LMPO
MAYBE
We consider the following Problem:
Strict Trs:
{ a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z))
, a__sel(0(), cons(X, Z)) -> mark(X)
, a__first(0(), Z) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__sel1(s(X), cons(Y, Z)) -> a__sel1(mark(X), mark(Z))
, a__sel1(0(), cons(X, Z)) -> a__quote(X)
, a__first1(0(), Z) -> nil1()
, a__first1(s(X), cons(Y, Z)) ->
cons1(a__quote(Y), a__first1(mark(X), mark(Z)))
, a__quote(0()) -> 01()
, a__quote1(cons(X, Z)) -> cons1(a__quote(X), a__quote1(Z))
, a__quote1(nil()) -> nil1()
, a__quote(s(X)) -> s1(a__quote(X))
, a__quote(sel(X, Z)) -> a__sel1(mark(X), mark(Z))
, a__quote1(first(X, Z)) -> a__first1(mark(X), mark(Z))
, a__unquote(01()) -> 0()
, a__unquote(s1(X)) -> s(a__unquote(mark(X)))
, a__unquote1(nil1()) -> nil()
, a__unquote1(cons1(X, Z)) ->
a__fcons(a__unquote(mark(X)), a__unquote1(mark(Z)))
, a__fcons(X, Z) -> cons(mark(X), Z)
, mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(mark(X))
, mark(sel1(X1, X2)) -> a__sel1(mark(X1), mark(X2))
, mark(quote(X)) -> a__quote(X)
, mark(first1(X1, X2)) -> a__first1(mark(X1), mark(X2))
, mark(quote1(X)) -> a__quote1(X)
, mark(unquote(X)) -> a__unquote(mark(X))
, mark(unquote1(X)) -> a__unquote1(mark(X))
, mark(fcons(X1, X2)) -> a__fcons(mark(X1), mark(X2))
, mark(s(X)) -> s(mark(X))
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(0()) -> 0()
, mark(nil()) -> nil()
, mark(nil1()) -> nil1()
, mark(cons1(X1, X2)) -> cons1(mark(X1), mark(X2))
, mark(01()) -> 01()
, mark(s1(X)) -> s1(mark(X))
, a__sel(X1, X2) -> sel(X1, X2)
, a__first(X1, X2) -> first(X1, X2)
, a__from(X) -> from(X)
, a__sel1(X1, X2) -> sel1(X1, X2)
, a__quote(X) -> quote(X)
, a__first1(X1, X2) -> first1(X1, X2)
, a__quote1(X) -> quote1(X)
, a__unquote(X) -> unquote(X)
, a__unquote1(X) -> unquote1(X)
, a__fcons(X1, X2) -> fcons(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: MAYBE
Proof:
The input cannot be shown compatible
Arrrr..
MPO
MAYBE
We consider the following Problem:
Strict Trs:
{ a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z))
, a__sel(0(), cons(X, Z)) -> mark(X)
, a__first(0(), Z) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__sel1(s(X), cons(Y, Z)) -> a__sel1(mark(X), mark(Z))
, a__sel1(0(), cons(X, Z)) -> a__quote(X)
, a__first1(0(), Z) -> nil1()
, a__first1(s(X), cons(Y, Z)) ->
cons1(a__quote(Y), a__first1(mark(X), mark(Z)))
, a__quote(0()) -> 01()
, a__quote1(cons(X, Z)) -> cons1(a__quote(X), a__quote1(Z))
, a__quote1(nil()) -> nil1()
, a__quote(s(X)) -> s1(a__quote(X))
, a__quote(sel(X, Z)) -> a__sel1(mark(X), mark(Z))
, a__quote1(first(X, Z)) -> a__first1(mark(X), mark(Z))
, a__unquote(01()) -> 0()
, a__unquote(s1(X)) -> s(a__unquote(mark(X)))
, a__unquote1(nil1()) -> nil()
, a__unquote1(cons1(X, Z)) ->
a__fcons(a__unquote(mark(X)), a__unquote1(mark(Z)))
, a__fcons(X, Z) -> cons(mark(X), Z)
, mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(mark(X))
