MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ sel(s(X), cons(Y, Z)) -> sel(X, Z)
, sel(0(), cons(X, Z)) -> X
, first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, first(0(), Z) -> nil()
, from(X) -> cons(X, from(s(X)))
, sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
, sel1(0(), cons(X, Z)) -> quote(X)
, quote(sel(X, Z)) -> sel1(X, Z)
, quote(s(X)) -> s1(quote(X))
, quote(0()) -> 01()
, first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, Z))
, first1(0(), Z) -> nil1()
, quote1(cons(X, Z)) -> cons1(quote(X), quote1(Z))
, quote1(first(X, Z)) -> first1(X, Z)
, quote1(nil()) -> nil1()
, unquote(01()) -> 0()
, unquote(s1(X)) -> s(unquote(X))
, unquote1(nil1()) -> nil()
, unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z))
, fcons(X, Z) -> cons(X, Z) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
None of the processors succeeded.
Details of failed attempt(s):
-----------------------------
1) 'empty' failed due to the following reason:
Empty strict component of the problem is NOT empty.
2) 'Best' failed due to the following reason:
None of the processors succeeded.
Details of failed attempt(s):
-----------------------------
1) 'WithProblem (timeout of 60 seconds)' failed due to the
following reason:
We add the following innermost weak dependency pairs:
Strict DPs:
{ sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z))
, sel^#(0(), cons(X, Z)) -> c_2()
, first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z))
, first^#(0(), Z) -> c_4()
, from^#(X) -> c_5(from^#(s(X)))
, sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z))
, sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X))
, quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z))
, quote^#(s(X)) -> c_9(quote^#(X))
, quote^#(0()) -> c_10()
, first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z))
, first1^#(0(), Z) -> c_12()
, quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z))
, quote1^#(first(X, Z)) -> c_14(first1^#(X, Z))
, quote1^#(nil()) -> c_15()
, unquote^#(01()) -> c_16()
, unquote^#(s1(X)) -> c_17(unquote^#(X))
, unquote1^#(nil1()) -> c_18()
, unquote1^#(cons1(X, Z)) -> c_19(fcons^#(unquote(X), unquote1(Z)))
, fcons^#(X, Z) -> c_20() }
and mark the set of starting terms.
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict DPs:
{ sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z))
, sel^#(0(), cons(X, Z)) -> c_2()
, first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z))
, first^#(0(), Z) -> c_4()
, from^#(X) -> c_5(from^#(s(X)))
, sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z))
, sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X))
, quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z))
, quote^#(s(X)) -> c_9(quote^#(X))
, quote^#(0()) -> c_10()
, first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z))
, first1^#(0(), Z) -> c_12()
, quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z))
, quote1^#(first(X, Z)) -> c_14(first1^#(X, Z))
, quote1^#(nil()) -> c_15()
, unquote^#(01()) -> c_16()
, unquote^#(s1(X)) -> c_17(unquote^#(X))
, unquote1^#(nil1()) -> c_18()
, unquote1^#(cons1(X, Z)) -> c_19(fcons^#(unquote(X), unquote1(Z)))
, fcons^#(X, Z) -> c_20() }
Strict Trs:
{ sel(s(X), cons(Y, Z)) -> sel(X, Z)
, sel(0(), cons(X, Z)) -> X
, first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, first(0(), Z) -> nil()
