [0]
[c_4](x1, x2) = [1] x1 + [1] x2 + [0]
[c_5](x1) = [1] x1 + [0]
[c_6](x1) = [1] x1 + [1]
The order satisfies the following ordering constraints:
[activate^#(n__s(X))] = [4] X + [0]
>= [4] X + [0]
= [c_3(activate^#(X))]
[activate^#(n__prod(X1, X2))] = [4] X1 + [4] X2 + [0]
>= [4] X1 + [4] X2 + [0]
= [c_4(activate^#(X1), activate^#(X2))]
[activate^#(n__fact(X))] = [4] X + [8]
> [4] X + [0]
= [c_5(activate^#(X))]
[activate^#(n__p(X))] = [4] X + [8]
> [4] X + [1]
= [c_6(activate^#(X))]
The strictly oriented rules are moved into the weak component.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ activate^#(n__s(X)) -> c_3(activate^#(X))
, activate^#(n__prod(X1, X2)) ->
c_4(activate^#(X1), activate^#(X2)) }
Weak DPs:
{ activate^#(n__fact(X)) -> c_5(activate^#(X))
, activate^#(n__p(X)) -> c_6(activate^#(X)) }
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.
DPs:
{ 1: activate^#(n__s(X)) -> c_3(activate^#(X))
, 2: activate^#(n__prod(X1, X2)) ->
c_4(activate^#(X1), activate^#(X2))
, 3: activate^#(n__fact(X)) -> c_5(activate^#(X))
, 4: activate^#(n__p(X)) -> c_6(activate^#(X)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_3) = {1}, Uargs(c_4) = {1, 2}, Uargs(c_5) = {1},
Uargs(c_6) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[n__s](x1) = [1] x1 + [4]
[n__prod](x1, x2) = [1] x1 + [1] x2 + [4]
[n__fact](x1) = [1] x1 + [4]
[n__p](x1) = [1] x1 + [4]
[activate^#](x1) = [2] x1 + [0]
[c_3](x1) = [1] x1 + [0]
[c_4](x1, x2) = [1] x1 + [1] x2 + [1]
[c_5](x1) = [1] x1 + [1]
[c_6](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[activate^#(n__s(X))] = [2] X + [8]
> [2] X + [0]
= [c_3(activate^#(X))]
[activate^#(n__prod(X1, X2))] = [2] X1 + [2] X2 + [8]
> [2] X1 + [2] X2 + [1]
= [c_4(activate^#(X1), activate^#(X2))]
[activate^#(n__fact(X))] = [2] X + [8]
> [2] X + [1]
= [c_5(activate^#(X))]
[activate^#(n__p(X))] = [2] X + [8]
> [2] X + [0]
= [c_6(activate^#(X))]
The strictly oriented rules are moved into the weak component.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak DPs:
{ activate^#(n__s(X)) -> c_3(activate^#(X))
, activate^#(n__prod(X1, X2)) ->
c_4(activate^#(X1), activate^#(X2))
, activate^#(n__fact(X)) -> c_5(activate^#(X))
, activate^#(n__p(X)) -> c_6(activate^#(X)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ activate^#(n__s(X)) -> c_3(activate^#(X))
, activate^#(n__prod(X1, X2)) ->
c_4(activate^#(X1), activate^#(X2))
, activate^#(n__fact(X)) -> c_5(activate^#(X))
, activate^#(n__p(X)) -> c_6(activate^#(X)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Rules: Empty
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^1))