On 4/25/2018 3:35 AM, Bruno Marchal wrote:
G proves that (p <-> ~ []p) is equivalent with (p <-> <>t), or equivalently (p <-> ~[]f). So consistency (<>t) is a solution to the (logical) equation x <-> ~[]x.
?? What does this proof look like? Why doesn't it prove f <->~[]f ? Brent -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