, from(X) -> cons(X, from(s(X)))
, sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
, sel1(0(), cons(X, Z)) -> quote(X)
, quote(sel(X, Z)) -> sel1(X, Z)
, quote(s(X)) -> s1(quote(X))
, quote(0()) -> 01()
, first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, Z))
, first1(0(), Z) -> nil1()
, quote1(cons(X, Z)) -> cons1(quote(X), quote1(Z))
, quote1(first(X, Z)) -> first1(X, Z)
, quote1(nil()) -> nil1()
, unquote(01()) -> 0()
, unquote(s1(X)) -> s(unquote(X))
, unquote1(nil1()) -> nil()
, unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z))
, fcons(X, Z) -> cons(X, Z) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
We replace rewrite rules by usable rules:
Strict Usable Rules:
{ unquote(01()) -> 0()
, unquote(s1(X)) -> s(unquote(X))
, unquote1(nil1()) -> nil()
, unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z))
, fcons(X, Z) -> cons(X, Z) }
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict DPs:
{ sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z))
, sel^#(0(), cons(X, Z)) -> c_2()
, first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z))
, first^#(0(), Z) -> c_4()
, from^#(X) -> c_5(from^#(s(X)))
, sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z))
, sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X))
, quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z))
, quote^#(s(X)) -> c_9(quote^#(X))
, quote^#(0()) -> c_10()
, first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z))
, first1^#(0(), Z) -> c_12()
, quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z))
, quote1^#(first(X, Z)) -> c_14(first1^#(X, Z))
, quote1^#(nil()) -> c_15()
, unquote^#(01()) -> c_16()
, unquote^#(s1(X)) -> c_17(unquote^#(X))
, unquote1^#(nil1()) -> c_18()
, unquote1^#(cons1(X, Z)) -> c_19(fcons^#(unquote(X), unquote1(Z)))
, fcons^#(X, Z) -> c_20() }
Strict Trs:
{ unquote(01()) -> 0()
, unquote(s1(X)) -> s(unquote(X))
, unquote1(nil1()) -> nil()
, unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z))
, fcons(X, Z) -> cons(X, Z) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The weightgap principle applies (using the following constant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(s) = {1}, Uargs(fcons) = {1, 2}, Uargs(c_1) = {1},
Uargs(c_3) = {1}, Uargs(c_5) = {1}, Uargs(c_6) = {1},
Uargs(c_7) = {1}, Uargs(c_8) = {1}, Uargs(c_9) = {1},
Uargs(c_11) = {1, 2}, Uargs(c_13) = {1, 2}, Uargs(c_14) = {1},
Uargs(c_17) = {1}, Uargs(c_19) = {1}, Uargs(fcons^#) = {1, 2}
TcT has computed the following constructor-restricted matrix
interpretation.
[sel](x1, x2) = [2 2] x1 + [2 2] x2 + [0]
[1 1] [1 1] [0]
[s](x1) = [1 0] x1 + [0]
[0 0] [0]
[cons](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
[0] = [0]
[0]
[first](x1, x2) = [2 2] x1 + [2 1] x2 + [0]
[1 1] [1 2] [0]
[nil] = [0]
[0]
[nil1] = [0]
[0]
[cons1](x1, x2) = [1 0] x1 + [1 0] x2 + [2]
[0 0] [0 0] [2]
[01] = [0]
[0]
[s1](x1) = [1 0] x1 + [2]
[0 0] [0]
[unquote](x1) = [2 0] x1 + [1]
[0 0] [0]
[unquote1](x1) = [2 0] x1 + [1]
[0 2] [0]
[fcons](x1, x2) = [1 0] x1 + [1 0] x2 + [2]
[0 0] [0 0] [0]
[sel^#](x1, x2) = [0 0] x1 + [0 0] x2 + [2]
[1 2] [2 2] [2]
[c_1](x1) = [1 0] x1 + [1]
[0 1] [1]
[c_2] = [2]
[1]
[first^#](x1, x2) = [0 0] x1 + [0 0] x2 + [2]
[1 1] [2 2] [2]
[c_3](x1) = [1 0] x1 + [1]
[0 1] [1]
[c_4] = [1]
[1]
[from^#](x1) = [1 2] x1 + [2]
[2 1] [2]
[c_5](x1) = [1 0] x1 + [2]
[0 1] [2]
