Hi.

Please help :)

How to prove:

1.p=>q - permission goal ~q=>~p

2. ~p=>q - permission and q=>r goal (~p=>~r)=>p

3. ~pI~q goal ~(p^q)

Thanks

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- April 20th 2013, 07:13 AManastazyaFitch :(
Hi.

Please help :)

How to prove:

1.p=>q - permission goal ~q=>~p

2. ~p=>q - permission and q=>r goal (~p=>~r)=>p

3. ~pI~q goal ~(p^q)

Thanks - April 20th 2013, 07:54 AMemakarovRe: Fitch :(
Since you did not edit your post to add a note that it has been solved, was the [SOLVED] tag added intentionally?

- April 20th 2013, 11:26 AManastazyaRe: Fitch :(
Thanks. I found a solution :)

Only the last item leaves me wrong, for example: ~ pI goal ~ q ~ (p ^ q) - I can do up to 17, then later I got wrongI do not know why? Maybe I do not understand something. From the pictures that I have noted: 1, 2-9, 10 -17 and or eliminations. - Only 18 leaves me wrong: ( - April 20th 2013, 11:46 AMemakarovRe: Fitch :(
Does "I" mean disjunction, i.e., "or"? Usually it is denoted by ∨. In ASCII one can also write \/ or the letter "v".

By "last", do you mean number 3 in the quote above?

This is hard to understand. I am not sure what "pI" is and why there is no connective between "~q" and "~(p ^ q)".

This is also hard to understand without seeing the first 17 steps. - April 21st 2013, 05:03 AMThejanieRe: Fitch :(
do u mean

premise ~p|~q

to prove ~(p&q)???

I'm also looking for the proof of this. - April 21st 2013, 06:54 AMThejanieRe: Fitch :(
1.~p | ~q Premise

2.~p Assumption

3.p & q Assumption

4.p And Elimination: 3

5.p & q => p Implication Introduction: 4

6.~p Reiteration: 2

7.p & q Assumption

8.~p Reiteration: 6

9.p & q => ~p Implication Introduction: 8

10.~(p & q) Negation Introduction: 5, 9

11.~p => ~(p & q) Implication Introduction: 10

12.~q Assumption

13.p & q Assumption

14.q And Elimination: 13

15.p & q => q Implication Introduction: 14

16.p & q Assumption

17.~q Reiteration: 12

18.p & q => ~q Implication Introduction: 17

19.~(p & q) Negation Introduction: 15, 18

20.~q => ~(p & q) Implication Introduction: 19

21.~(p & q) Or Elimination: 1, 11, 20

This is the solution I got. But it's too long. :(

Still it works :) - April 21st 2013, 10:53 AMemakarovRe: Fitch :(
I believe this is the correct derivation. I don't see how it can be shortened except by removing step 6 and making step 8 a reiteration of 2.

A more standard variant of natural deduction (of which Fitch calculus is a particular notation) has a symbol ⊥ for contradiction. Then negation introduction is shorter: you assume ~p and p & q, derive p, derive ⊥ from ~p and p, then close the assumption p & q and derive ~(p & q) in one step. There is no need to derive p & q => p and p & q => ~p. Similarly, or elimination does not require implications ~p => ~(p & q) and ~q => ~(p & q): you just derive ~(p & q) two times from assumptions ~p and ~q, respectively, and or elimination derives ~(p & q) and closes the ~p and ~q in one step. - April 21st 2013, 11:12 AMThejanieRe: Fitch :(
The program I used doesn,t have contradiction. BTW can you help me to prove this the other way around? I mean starting from ~(p & q) as the premise to prove ~p | ~q.

- April 21st 2013, 11:42 AMemakarovRe: Fitch :(
- April 21st 2013, 12:01 PMThejanieRe: Fitch :(
- April 21st 2013, 12:28 PMemakarovRe: Fitch :(
I see that you don't have the law of excluded middle, but I meant to ask what the negation elimination rule looks like. What formulas does it take and produce? I would guess it takes ~~A and produce A. Is this correct?

- April 21st 2013, 12:39 PMThejanieRe: Fitch :(
Yeah... :)

- April 21st 2013, 12:58 PMemakarovRe: Fitch :(
We have a premise ~(p & q). The proof of ~p | ~q is by contradiction, i.e., we assume ~(~p | ~q) and prove p & q (described below), which contradicts the premise. This gives ~~(~p | ~q) by negation introduction and then ~p | ~q by negation elimination.

So, assume ~(~p | ~q). We need to prove p and q. These two subderivations are similar. To prove p, we assume ~p and derive ~p | ~q, which contradicts the assumption. Therefore, ~~p and we conclude p by negation elimination. The second part, q, is proved similarly. - April 22nd 2013, 07:16 AMThejanieRe: Fitch :(
Thank you so much!!! :)

- October 23rd 2013, 06:44 PMR2013Re: Fitch :(
hi emakarov :)

I'm struggling with this one still ...

this is what I have so far:

1. ~(p & q) Premise

2. ~(~p | ~q) Assumption

3. ~p Assumption

4. ~p | ~q Or Introduction: 3

5. ~p => ~p | ~q Implication Introduction: 4

6. ~p Assumption

Now I'm stuck as to how to get to the conclusion of ~p|~q.

Can you please help?

Also ... please bear in mind I don't have the necessary background in discrete mathematics .. so is there a specific textbook or site you can recommend also so I can understand better and learn?

Thank you so much in advance :)