Language Proof and Logic 13.4 13.5 •yvanrc51 Newbie Usergroup: Members Joined: Jan 30, 2011 Total Topics: 1 Total Posts: 9 #11 - Quote - Permalink 1 of 1 people found this post helpful Posted Apr 25, 2011 - 8:44 PM: 13.4 Attached Files: Screen shot 13.4.png (120 KB, 150 downloads) •yvanrc51 Newbie Usergroup: Members Joined: Jan 30, 2011 Total Topics: 1 Total Posts: 9 #12 - Quote - Permalink 1 of 1 people found this post helpful Posted Apr 25, 2011 - 8:46 PM: 13.5 Attached Files: Screen shot 13.5.png (132 KB, 246 downloads) •yvanrc51 Newbie Usergroup: Members Joined: Jan 30, 2011 Total Topics: 1 Total Posts: 9 #13 - Quote - Permalink Posted Apr 25, 2011 - 8:46 PM: Now if some could help me out with 13.7, that would awesome. Anyone? •bartolmarin Initiate Usergroup: Members Joined: Dec 23, 2010 Total Topics: 0 Total Posts: 73 #14 - Quote - Permalink Posted Apr 26, 2011 - 1:59 AM: First try from (P & Q) ->R deduce (P->(Q->R)). The procedure is same. In 13.5: step 3. and 4. can be one step (ab - side by side), step 6. and 7. can be one step, 9. and 10. You can input universal generalization and implication together in one step (rule is: universal generalization ) Edited by bartolmarin on Apr 26, 2011 - 2:05 AM •Spoonwood Initiate Usergroup: Members Joined: Apr 01, 2011 Total Topics: 0 Total Posts: 57 #15 - Quote - Permalink 1 of 1 people found this post helpful Posted Apr 26, 2011 - 6:46 PM: deseraimp wrote:HEY ASSHOLE. No one cares that you have your PhD in first-order logic. Don't be a condescending prick. If someone needs help, help him. I promise you do not need an egoboost. Interesting enough he had a vacuous quantifier in step 5 also. So easy? Then why isn't every line of your proof perfect? •Axiomatica Newbie Usergroup: Members Joined: Feb 22, 2012 Total Topics: 0 Total Posts: 3 #16 - Quote - Permalink 1 of 1 people found this post helpful Posted Apr 21, 2012 - 5:17 PM: You can do the proof for 13.5 in 8 steps instead of 14 steps. I think this is worthy of "resurrecting" the thread for. Here's why: The solutions to this problem turns out to be very easy if one remembers that when one introduces one or more individual constants (two in the case of these two problems), these constants are representative of arbitrary individuals in the domain of discourse. They are not tied to particular variables.Thus, in 13.5, having introduced a subproof with boxed c and d, the first premise can be instantiated as ‘(Cube(c) ˄ Dodec (d)) →Larger (d,c)’ and the second premise can be instantiated as ‘Larger (d,c) ↔ LeftOf (d,c)’ . It doesn’t have to be instantiated as ‘Larger (c,d) ↔ LeftOf (c,d)’. The proof follows immediately (and easily).1. ∀x ∀y ((Cube(x) ∧ Dodec(y)) → Larger(y, x))2. ∀x ∀y (Larger(x, y) ↔ LeftOf(x, y))3. Assume Cube(c) ^ Dodec(d)4. ((Cube(c) ^ Dodec(d)) → Larger(d,c)) ∀Elim 15. (Larger(d,c) ↔ LeftOf(d,c)) ∀Elim 26. Larger(d,c) -> Elim 3,47. LeftOf(d,c) <-> Elim 6,5 End assumption8. ∀x ∀y ((Cube(x) ^ Dodec(y)) → LeftOf(y, x)) ∀Intro 3-7Hope this helps.
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