[sel1^#](x1, x2) = [0 0] x1 + [0 0] x2 + [2]
[1 2] [1 2] [2]
[c_6](x1) = [1 0] x1 + [1]
[0 1] [1]
[c_7](x1) = [1 0] x1 + [2]
[0 1] [2]
[quote^#](x1) = [0]
[0]
[c_8](x1) = [1 0] x1 + [1]
[0 1] [1]
[c_9](x1) = [1 0] x1 + [2]
[0 1] [2]
[c_10] = [1]
[1]
[first1^#](x1, x2) = [0 0] x1 + [0 0] x2 + [2]
[1 1] [2 2] [2]
[c_11](x1, x2) = [1 0] x1 + [1 0] x2 + [1]
[0 1] [0 1] [1]
[c_12] = [1]
[1]
[quote1^#](x1) = [0]
[0]
[c_13](x1, x2) = [1 0] x1 + [1 0] x2 + [2]
[0 1] [0 1] [2]
[c_14](x1) = [1 0] x1 + [1]
[0 1] [1]
[c_15] = [1]
[1]
[unquote^#](x1) = [2 0] x1 + [2]
[1 1] [2]
[c_16] = [1]
[1]
[c_17](x1) = [1 0] x1 + [1]
[0 1] [1]
[unquote1^#](x1) = [2 1] x1 + [1]
[1 1] [2]
[c_18] = [0]
[1]
[c_19](x1) = [1 0] x1 + [2]
[0 1] [2]
[fcons^#](x1, x2) = [1 0] x1 + [1 0] x2 + [2]
[0 0] [0 0] [2]
[c_20] = [1]
[1]
The following symbols are considered usable
{unquote, unquote1, fcons, sel^#, first^#, from^#, sel1^#, quote^#,
first1^#, quote1^#, unquote^#, unquote1^#, fcons^#}
The order satisfies the following ordering constraints:
[unquote(01())] = [1]
[0]
> [0]
[0]
= [0()]
[unquote(s1(X))] = [2 0] X + [5]
[0 0] [0]
> [2 0] X + [1]
[0 0] [0]
= [s(unquote(X))]
[unquote1(nil1())] = [1]
[0]
> [0]
[0]
= [nil()]
[unquote1(cons1(X, Z))] = [2 0] X + [2 0] Z + [5]
[0 0] [0 0] [4]
> [2 0] X + [2 0] Z + [4]
[0 0] [0 0] [0]
= [fcons(unquote(X), unquote1(Z))]
[fcons(X, Z)] = [1 0] X + [1 0] Z + [2]
[0 0] [0 0] [0]
> [1 0] X + [1 0] Z + [0]
[0 0] [0 0] [0]
= [cons(X, Z)]
[sel^#(s(X), cons(Y, Z))] = [0 0] X + [0 0] Y + [0 0] Z + [2]
[1 0] [2 0] [2 0] [2]
? [0 0] X + [0 0] Z + [3]
[1 2] [2 2] [3]
= [c_1(sel^#(X, Z))]
[sel^#(0(), cons(X, Z))] = [0 0] X + [0 0] Z + [2]
[2 0] [2 0] [2]
>= [2]
[1]
= [c_2()]
[first^#(s(X), cons(Y, Z))] = [0 0] X + [0 0] Y + [0 0] Z + [2]
[1 0] [2 0] [2 0] [2]
? [0 0] X + [0 0] Z + [3]
[1 1] [2 2] [3]
= [c_3(first^#(X, Z))]
[first^#(0(), Z)] = [0 0] Z + [2]
[2 2] [2]
> [1]
[1]
= [c_4()]
[from^#(X)] = [1 2] X + [2]
[2 1] [2]
? [1 0] X + [4]
[2 0] [4]
= [c_5(from^#(s(X)))]
[sel1^#(s(X), cons(Y, Z))] = [0 0] X + [0 0] Y + [0 0] Z + [2]
[1 0] [1 0] [1 0] [2]
? [0 0] X + [0 0] Z + [3]
[1 2] [1 2] [3]
= [c_6(sel1^#(X, Z))]
[sel1^#(0(), cons(X, Z))] = [0 0] X + [0 0] Z + [2]
[1 0] [1 0] [2]
>= [2]
[2]
= [c_7(quote^#(X))]
[quote^#(sel(X, Z))] = [0]
[0]
? [0 0] X + [0 0] Z + [3]
[1 2] [1 2] [3]
= [c_8(sel1^#(X, Z))]
[quote^#(s(X))] = [0]
[0]
? [2]
[2]
= [c_9(quote^#(X))]
[quote^#(0())] = [0]
[0]
? [1]
[1]
= [c_10()]
[first1^#(s(X), cons(Y, Z))] = [0 0] X + [0 0] Y + [0 0] Z + [2]
[1 0] [2 0] [2 0] [2]
? [0 0] X + [0 0] Z + [3]
[1 1] [2 2] [3]
= [c_11(quote^#(Y), first1^#(X, Z))]
[first1^#(0(), Z)] = [0 0] Z + [2]
[2 2] [2]
> [1]
[1]
= [c_12()]
[quote1^#(cons(X, Z))] = [0]
[0]
? [2]
[2]
= [c_13(quote^#(X), quote1^#(Z))]
[quote1^#(first(X, Z))] = [0]
[0]
? [0 0] X + [0 0] Z + [3]
[1 1] [2 2] [3]
= [c_14(first1^#(X, Z))]
[quote1^#(nil())] = [0]
[0]
? [1]
[1]
= [c_15()]
[unquote^#(01())] = [2]
[2]
> [1]
[1]
= [c_16()]
[unquote^#(s1(X))] = [2 0] X + [6]
[1 0] [4]
? [2 0] X + [3]
[1 1] [3]
= [c_17(unquote^#(X))]
[unquote1^#(nil1())] = [1]
[2]
> [0]
[1]
= [c_18()]
[unquote1^#(cons1(X, Z))] = [2 0] X + [2 0] Z + [7]
[1 0] [1 0] [6]
> [2 0] X + [2 0] Z + [6]
[0 0] [0 0] [4]
= [c_19(fcons^#(unquote(X), unquote1(Z)))]
[fcons^#(X, Z)] = [1 0] X + [1 0] Z + [2]
[0 0] [0 0] [2]
> [1]
[1]
= [c_20()]
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 MAYBE.