, mark(sel1(X1, X2)) -> a__sel1(mark(X1), mark(X2))
, mark(quote(X)) -> a__quote(X)
, mark(first1(X1, X2)) -> a__first1(mark(X1), mark(X2))
, mark(quote1(X)) -> a__quote1(X)
, mark(unquote(X)) -> a__unquote(mark(X))
, mark(unquote1(X)) -> a__unquote1(mark(X))
, mark(fcons(X1, X2)) -> a__fcons(mark(X1), mark(X2))
, mark(s(X)) -> s(mark(X))
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(0()) -> 0()
, mark(nil()) -> nil()
, mark(nil1()) -> nil1()
, mark(cons1(X1, X2)) -> cons1(mark(X1), mark(X2))
, mark(01()) -> 01()
, mark(s1(X)) -> s1(mark(X))
, a__sel(X1, X2) -> sel(X1, X2)
, a__first(X1, X2) -> first(X1, X2)
, a__from(X) -> from(X)
, a__sel1(X1, X2) -> sel1(X1, X2)
, a__quote(X) -> quote(X)
, a__first1(X1, X2) -> first1(X1, X2)
, a__quote1(X) -> quote1(X)
, a__unquote(X) -> unquote(X)
, a__unquote1(X) -> unquote1(X)
, a__fcons(X1, X2) -> fcons(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: MAYBE
Proof:
The input cannot be shown compatible
Arrrr..
POP*
MAYBE
We consider the following Problem:
Strict Trs:
{ a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z))
, a__sel(0(), cons(X, Z)) -> mark(X)
, a__first(0(), Z) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__sel1(s(X), cons(Y, Z)) -> a__sel1(mark(X), mark(Z))
, a__sel1(0(), cons(X, Z)) -> a__quote(X)
, a__first1(0(), Z) -> nil1()
, a__first1(s(X), cons(Y, Z)) ->
cons1(a__quote(Y), a__first1(mark(X), mark(Z)))
, a__quote(0()) -> 01()
, a__quote1(cons(X, Z)) -> cons1(a__quote(X), a__quote1(Z))
, a__quote1(nil()) -> nil1()
, a__quote(s(X)) -> s1(a__quote(X))
, a__quote(sel(X, Z)) -> a__sel1(mark(X), mark(Z))
, a__quote1(first(X, Z)) -> a__first1(mark(X), mark(Z))
, a__unquote(01()) -> 0()
, a__unquote(s1(X)) -> s(a__unquote(mark(X)))
, a__unquote1(nil1()) -> nil()
, a__unquote1(cons1(X, Z)) ->
a__fcons(a__unquote(mark(X)), a__unquote1(mark(Z)))
, a__fcons(X, Z) -> cons(mark(X), Z)
, mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(mark(X))
, mark(sel1(X1, X2)) -> a__sel1(mark(X1), mark(X2))
, mark(quote(X)) -> a__quote(X)
, mark(first1(X1, X2)) -> a__first1(mark(X1), mark(X2))
, mark(quote1(X)) -> a__quote1(X)
, mark(unquote(X)) -> a__unquote(mark(X))
, mark(unquote1(X)) -> a__unquote1(mark(X))
, mark(fcons(X1, X2)) -> a__fcons(mark(X1), mark(X2))
, mark(s(X)) -> s(mark(X))
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(0()) -> 0()
, mark(nil()) -> nil()
, mark(nil1()) -> nil1()
, mark(cons1(X1, X2)) -> cons1(mark(X1), mark(X2))
, mark(01()) -> 01()
, mark(s1(X)) -> s1(mark(X))
, a__sel(X1, X2) -> sel(X1, X2)
, a__first(X1, X2) -> first(X1, X2)
, a__from(X) -> from(X)
, a__sel1(X1, X2) -> sel1(X1, X2)
, a__quote(X) -> quote(X)
, a__first1(X1, X2) -> first1(X1, X2)
, a__quote1(X) -> quote1(X)
, a__unquote(X) -> unquote(X)
, a__unquote1(X) -> unquote1(X)
, a__fcons(X1, X2) -> fcons(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: MAYBE
Proof:
The input cannot be shown compatible
Arrrr..
