------------------------------------------------------------ a | predicate | is a proposition with parameters ------------------------------------------------------------ then P(2) is | true | , P(-8) is | false | ------------------------------------------------------------ | domain of discourse: D | ------------------------------------------------------------ | universal quantifier | ------------------------------------------------------------ the value of "(∀x)[P(x)]" is | true | if D is {1,2,3,...} ------------------------------------------------------------ the value of "(∀x)[P(x)]" is | false | if D is real numbers ------------------------------------------------------------ | existential quantifier | ------------------------------------------------------------ |(∃z)[R(z)]| is | false | if D is {1,2,3,...} ------------------------------------------------------------ but would be true if D were | complex numbers | ------------------------------------------------------------ |(∃x)[x < 3]| is | true | ------------------------------------------------------------ can substitute | 2 | for |x| in the subformula, yielding: ------------------------------------------------------------ |( | 2 | > 1 → | 2 | ² > | 2 | )| ------------------------------------------------------------ |( | true | → | 4 | > 2)| ------------------------------------------------------------ |( | true | → | true | )| ------------------------------------------------------------ |( | true | )| ------------------------------------------------------------ incidentally, could also substitute | -5 | for |x| in the subformula, yielding: ------------------------------------------------------------ |( | -5 | > 1 → ( | -5 | )² > | -5 | )| ------------------------------------------------------------ |( | false | → | 25 | > -5)| ------------------------------------------------------------ |( | false | → | true | )| ------------------------------------------------------------ |( | true | )| ------------------------------------------------------------ such a value @v@ called | counterexample | ------------------------------------------------------------ |( | 0 | > 0)| ------------------------------------------------------------ |( | false | )| ------------------------------------------------------------ |( | 1/4 | > ½)| ------------------------------------------------------------ |( | false | )| ------------------------------------------------------------ anyway, do you think it's true or false? | true | ------------------------------------------------------------ this is case 1 just need to find even just one expression to plug in for y making subformula [r < y] true ------------------------------------------------------------ use | r + 1 | for this value ------------------------------------------------------------ if we try this value, is [r < | r + 1 | ] true? yes, we know that! ------------------------------------------------------------ first, do you think it's true or false? | false | ------------------------------------------------------------ to show (∀... is false, just need to find one | counterexample | ------------------------------------------------------------ try | 4 | for v ------------------------------------------------------------ plugging that in for x, next want (∀y)[ | 4 | < y] false ------------------------------------------------------------ how show (∀y)[ | 4 | < y] false? we're in case 2: counterexample y := | 1 | ------------------------------------------------------------ then check [ | 4 | < | 1 | ] false? yes it's false ------------------------------------------------------------ first, do you think it's true or false? | true | ------------------------------------------------------------ try x := | 6 | ------------------------------------------------------------ so next want (∀y)[ | 6 | < y] true ------------------------------------------------------------ and show that [ | 6 | < r] is true ------------------------------------------------------------ rethink: do you think (∃x)(∀y)[x < y] is true or false? | false | ------------------------------------------------------------ case 4 say |r| is any arbitrary value from domain~D with no assumptions made about it plug r in for x then try to show that subformula is false ------------------------------------------------------------ how show (forall... formula is false? | case 2: find one counterexample | ------------------------------------------------------------ try y := | 0 / r itself | ------------------------------------------------------------ check |[r < | 0 | ]| false? | yes it's false | ------------------------------------------------------------ first, do you think it's true or false? | true | ------------------------------------------------------------ let's try showing (∀y)(∃x)[x < y] is | false | ------------------------------------------------------------ | case 2: counterexample y := 0 | ------------------------------------------------------------ case 4: say |r| is any arbitrary value from domain~D with no assumptions made about it plug in r for x then try to show that @subformula@ is false ------------------------------------------------------------ | true: choose y := x | ------------------------------------------------------------ | still true: choose y := x + 1 | ------------------------------------------------------------ in this case it's false: counterexample x := Integer.MAX_VALUE which is 2^31 - 1 which is 2147483647 then "(∃y)[¬(2147483647 ≥ y)]" should be false plug in arbitrary r for y: "[¬(2147483647 >= r)]" should be false i.e. "(2147483647 >= r)" should be true which it is according to documentation ------------------------------------------------------------ | not true in this case either | ------------------------------------------------------------ ====================================================================

(Copyright © 2009 by Hugh McGuire   — for thoughts about this, see   http://www.cis.gvsu.edu/~mcguire/teaching/copyright_thoughts.html .)