Post History
#3: Post edited
by
clemens
·
2026-05-22T11:53:40Z (4 months ago)
- I confess I don't fully understand this question (I await further clarification), but in general dualization is not that complicated. If $P$ is the dual of $Q$ it just means that it acts on $\bot$ the same way that $Q$ acts on $\top$, and vice versa. So whenever you have $P$ (say, a quantifier), it's easy to find its dual: just define $Q(x_1,x_2,…) := ¬P(¬x_1,¬x_2,…)$.
E.g. let's say we want to find the dual of "there exist countably many $x$ such that…" ($\exists^{\aleph_0} x…$). The dual would be $¬\exists^{\aleph_0}x ¬…$ "there do not exist countably many $x$ such that not…". This expression expands easily to "for all $x$ (except perhaps either a finite or uncountable number of different $x$), …". Thus we have found the dual of the quantifier $\exists^{\aleph_0} x$!I hope this clarifies the particular applications under consideration in the OP.
- I confess I don't fully understand this question (I await further clarification), but in general dualization is not that complicated. If $P$ is the dual of $Q$ it just means that it acts on $\bot$ the same way that $Q$ acts on $\top$, and vice versa. So whenever you have $P$ (say, a quantifier), it's easy to find its dual: just define $Q(x_1,x_2,…) := ¬P(¬x_1,¬x_2,…)$.
- E.g. let's say we want to find the dual of "there exist countably many $x$ such that…" ($\exists^{\aleph_0} x…$). The dual would be $¬\exists^{\aleph_0}x ¬…$ "there do not exist countably many $x$ such that not…". This expression expands easily to "for all $x$ (except perhaps either a finite or uncountable number of different $x$), …". Thus we have found the dual of the quantifier $\exists^{\aleph_0} x$!
#2: Post edited
by
clemens
·
2026-05-22T11:52:44Z (4 months ago)
I confess I don't fully understand this question (I await clarification), but in general dualization is not that complicated. If $P$ is the dual of $Q$ it just means that it acts on $\bot$ the same way that $Q$ acts on $\top$, and vice versa. So whenever you have $P$ (say, a quantifier), it's easy to find its dual: just define $Q(x_1,x_2,…) := ¬P(¬x_1,¬x_2,…)$.E.g. let's say we want to find the dual of "there exist countably many $x$ such that…" ($\exists^{\aleph_0} x…$). The dual would be $¬\exists^{\aleph_0}x ¬…$ "there do not exist countably many $x$ such that not…". This expression expands easily to "for all $x$ (except perhaps either a finite or uncountable number of different $x$), …". Thus we have found the dual of the quantifier $\exists^{\aleph_0} x$!
- I confess I don't fully understand this question (I await further clarification), but in general dualization is not that complicated. If $P$ is the dual of $Q$ it just means that it acts on $\bot$ the same way that $Q$ acts on $\top$, and vice versa. So whenever you have $P$ (say, a quantifier), it's easy to find its dual: just define $Q(x_1,x_2,…) := ¬P(¬x_1,¬x_2,…)$.
- E.g. let's say we want to find the dual of "there exist countably many $x$ such that…" ($\exists^{\aleph_0} x…$). The dual would be $¬\exists^{\aleph_0}x ¬…$ "there do not exist countably many $x$ such that not…". This expression expands easily to "for all $x$ (except perhaps either a finite or uncountable number of different $x$), …". Thus we have found the dual of the quantifier $\exists^{\aleph_0} x$!
- I hope this clarifies the particular applications under consideration in the OP.
#1: Initial revision
by
clemens
·
2026-05-22T11:51:08Z (4 months ago)
I confess I don't fully understand this question (I await clarification), but in general dualization is not that complicated. If $P$ is the dual of $Q$ it just means that it acts on $\bot$ the same way that $Q$ acts on $\top$, and vice versa. So whenever you have $P$ (say, a quantifier), it's easy to find its dual: just define $Q(x_1,x_2,…) := ¬P(¬x_1,¬x_2,…)$.
E.g. let's say we want to find the dual of "there exist countably many $x$ such that…" ($\exists^{\aleph_0} x…$). The dual would be $¬\exists^{\aleph_0}x ¬…$ "there do not exist countably many $x$ such that not…". This expression expands easily to "for all $x$ (except perhaps either a finite or uncountable number of different $x$), …". Thus we have found the dual of the quantifier $\exists^{\aleph_0} x$!