POP* (PS)
MAYBE
We consider the following Problem:
Strict Trs:
{ a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z))
, a__sel(0(), cons(X, Z)) -> mark(X)
, a__first(0(), Z) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__sel1(s(X), cons(Y, Z)) -> a__sel1(mark(X), mark(Z))
, a__sel1(0(), cons(X, Z)) -> a__quote(X)
, a__first1(0(), Z) -> nil1()
, a__first1(s(X), cons(Y, Z)) ->
cons1(a__quote(Y), a__first1(mark(X), mark(Z)))
, a__quote(0()) -> 01()
, a__quote1(cons(X, Z)) -> cons1(a__quote(X), a__quote1(Z))
, a__quote1(nil()) -> nil1()
, a__quote(s(X)) -> s1(a__quote(X))
, a__quote(sel(X, Z)) -> a__sel1(mark(X), mark(Z))
, a__quote1(first(X, Z)) -> a__first1(mark(X), mark(Z))
, a__unquote(01()) -> 0()
, a__unquote(s1(X)) -> s(a__unquote(mark(X)))
, a__unquote1(nil1()) -> nil()
, a__unquote1(cons1(X, Z)) ->
a__fcons(a__unquote(mark(X)), a__unquote1(mark(Z)))
, a__fcons(X, Z) -> cons(mark(X), Z)
, mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(mark(X))
, mark(sel1(X1, X2)) -> a__sel1(mark(X1), mark(X2))
, mark(quote(X)) -> a__quote(X)
, mark(first1(X1, X2)) -> a__first1(mark(X1), mark(X2))
, mark(quote1(X)) -> a__quote1(X)
, mark(unquote(X)) -> a__unquote(mark(X))
, mark(unquote1(X)) -> a__unquote1(mark(X))
, mark(fcons(X1, X2)) -> a__fcons(mark(X1), mark(X2))
, mark(s(X)) -> s(mark(X))
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(0()) -> 0()
, mark(nil()) -> nil()
, mark(nil1()) -> nil1()
, mark(cons1(X1, X2)) -> cons1(mark(X1), mark(X2))
, mark(01()) -> 01()
, mark(s1(X)) -> s1(mark(X))
, a__sel(X1, X2) -> sel(X1, X2)
, a__first(X1, X2) -> first(X1, X2)
, a__from(X) -> from(X)
, a__sel1(X1, X2) -> sel1(X1, X2)
, a__quote(X) -> quote(X)
, a__first1(X1, X2) -> first1(X1, X2)
, a__quote1(X) -> quote1(X)
, a__unquote(X) -> unquote(X)
, a__unquote1(X) -> unquote1(X)
, a__fcons(X1, X2) -> fcons(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: MAYBE
Proof:
The input cannot be shown compatible
Arrrr..
Small POP*
MAYBE
We consider the following Problem:
Strict Trs:
{ a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z))
, a__sel(0(), cons(X, Z)) -> mark(X)
, a__first(0(), Z) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__sel1(s(X), cons(Y, Z)) -> a__sel1(mark(X), mark(Z))
, a__sel1(0(), cons(X, Z)) -> a__quote(X)
, a__first1(0(), Z) -> nil1()
, a__first1(s(X), cons(Y, Z)) ->
cons1(a__quote(Y), a__first1(mark(X), mark(Z)))
, a__quote(0()) -> 01()
, a__quote1(cons(X, Z)) -> cons1(a__quote(X), a__quote1(Z))
, a__quote1(nil()) -> nil1()
, a__quote(s(X)) -> s1(a__quote(X))
, a__quote(sel(X, Z)) -> a__sel1(mark(X), mark(Z))
, a__quote1(first(X, Z)) -> a__first1(mark(X), mark(Z))
, a__unquote(01()) -> 0()
, a__unquote(s1(X)) -> s(a__unquote(mark(X)))
, a__unquote1(nil1()) -> nil()
, a__unquote1(cons1(X, Z)) ->
a__fcons(a__unquote(mark(X)), a__unquote1(mark(Z)))
, a__fcons(X, Z) -> cons(mark(X), Z)
, mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(mark(X))
