tag:blogger.com,1999:blog-23478689.post4265454780694416754..comments2008-07-19T13:53:57.804ZComments on Logic Matters: What language is this?Peter Smithhttp://www.blogger.com/profile/03957579588136008664noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-23478689.post-39573352477664333572008-07-19T13:53:00.000Z2008-07-19T13:53:00.000ZThank you!Thank you!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-23478689.post-8728219406913385922008-07-18T23:16:00.000Z2008-07-18T23:16:00.000ZThat makes "F" denote a one-one or injective funct...That makes "F" denote a one-one or injective function. See http://en.wikipedia.org/wiki/Injective_functionPeter Smithhttp://www.blogger.com/profile/03957579588136008664noreply@blogger.comtag:blogger.com,1999:blog-23478689.post-69747220832888214442008-07-17T16:10:00.000Z2008-07-17T16:10:00.000ZHi!I have a question which is totally unrelated to...Hi!<BR/><BR/>I have a question which is totally unrelated to your post; but since you seem to be both a kind and able logician, I thought you might be willing to help.<BR/><BR/>My question: Is there a special term for a functional expression (i.e. a singular term forming operator on singular terms) F, which fulfils the following condition:<BR/><BR/>AxAy [(F(x)=F(y)) -> x=y]<BR/><BR/>Thanks in advance! (By the way, I really enjoy your blog!)Anonymousnoreply@blogger.com