Strict DPs:
{ sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z))
, sel^#(0(), cons(X, Z)) -> c_2()
, first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z))
, from^#(X) -> c_5(from^#(s(X)))
, sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z))
, sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X))
, quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z))
, quote^#(s(X)) -> c_9(quote^#(X))
, quote^#(0()) -> c_10()
, first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z))
, quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z))
, quote1^#(first(X, Z)) -> c_14(first1^#(X, Z))
, quote1^#(nil()) -> c_15()
, unquote^#(s1(X)) -> c_17(unquote^#(X)) }
Weak DPs:
{ first^#(0(), Z) -> c_4()
, first1^#(0(), Z) -> c_12()
, unquote^#(01()) -> c_16()
, unquote1^#(nil1()) -> c_18()
, unquote1^#(cons1(X, Z)) -> c_19(fcons^#(unquote(X), unquote1(Z)))
, fcons^#(X, Z) -> c_20() }
Weak Trs:
{ unquote(01()) -> 0()
, unquote(s1(X)) -> s(unquote(X))
, unquote1(nil1()) -> nil()
, unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z))
, fcons(X, Z) -> cons(X, Z) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
We estimate the number of application of {2,9,13} by applications
of Pre({2,9,13}) = {1,6,8,10,11}. Here rules are labeled as
follows:
DPs:
{ 1: sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z))
, 2: sel^#(0(), cons(X, Z)) -> c_2()
, 3: first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z))
, 4: from^#(X) -> c_5(from^#(s(X)))
, 5: sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z))
, 6: sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X))
, 7: quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z))
, 8: quote^#(s(X)) -> c_9(quote^#(X))
, 9: quote^#(0()) -> c_10()
, 10: first1^#(s(X), cons(Y, Z)) ->
c_11(quote^#(Y), first1^#(X, Z))
, 11: quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z))
, 12: quote1^#(first(X, Z)) -> c_14(first1^#(X, Z))
, 13: quote1^#(nil()) -> c_15()
, 14: unquote^#(s1(X)) -> c_17(unquote^#(X))
, 15: first^#(0(), Z) -> c_4()
, 16: first1^#(0(), Z) -> c_12()
, 17: unquote^#(01()) -> c_16()
, 18: unquote1^#(nil1()) -> c_18()
, 19: unquote1^#(cons1(X, Z)) ->
c_19(fcons^#(unquote(X), unquote1(Z)))
, 20: fcons^#(X, Z) -> c_20() }
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict DPs:
{ sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z))
, first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z))
, from^#(X) -> c_5(from^#(s(X)))
, sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z))
, sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X))
, quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z))
, quote^#(s(X)) -> c_9(quote^#(X))
, first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z))
, quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z))
, quote1^#(first(X, Z)) -> c_14(first1^#(X, Z))
, unquote^#(s1(X)) -> c_17(unquote^#(X)) }
Weak DPs:
{ sel^#(0(), cons(X, Z)) -> c_2()
, first^#(0(), Z) -> c_4()
, quote^#(0()) -> c_10()
, first1^#(0(), Z) -> c_12()
, quote1^#(nil()) -> c_15()
, unquote^#(01()) -> c_16()
, unquote1^#(nil1()) -> c_18()
, unquote1^#(cons1(X, Z)) -> c_19(fcons^#(unquote(X), unquote1(Z)))
, fcons^#(X, Z) -> c_20() }
Weak Trs:
{ unquote(01()) -> 0()
, unquote(s1(X)) -> s(unquote(X))
, unquote1(nil1()) -> nil()
, unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z))
, fcons(X, Z) -> cons(X, Z) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ sel^#(0(), cons(X, Z)) -> c_2()
, first^#(0(), Z) -> c_4()
, quote^#(0()) -> c_10()
, first1^#(0(), Z) -> c_12()
, quote1^#(nil()) -> c_15()
, unquote^#(01()) -> c_16()
, unquote1^#(nil1()) -> c_18()
, unquote1^#(cons1(X, Z)) -> c_19(fcons^#(unquote(X), unquote1(Z)))
, fcons^#(X, Z) -> c_20() }
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict DPs:
{ sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z))
, first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z))
, from^#(X) -> c_5(from^#(s(X)))
, sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z))
, sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X))
, quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z))
, quote^#(s(X)) -> c_9(quote^#(X))
, first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z))
, quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z))
, quote1^#(first(X, Z)) -> c_14(first1^#(X, Z))
, unquote^#(s1(X)) -> c_17(unquote^#(X)) }
Weak Trs:
{ unquote(01()) -> 0()
, unquote(s1(X)) -> s(unquote(X))
, unquote1(nil1()) -> nil()
, unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z))
, fcons(X, Z) -> cons(X, Z) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
No rule is usable, rules are removed from the input problem.