, mark(sel1(X1, X2)) -> a__sel1(mark(X1), mark(X2))
, mark(quote(X)) -> a__quote(X)
, mark(first1(X1, X2)) -> a__first1(mark(X1), mark(X2))
, mark(quote1(X)) -> a__quote1(X)
, mark(unquote(X)) -> a__unquote(mark(X))
, mark(unquote1(X)) -> a__unquote1(mark(X))
, mark(fcons(X1, X2)) -> a__fcons(mark(X1), mark(X2))
, mark(s(X)) -> s(mark(X))
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(0()) -> 0()
, mark(nil()) -> nil()
, mark(nil1()) -> nil1()
, mark(cons1(X1, X2)) -> cons1(mark(X1), mark(X2))
, mark(01()) -> 01()
, mark(s1(X)) -> s1(mark(X))
, a__sel(X1, X2) -> sel(X1, X2)
, a__first(X1, X2) -> first(X1, X2)
, a__from(X) -> from(X)
, a__sel1(X1, X2) -> sel1(X1, X2)
, a__quote(X) -> quote(X)
, a__first1(X1, X2) -> first1(X1, X2)
, a__quote1(X) -> quote1(X)
, a__unquote(X) -> unquote(X)
, a__unquote1(X) -> unquote1(X)
, a__fcons(X1, X2) -> fcons(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: MAYBE
Proof:
The input cannot be shown compatible
Arrrr..
Small POP* (PS)
MAYBE
We consider the following Problem:
Strict Trs:
{ a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z))
, a__sel(0(), cons(X, Z)) -> mark(X)
, a__first(0(), Z) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__sel1(s(X), cons(Y, Z)) -> a__sel1(mark(X), mark(Z))
, a__sel1(0(), cons(X, Z)) -> a__quote(X)
, a__first1(0(), Z) -> nil1()
, a__first1(s(X), cons(Y, Z)) ->
cons1(a__quote(Y), a__first1(mark(X), mark(Z)))
, a__quote(0()) -> 01()
, a__quote1(cons(X, Z)) -> cons1(a__quote(X), a__quote1(Z))
, a__quote1(nil()) -> nil1()
, a__quote(s(X)) -> s1(a__quote(X))
, a__quote(sel(X, Z)) -> a__sel1(mark(X), mark(Z))
, a__quote1(first(X, Z)) -> a__first1(mark(X), mark(Z))
, a__unquote(01()) -> 0()
, a__unquote(s1(X)) -> s(a__unquote(mark(X)))
, a__unquote1(nil1()) -> nil()
, a__unquote1(cons1(X, Z)) ->
a__fcons(a__unquote(mark(X)), a__unquote1(mark(Z)))
, a__fcons(X, Z) -> cons(mark(X), Z)
, mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(mark(X))
, mark(sel1(X1, X2)) -> a__sel1(mark(X1), mark(X2))
, mark(quote(X)) -> a__quote(X)
, mark(first1(X1, X2)) -> a__first1(mark(X1), mark(X2))
, mark(quote1(X)) -> a__quote1(X)
, mark(unquote(X)) -> a__unquote(mark(X))
, mark(unquote1(X)) -> a__unquote1(mark(X))
, mark(fcons(X1, X2)) -> a__fcons(mark(X1), mark(X2))
, mark(s(X)) -> s(mark(X))
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(0()) -> 0()
, mark(nil()) -> nil()
, mark(nil1()) -> nil1()
, mark(cons1(X1, X2)) -> cons1(mark(X1), mark(X2))
, mark(01()) -> 01()
, mark(s1(X)) -> s1(mark(X))
, a__sel(X1, X2) -> sel(X1, X2)
, a__first(X1, X2) -> first(X1, X2)
, a__from(X) -> from(X)
, a__sel1(X1, X2) -> sel1(X1, X2)
, a__quote(X) -> quote(X)
, a__first1(X1, X2) -> first1(X1, X2)
, a__quote1(X) -> quote1(X)
, a__unquote(X) -> unquote(X)
, a__unquote1(X) -> unquote1(X)
, a__fcons(X1, X2) -> fcons(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: MAYBE
Proof:
The input cannot be shown compatible
Arrrr..