i'm at a loss with this one.....can anyone give me a hint to solve it with Fitch system:

Given (p ⇒ q) and (r ⇒ s), use the Fitch System to prove (p ∨ r ⇒ q ∨ s).

Printable View

- October 13th 2012, 07:26 AMrai009NEW QUESTION about Fitch
i'm at a loss with this one.....can anyone give me a hint to solve it with Fitch system:

Given (p ⇒ q) and (r ⇒ s), use the Fitch System to prove (p ∨ r ⇒ q ∨ s). - October 13th 2012, 08:52 AMrai009Re: Abstract reasoning/natural deduction with Fitch
Given (¬p ∨ ¬q), use the Fitch System to prove ¬(p ∧ q).

- October 13th 2012, 08:55 AMrai009Re: Abstract reasoning/natural deduction with Fitch
Given (p ⇒ q) and (r ⇒ s), use the Fitch System to prove (p ∨ r ⇒ q ∨ s).

- October 13th 2012, 08:58 AMrai009Re: Abstract reasoning/natural deduction with Fitch
Starting from the premise (p ⇒ q), use the Fitch System to prove the conclusion (¬q ⇒ ¬p).

- October 13th 2012, 09:33 AMHallsofIvyRe: NEW QUESTION about Fitch
You might get

**some**response if, instead of just listing more problems, you explained**what**the "Fitch system" is! - October 13th 2012, 12:58 PMemakarovRe: Abstract reasoning/natural deduction with Fitch
I'll describe the derivation with words and let you translate it to the Fitch form.

Assume p ∨ r. Use disjunction elimination. If p, then p ⇒ q gives q and therefore q ∨ s. Similarly, if r, then

r ⇒ s gives s and therefore q ∨ s. In both cases, we get q ∨ s, which concludes the disjunction elimination rule.

Use disjunction elimination on ¬p ∨ ¬q. Suppose ¬p. Assume p ∧ q, derive p. Together with ¬p this gives a contradiction; therefore, ¬(p ∧ q). Derive ¬(p ∧ q) in a similar way if ¬q.

Assume ¬q and p. Then p together with p ⇒ q this gives q, and q with ¬q gives a contradiction. Therefore, ¬p. - October 13th 2012, 02:16 PMHallsofIvyRe: NEW QUESTION about Fitch
Yes, that is how I would have done it. But what does that have to do with "the Fitch system"?

- October 14th 2012, 02:46 AMemakarovRe: NEW QUESTION about Fitch
Fitch calculus, or flag notation, is a notation for derivations in natural deduction (ND). In my English description, I used only inference rules available in ND. For example, deriving B from an assumption A in order to derive A -> B is an ND rule, but De Morgan's law is not.

- October 16th 2012, 07:21 PMrai009Re: NEW QUESTION about Fitch
Can't i solve the De Morgan's law without using the contradiction introduction tool?

only using the "implication introduction", "implication elimination", "and introduction", and elimination", or introduction, or elimination, negation introduction, negation elimination & biconditional introduction & elimination......

what do i do if i have to use this tools only to prove it with Fitch.... - October 17th 2012, 01:13 AMemakarovRe: NEW QUESTION about Fitch
Could you write the exact version of the De Morgan's law you are talking about? Also, please write all rules in your system dealing with contradiction and negation. I am asking this because different sources give different names to such rules. In particular, sometimes negation elimination is called contradiction introduction. Unlike for other connectives, it would be more correct to have only the elimination rule for contradiction, which derives any formula from contradiction. Dually, if there is a connective denoting truth, it should have only the introduction rule.

A couple of remarks about the terminology. De Morgan's law can be derived, or deduced, or proved, but not solved. Also, implication introduction and similar things may be called tools in a particular computer program, but in logic they are called inference rules. - October 17th 2012, 07:26 AMrai009Re: NEW QUESTION about Fitch
the version of De Morgan's law i am talking about is: Given (¬p ∨ ¬q), use the Fitch System to prove ¬(p ∧ q).

and the 'negation elimination' in this case does not mean contradiction introduction.....it only eliminates double negations such as: ~~p by negation elimination we get p.

negation introduction does something like this: when i find p&q=>q ,p&q=>~q then by negation introduction i get ~(p&q)....i reached to this conclusion but i have a dependence line in my program.....that's why it's saying the proof is incomplete........and when i try to use it on p, ~p it doesn't do anything....so i am not sure whether NI can be called contradiction intro in this case.......

and there is no contradiction tool in my program to use....that's why i can't use contradiction introduction rule....

implication introduction and elimination rules are what they're supposed to do...........though i'm not sure about biconditional introduction and elimination because i don't understand them well enough.......

i'm still new to the use of inference rules in propositional logic. - October 17th 2012, 08:07 AMemakarovRe: NEW QUESTION about Fitch
What about the following derivation?

Code:`1. ~p \/ ~q Assumption`

2. ~p Assumption

3. p /\ q Assumption

4. p 3, /\E

5. p /\ q -> p 3-4, ->I

6. p /\ q Assumption

7. ~p 2

8. p /\ q -> ~p 6-7, ->I

9. ~(p /\ q) 5, 8, ~I

10. ~q Assumption

11. p /\ q Assumption

12. q 11, /\E

13. p /\ q -> q 11-12, ->I

14. p /\ q Assumption

15. ~q 10

16. p /\ q -> ~q 14-15, ->I

17. ~(p /\ q) 13, 16, ~I

18. ~(p /\ q) 1, 2-9, 10-17, \/E

19. ~p \/ ~q -> ~(p /\ q) 1-18, ->I

- October 17th 2012, 09:26 AMrai009Re: NEW QUESTION about Fitch
yes it works..........but how do we prove ~p|~q from the given ~(p&q)

- October 17th 2012, 09:28 AMemakarovRe: NEW QUESTION about Fitch
By | in ~p|~q do you mean disjunction? Why not use the same notation as in the previous posts?

- October 18th 2012, 08:08 AMrai009Re: NEW QUESTION about Fitch
oh yes ..........sorry for that.

Given ¬(p∧q), use the Fitch system to prove (¬p∨¬q)