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict DPs:
{ sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z))
, first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z))
, from^#(X) -> c_5(from^#(s(X)))
, sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z))
, sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X))
, quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z))
, quote^#(s(X)) -> c_9(quote^#(X))
, first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z))
, quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z))
, quote1^#(first(X, Z)) -> c_14(first1^#(X, Z))
, unquote^#(s1(X)) -> c_17(unquote^#(X)) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
None of the processors succeeded.
Details of failed attempt(s):
-----------------------------
1) 'empty' failed due to the following reason:
Empty strict component of the problem is NOT empty.
2) 'WithProblem' failed due to the following reason:
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
DPs:
{ 4: sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z))
, 5: sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X))
, 7: quote^#(s(X)) -> c_9(quote^#(X))
, 8: first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z))
, 11: unquote^#(s1(X)) -> c_17(unquote^#(X)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_1) = {1}, Uargs(c_3) = {1}, Uargs(c_5) = {1},
Uargs(c_6) = {1}, Uargs(c_7) = {1}, Uargs(c_8) = {1},
Uargs(c_9) = {1}, Uargs(c_11) = {1, 2}, Uargs(c_13) = {1, 2},
Uargs(c_14) = {1}, Uargs(c_17) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[sel](x1, x2) = [4] x1 + [4] x2 + [0]
[s](x1) = [1] x1 + [3]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[0] = [3]
[first](x1, x2) = [4] x1 + [1] x2 + [0]
[nil] = [0]
[from](x1) = [7] x1 + [0]
[sel1](x1, x2) = [7] x1 + [7] x2 + [0]
[quote](x1) = [7] x1 + [0]
[first1](x1, x2) = [7] x1 + [7] x2 + [0]
[nil1] = [0]
[cons1](x1, x2) = [1] x1 + [1] x2 + [0]
[01] = [0]
[quote1](x1) = [7] x1 + [0]
[s1](x1) = [1] x1 + [4]
[unquote](x1) = [7] x1 + [0]
[unquote1](x1) = [7] x1 + [0]
[fcons](x1, x2) = [7] x1 + [7] x2 + [0]
[sel^#](x1, x2) = [0]
[c_1](x1) = [1] x1 + [0]
[c_2] = [0]
[first^#](x1, x2) = [0]
[c_3](x1) = [4] x1 + [0]
[c_4] = [0]
[from^#](x1) = [0]
[c_5](x1) = [1] x1 + [0]
[sel1^#](x1, x2) = [3] x1 + [4] x2 + [0]
[c_6](x1) = [1] x1 + [7]
[c_7](x1) = [2] x1 + [3]
[quote^#](x1) = [2] x1 + [0]
[c_8](x1) = [2] x1 + [0]
[c_9](x1) = [1] x1 + [3]
[c_10] = [0]
[first1^#](x1, x2) = [3] x1 + [2] x2 + [0]
[c_11](x1, x2) = [1] x1 + [1] x2 + [7]
[c_12] = [0]
[quote1^#](x1) = [2] x1 + [0]
[c_13](x1, x2) = [1] x1 + [1] x2 + [0]
[c_14](x1) = [1] x1 + [0]
[c_15] = [0]
[unquote^#](x1) = [2] x1 + [0]
[c_16] = [0]
[c_17](x1) = [1] x1 + [5]
[unquote1^#](x1) = [7] x1 + [0]
[c_18] = [0]
[c_19](x1) = [7] x1 + [0]
[fcons^#](x1, x2) = [7] x1 + [7] x2 + [0]
[c_20] = [0]
The following symbols are considered usable
{sel^#, first^#, from^#, sel1^#, quote^#, first1^#, quote1^#,
unquote^#}
The order satisfies the following ordering constraints:
[sel^#(s(X), cons(Y, Z))] = [0]
>= [0]
= [c_1(sel^#(X, Z))]
[first^#(s(X), cons(Y, Z))] = [0]
>= [0]
= [c_3(first^#(X, Z))]
[from^#(X)] = [0]
>= [0]
= [c_5(from^#(s(X)))]
[sel1^#(s(X), cons(Y, Z))] = [3] X + [4] Y + [4] Z + [9]
> [3] X + [4] Z + [7]
= [c_6(sel1^#(X, Z))]
[sel1^#(0(), cons(X, Z))] = [4] X + [4] Z + [9]
> [4] X + [3]
= [c_7(quote^#(X))]
[quote^#(sel(X, Z))] = [8] X + [8] Z + [0]
>= [6] X + [8] Z + [0]
= [c_8(sel1^#(X, Z))]
[quote^#(s(X))] = [2] X + [6]
> [2] X + [3]
