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Well, indeed I have the answer to this exercise but I don't understand some steps. From 6 to 17 are OK for me but from 2 to 5 and then when step 5 is again called in step 10 is something I don't get very well.

I mean, could you re use a formula from a conditional box from above in other one? I would never have thought that. enter image description here

  • 2-5) assumed "p to q", derived a contradiction and concluded with "not (p to q)". – Mauro ALLEGRANZA Nov 24 '19 at 9:03
  • 6-13) assumed q and derived "p to q" (this can be dona more simply by to-intro). Now this contradicts "not (p to q)" derived above in step 5) and we can conclude with p, and from it with "q to p" by to-intro. – Mauro ALLEGRANZA Nov 24 '19 at 9:06
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The symbol F stands for 'False' or 'Contradiction'. Negation is at times defined as "this implies a contradiction".

~P is equivalent to P => F

Your system reuses the Conditional Elimination and Introduction symbols (=>E, =>I), but the inferences are more commonly known as Negation Elimination and Introduction.

Just as we may infer F from P and P => F by Conditional Elimination, so too may we infer F from P and ~P by Negation Elimination.

|  P
|  P=>F
|  F     Conditional Elimination (=>E)

|  P
|  ~P
|  F     Negation Elimination  

Or basically: We may derive F from any two statements which contradict each other. Such as lines 5 and 10, ~(p->q) and p->q. Also line 1 ~((p->q)+(q->p) and 4 (and later 15) (p->q)+(q->p)


Likewise: Just as we may infer P => F when we can derive F under assumed P by Conditional Introduction, so similarly we may infer ~P when we can derive F under assumption of P by Negation Introduction. These are assumption discharge rules

|  |_ P    Assumption
|  |  :
|  |  F
|  P => F  Conditional Introduction (=>I)

|  |_ P    Assumption
|  |  :
|  |  F
|  ~P      Negation Introduction

Note: what your system calls RAA is more commonly known as Double Negation Elimination (DNE). (RAA is more usually where: we may infer P when we can derive F under assumption of ~P).


   |_
 1.|  |_ ~((p -> q) + (q -> p))    Assumption
 2.|  |  |_ (p -> q)               Assumption
 3.|  |  |  (p -> q) + (q -> p)    2, Disjunction Introduction
 4.|  |  |  F                      1,3 Negation Elimination 
 5.|  |  ~(p -> q)                 2-4 Negation Introduction
 6.|  |  |_ q                      Assumption
 7.|  |  |  |_ ~p                  Assumption
 8.|  |  |  |  |_ p                Assumption
 9.|  |  |  |  |  q                6 Reiteration
10.|  |  |  |  p -> q              8-9 Conditional Introduction
11.|  |  |  |  F.                  5,10 Negation Elimination
12.|  |  |  ~~p                    7-11 Negation Introduction
13.|  |  |  p                      12   Double Negation Elimination
14.|  |  q -> p                    6-13 Conditional Introduction
15.|  |  (p -> q) + (q -> p)       14 Disjunction Introduction
16.|  |  F                         1,15 Negation Elimination
17.|  ~~((p -> q) + (q -> p))      1-16 Negation Introduction
18.|  (p -> q) + (q -> p)          17 Double Negation Elimination

I mean, could you re use a formula from a conditional box from above in other one? I would never have thought that.

Yes. This is the Rule of Reiteration used on line 8. You can restate something from a lower context. Think of it as 'moving into the box', if you like.

Usually Reiteration is not needed to be explicitly used; you may just refer to statements from ancestral contexts.

Remember you cannot do the reverse. You cannot 'move out of the box'.

| improve this answer | |
0

Assume ~ [ (p-->q) v (q -->p)].

Using DeMorgan, infer : ~ (p-->q) & ~ (q-->p)

Using &-elim, infer

(1) ~ (p-->q)

(2) ~ (q-->p)

Using ~ ( A& ~B) as a definition of material implication, and Double Negation infer

  • from (1) : p & ~Q

  • from (2) : q& ~p

Using &-intro, & commutativity and &-associativity, and finally &-elim, rearrange (p & ~q)&( q& ~p) to get the contradiction (p&~p).

Finally, using ~ intro, infer that the original assumption was false.

Double negation will then allow you to infer that the goal-statement was true.

| improve this answer | |
  • Hi and thanks for replying but I am not allowed to use DeMorgan or similars. What I mean is that any further than MP, MT, &E, &I, vE, vI, ->, <->, RAA and Hyp rules are forbidden. That's why I still don't get the solution given. For me there's no point of 2-5 and from steps 7 and 8 I could get easlily a contradiction suitable for "p". Literal later used in 13 for "q->p". – daenius Nov 24 '19 at 16:18
  • @daenius.-are you allowed to use an equivalence for (X--> Y) ? – user39744 Nov 24 '19 at 22:01
  • 1
    what do you mean? "-X v Y" ? This kind of "quick" transformations or equivalences are not allowed. For example, if we use Modus Ponens (MP as above) is because we have this scenario: p, p->q, so we have q – daenius Nov 25 '19 at 8:32
  • @daenius.- This is a rather demanding logical system. How is it called? – user39744 Nov 25 '19 at 8:37
  • don't worry. I've found the solution. Indeed It was a little different from the image from above. Thanks anyway – daenius Nov 26 '19 at 17:41

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