Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Oct 13th 2012, 07:26 AM
rai009
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).
• Oct 13th 2012, 08:52 AM
rai009
Re: Abstract reasoning/natural deduction with Fitch
Given (¬p ∨ ¬q), use the Fitch System to prove ¬(p ∧ q).
• Oct 13th 2012, 08:55 AM
rai009
Re: Abstract reasoning/natural deduction with Fitch
Given (p ⇒ q) and (r ⇒ s), use the Fitch System to prove (p ∨ r ⇒ q ∨ s).
• Oct 13th 2012, 08:58 AM
rai009
Re: Abstract reasoning/natural deduction with Fitch
Starting from the premise (p ⇒ q), use the Fitch System to prove the conclusion (¬q ⇒ ¬p).
• Oct 13th 2012, 09:33 AM
HallsofIvy
You might get some response if, instead of just listing more problems, you explained what the "Fitch system" is!
• Oct 13th 2012, 12:58 PM
emakarov
Re: Abstract reasoning/natural deduction with Fitch
I'll describe the derivation with words and let you translate it to the Fitch form.
Quote:

Originally Posted by rai009
Given (p ⇒ q) and (r ⇒ s), use the Fitch System to prove (p ∨ r ⇒ q ∨ s).

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.

Quote:

Originally Posted by rai009
Given (¬p ∨ ¬q), use the Fitch System to prove ¬(p ∧ q).

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.

Quote:

Originally Posted by rai009
Starting from the premise (p ⇒ q), use the Fitch System to prove the conclusion (¬q ⇒ ¬p).

Assume ¬q and p. Then p together with p ⇒ q this gives q, and q with ¬q gives a contradiction. Therefore, ¬p.
• Oct 13th 2012, 02:16 PM
HallsofIvy
Yes, that is how I would have done it. But what does that have to do with "the Fitch system"?
• Oct 14th 2012, 02:46 AM
emakarov
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.
• Oct 16th 2012, 07:21 PM
rai009
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....
• Oct 17th 2012, 01:13 AM
emakarov
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.
• Oct 17th 2012, 07:26 AM
rai009
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.
• Oct 17th 2012, 08:07 AM
emakarov

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```
• Oct 17th 2012, 09:26 AM
rai009