= [c_9(quote^#(X))]
[first1^#(s(X), cons(Y, Z))] = [3] X + [2] Y + [2] Z + [9]
> [3] X + [2] Y + [2] Z + [7]
= [c_11(quote^#(Y), first1^#(X, Z))]
[quote1^#(cons(X, Z))] = [2] X + [2] Z + [0]
>= [2] X + [2] Z + [0]
= [c_13(quote^#(X), quote1^#(Z))]
[quote1^#(first(X, Z))] = [8] X + [2] Z + [0]
>= [3] X + [2] Z + [0]
= [c_14(first1^#(X, Z))]
[unquote^#(s1(X))] = [2] X + [8]
> [2] X + [5]
= [c_17(unquote^#(X))]
The strictly oriented rules are moved into the weak component.
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict DPs:
{ sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z))
, first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z))
, from^#(X) -> c_5(from^#(s(X)))
, quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z))
, quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z))
, quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) }
Weak DPs:
{ sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z))
, sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X))
, quote^#(s(X)) -> c_9(quote^#(X))
, first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z))
, unquote^#(s1(X)) -> c_17(unquote^#(X)) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ unquote^#(s1(X)) -> c_17(unquote^#(X)) }
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict DPs:
{ sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z))
, first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z))
, from^#(X) -> c_5(from^#(s(X)))
, quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z))
, quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z))
, quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) }
Weak DPs:
{ sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z))
, sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X))
, quote^#(s(X)) -> c_9(quote^#(X))
, first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
DPs:
{ 4: quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z))
, 5: quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z))
, 6: quote1^#(first(X, Z)) -> c_14(first1^#(X, Z))
, 7: sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z))
, 8: sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X))
, 9: quote^#(s(X)) -> c_9(quote^#(X))
, 10: first1^#(s(X), cons(Y, Z)) ->
c_11(quote^#(Y), first1^#(X, Z)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_1) = {1}, Uargs(c_3) = {1}, Uargs(c_5) = {1},
Uargs(c_6) = {1}, Uargs(c_7) = {1}, Uargs(c_8) = {1},
Uargs(c_9) = {1}, Uargs(c_11) = {1, 2}, Uargs(c_13) = {1, 2},
Uargs(c_14) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[sel](x1, x2) = [4] x2 + [4]
[s](x1) = [1] x1 + [4]
[cons](x1, x2) = [1] x1 + [1] x2 + [2]
[0] = [7]
[first](x1, x2) = [2] x1 + [2] x2 + [2]
[nil] = [0]
[from](x1) = [7] x1 + [0]
[sel1](x1, x2) = [7] x1 + [7] x2 + [0]
[quote](x1) = [7] x1 + [0]
[first1](x1, x2) = [7] x1 + [7] x2 + [0]
[nil1] = [0]
[cons1](x1, x2) = [1] x1 + [1] x2 + [0]
[01] = [0]
[quote1](x1) = [7] x1 + [0]
[s1](x1) = [1] x1 + [0]
[unquote](x1) = [7] x1 + [0]
[unquote1](x1) = [7] x1 + [0]
[fcons](x1, x2) = [7] x1 + [7] x2 + [0]
[sel^#](x1, x2) = [0]
[c_1](x1) = [4] x1 + [0]
[c_2] = [0]
[first^#](x1, x2) = [0]
[c_3](x1) = [4] x1 + [0]
[c_4] = [0]
[from^#](x1) = [0]
[c_5](x1) = [1] x1 + [0]
[sel1^#](x1, x2) = [4] x2 + [0]
[c_6](x1) = [1] x1 + [1]
[c_7](x1) = [2] x1 + [1]
[quote^#](x1) = [2] x1 + [0]
[c_8](x1) = [1] x1 + [7]
[c_9](x1) = [1] x1 + [1]
[c_10] = [0]
[first1^#](x1, x2) = [4] x2 + [0]
[c_11](x1, x2) = [1] x1 + [1] x2 + [1]
[c_12] = [0]
[quote1^#](x1) = [4] x1 + [0]
