Search results
Jump to navigation
Jump to search
- ...or that is not that the universal quantifier is not importing but that the Fx is under both a negation and a disjunction, either of which would block the ...ng). The rule you cite is for unrestricted quantification, but in that (x)(Fx => Gx) is true. So your argument confuses two different things and fails2.07 MB (380,774 words) - 07:40, 7 March 2020
- ...or that is not that the universal quantifier is not importing but that the Fx is under both a negation and a disjunction, either of which would block the ...ng). The rule you cite is for unrestricted quantification, but in that (x)(Fx => Gx) is true. So your argument confuses two different things and fails.4.12 MB (665,740 words) - 16:32, 2 March 2020
- ...ol of the language but rather an abbreviation: “GixFx” was short for “[[Ix|Ix]]( Gx & [[Ay|Ay]](Fy <=> y=x].” If there was not a unique F, the par “~[[Ix|Ix]](Gx & [[Ay|Ay]](Fy <=> y=x]” and “[[Ix|Ix]](~Gx & [[Ay|Ay]](Fy <=> y=x].” Most subsequent – and previous –24 KB (4,143 words) - 08:13, 30 June 2014
- ...For every predicate, F, there is a d-predicate, F*, defined as a F* iff [[Ix: x bunch & xCa]] x F (Clearly, for d-F, F* is just F)… [[Ix: x bunch]] x F* & [[Ay: y F*]] x C y & xQn / xQf of the existents20 KB (3,778 words) - 08:22, 30 June 2014
- > [Ix: x group & xF* & [[Az: zF*] xCz] x G & xQn/ xQf of the F*s/xQr > [Ix: x group & Fx] {y: xCy}G52 KB (8,250 words) - 08:12, 27 January 2015
- you can replace "Qx:Fx" with "Fx &" if you like n+1-ad is [ix: x among a Ix: x among a](x n-ad & [ay: y among<br />a & y not among x Ay: y among608 KB (95,583 words) - 09:14, 27 January 2015