tag:blogger.com,1999:blog-8342347821869463061.post2962716835166130397..comments2023-11-02T17:30:26.989+05:30Comments on My Technical Scratch Pad.: how to prove it - ch2, sec2.2(Equivalence Involving Quantifiers) exhimanshuhttp://www.blogger.com/profile/02909790425038294533noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-8342347821869463061.post-48894728048019116432019-01-12T12:16:00.875+05:302019-01-12T12:16:00.875+05:30I don't have 12(f) and 12(e) the same. Anyone...I don't have 12(f) and 12(e) the same. Anyone else came to the same conclusion or am I wrong? fesodeshttps://www.blogger.com/profile/17761308025984234595noreply@blogger.comtag:blogger.com,1999:blog-8342347821869463061.post-4132349717679458462018-08-26T10:33:18.115+05:302018-08-26T10:33:18.115+05:30I second you. L(x,y) was defined as x likes y. I second you. L(x,y) was defined as x likes y. fesodeshttps://www.blogger.com/profile/17761308025984234595noreply@blogger.comtag:blogger.com,1999:blog-8342347821869463061.post-26635209214085907252016-09-09T19:26:01.448+05:302016-09-09T19:26:01.448+05:30Thanks for uploading this fantastic resource. I ju...Thanks for uploading this fantastic resource. I just wanted to point out that the English translation given in exercises 2.2 2b is incorrect. It is 'either there is someone who dislikes everyone, or there is someone who likes everyone.' Anonymoushttps://www.blogger.com/profile/15105799050184106674noreply@blogger.comtag:blogger.com,1999:blog-8342347821869463061.post-33576640467230041542016-02-21T12:42:01.084+05:302016-02-21T12:42:01.084+05:30This comment has been removed by the author.Anonymoushttps://www.blogger.com/profile/00581142647340578090noreply@blogger.comtag:blogger.com,1999:blog-8342347821869463061.post-6308112182971281232016-01-02T07:26:59.241+05:302016-01-02T07:26:59.241+05:30In 10)b), one could say that if A⋂B = ∅, then noth...In 10)b), one could say that if A⋂B = ∅, then nothing stops the first statement from being true in certain cases, while ∃x∈(A ⋂ B)P(x) is equivalent to ∃x(x∈A ⋂ B∧P(x)), which in this case would be ∃x(x∈∅∧P(x)), where it is clearly always false, therefore the two statements aren't equivalent.Anonymoushttps://www.blogger.com/profile/02955589736072906593noreply@blogger.comtag:blogger.com,1999:blog-8342347821869463061.post-81725605294577234212012-08-30T17:16:51.359+05:302012-08-30T17:16:51.359+05:30Hello. I just wanted to thank you for publishing ...Hello. I just wanted to thank you for publishing this information. I'm working through Velleman's book now and really appreciate being able to check your solutions against mine, or just look to yours for a hint when I'm stuck on an exercise.Andrewhttps://www.blogger.com/profile/05916035858567993392noreply@blogger.com