[c_13](x1, x2) = [1] x1 + [1] x2 + [1]
[c_14](x1) = [1] x1 + [1]
[c_15] = [0]
[unquote^#](x1) = [7] x1 + [0]
[c_16] = [0]
[c_17](x1) = [7] x1 + [0]
[unquote1^#](x1) = [7] x1 + [0]
[c_18] = [0]
[c_19](x1) = [7] x1 + [0]
[fcons^#](x1, x2) = [7] x1 + [7] x2 + [0]
[c_20] = [0]
The following symbols are considered usable
{sel^#, first^#, from^#, sel1^#, quote^#, first1^#, quote1^#}
The order satisfies the following ordering constraints:
[sel^#(s(X), cons(Y, Z))] = [0]
>= [0]
= [c_1(sel^#(X, Z))]
[first^#(s(X), cons(Y, Z))] = [0]
>= [0]
= [c_3(first^#(X, Z))]
[from^#(X)] = [0]
>= [0]
= [c_5(from^#(s(X)))]
[sel1^#(s(X), cons(Y, Z))] = [4] Y + [4] Z + [8]
> [4] Z + [1]
= [c_6(sel1^#(X, Z))]
[sel1^#(0(), cons(X, Z))] = [4] X + [4] Z + [8]
> [4] X + [1]
= [c_7(quote^#(X))]
[quote^#(sel(X, Z))] = [8] Z + [8]
> [4] Z + [7]
= [c_8(sel1^#(X, Z))]
[quote^#(s(X))] = [2] X + [8]
> [2] X + [1]
= [c_9(quote^#(X))]
[first1^#(s(X), cons(Y, Z))] = [4] Y + [4] Z + [8]
> [2] Y + [4] Z + [1]
= [c_11(quote^#(Y), first1^#(X, Z))]
[quote1^#(cons(X, Z))] = [4] X + [4] Z + [8]
> [2] X + [4] Z + [1]
= [c_13(quote^#(X), quote1^#(Z))]
[quote1^#(first(X, Z))] = [8] X + [8] Z + [8]
> [4] Z + [1]
= [c_14(first1^#(X, Z))]
The strictly oriented rules are moved into the weak component.
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict DPs:
{ sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z))
, first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z))
, from^#(X) -> c_5(from^#(s(X))) }
Weak DPs:
{ sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z))
, sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X))
, quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z))
, quote^#(s(X)) -> c_9(quote^#(X))
, first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z))
, quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z))
, quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z))
, sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X))
, quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z))
, quote^#(s(X)) -> c_9(quote^#(X))
, first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z))
, quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z))
, quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) }
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict DPs:
{ sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z))
, first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z))
, from^#(X) -> c_5(from^#(s(X))) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
DPs:
{ 1: sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z))
, 2: first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_1) = {1}, Uargs(c_3) = {1}, Uargs(c_5) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[sel](x1, x2) = [7] x1 + [7] x2 + [0]
[s](x1) = [1] x1 + [2]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[0] = [0]
[first](x1, x2) = [7] x1 + [7] x2 + [0]
[nil] = [0]
[from](x1) = [7] x1 + [0]
[sel1](x1, x2) = [7] x1 + [7] x2 + [0]
[quote](x1) = [7] x1 + [0]
[first1](x1, x2) = [7] x1 + [7] x2 + [0]
[nil1] = [0]
[cons1](x1, x2) = [1] x1 + [1] x2 + [0]
[01] = [0]
[quote1](x1) = [7] x1 + [0]
[s1](x1) = [1] x1 + [0]
[unquote](x1) = [7] x1 + [0]
[unquote1](x1) = [7] x1 + [0]
[fcons](x1, x2) = [7] x1 + [7] x2 + [0]
[sel^#](x1, x2) = [4] x1 + [0]
[c_1](x1) = [1] x1 + [3]
[c_2] = [0]
[first^#](x1, x2) = [4] x1 + [0]
[c_3](x1) = [1] x1 + [5]
[c_4] = [0]
[from^#](x1) = [0]
[c_5](x1) = [4] x1 + [0]
[sel1^#](x1, x2) = [7] x1 + [7] x2 + [0]
[c_6](x1) = [7] x1 + [0]
[c_7](x1) = [7] x1 + [0]
[quote^#](x1) = [7] x1 + [0]
[c_8](x1) = [7] x1 + [0]
[c_9](x1) = [7] x1 + [0]
[c_10] = [0]
[first1^#](x1, x2) = [7] x1 + [7] x2 + [0]
[c_11](x1, x2) = [7] x1 + [7] x2 + [0]
[c_12] = [0]
[quote1^#](x1) = [7] x1 + [0]
[c_13](x1, x2) = [7] x1 + [7] x2 + [0]
[c_14](x1) = [7] x1 + [0]
[c_15] = [0]
[unquote^#](x1) = [7] x1 + [0]
[c_16] = [0]
[c_17](x1) = [7] x1 + [0]
[unquote1^#](x1) = [7] x1 + [0]
[c_18] = [0]
[c_19](x1) = [7] x1 + [0]
[fcons^#](x1, x2) = [7] x1 + [7] x2 + [0]
[c_20] = [0]
The following symbols are considered usable
{sel^#, first^#, from^#}
The order satisfies the following ordering constraints:
[sel^#(s(X), cons(Y, Z))] = [4] X + [8]
> [4] X + [3]
= [c_1(sel^#(X, Z))]
[first^#(s(X), cons(Y, Z))] = [4] X + [8]
> [4] X + [5]
= [c_3(first^#(X, Z))]
[from^#(X)] = [0]
>= [0]
= [c_5(from^#(s(X)))]
The strictly oriented rules are moved into the weak component.
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict DPs: { from^#(X) -> c_5(from^#(s(X))) }
Weak DPs:
{ sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z))
, first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z)) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z))
, first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z)) }
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict DPs: { from^#(X) -> c_5(from^#(s(X))) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
None of the processors succeeded.
Details of failed attempt(s):
-----------------------------
1) 'empty' failed due to the following reason:
Empty strict component of the problem is NOT empty.
2) 'Fastest' failed due to the following reason:
None of the processors succeeded.
Details of failed attempt(s):
-----------------------------
1) 'WithProblem' failed due to the following reason:
None of the processors succeeded.
Details of failed attempt(s):
-----------------------------
1) 'empty' failed due to the following reason:
Empty strict component of the problem is NOT empty.
2) 'Polynomial Path Order (PS)' failed due to the following reason:
The input cannot be shown compatible
2) 'Fastest (timeout of 5 seconds)' failed due to the following
reason:
Computation stopped due to timeout after 5.0 seconds.
3) 'Polynomial Path Order (PS)' failed due to the following reason:
The input cannot be shown compatible
2) 'Best' failed due to the following reason:
None of the processors succeeded.
Details of failed attempt(s):
-----------------------------
1) 'WithProblem (timeout of 30 seconds) (timeout of 60 seconds)'
failed due to the following reason:
Computation stopped due to timeout after 30.0 seconds.
2) 'Best' failed due to the following reason:
None of the processors succeeded.
Details of failed attempt(s):
-----------------------------
1) 'Polynomial Path Order (PS) (timeout of 60 seconds)' failed due
to the following reason:
The input cannot be shown compatible
2) 'bsearch-popstar (timeout of 60 seconds)' failed due to the
following reason:
The input cannot be shown compatible
3) 'Fastest (timeout of 5 seconds) (timeout of 60 seconds)' failed
due to the following reason:
None of the processors succeeded.
Details of failed attempt(s):
-----------------------------
1) 'Bounds with minimal-enrichment and initial automaton 'match''
failed due to the following reason:
match-boundness of the problem could not be verified.
2) 'Bounds with perSymbol-enrichment and initial automaton 'match''
failed due to the following reason:
match-boundness of the problem could not be verified.
Arrrr..