BadgerJelly » September 29th, 2017, 2:09 am wrote:I have written 3 random premises and a random thing to prove.
BadgerJelly » September 29th, 2017, 2:09 am wrote:Premises:
1- p -> (m ^ p) v (~m v s)
2- (p ^ m) v r
3- q v r
I want to prove (s ^ m)
BadgerJelly » October 1st, 2017, 8:06 am wrote:I have no idea what why you are calling (s ^ m) a "conclusion"?
BadgerJelly » October 1st, 2017, 8:06 am wrote:I would have been better with double brackets,
BadgerJelly » October 1st, 2017, 1:13 pm wrote:I meant there is no need for the use of parenthesis you mentiones if you understand the hierarchy of the symbols.
BadgerJelly » October 1st, 2017, 1:13 pm wrote:I
I am assuming the premises are true not the "conclusion". If there is no way for me to prove or disprove (s ^ m) would be kind of strange. If that is the case then why is this?
BadgerJelly » October 1st, 2017, 8:08 pm wrote:I am playing with this. That is how I learn.
BadgerJelly » October 1st, 2017, 8:08 pm wrote:The only premise that matters is (1) as far as I can see.
BadgerJelly » October 1st, 2017, 8:08 pm wrote:Can I do this :
p -> (m ^ p) v (~m v s) <=> (p -> m) v (p -> (~m v s))
then
(~m v s) <=> (m -> s)
so
(p -> m) v (p -> (m -> s))
Which then goes to
(~p -> m) -> (p -> (m ->s))
Of exactly what use this is I don't honestly know!
BadgerJelly » October 1st, 2017, 8:08 pm wrote:haha! It looks to me to at least show something of a loop?
BadgerJelly » October 1st, 2017, 8:08 pm wrote:If I have a premise with both m and ~m what does this tell me straight of the bat?
BadgerJelly » October 1st, 2017, 8:08 pm wrote:note I can just as easily use this ... (~s -> ~m) instead of ... (m -> s)
BadgerJelly » October 1st, 2017, 8:08 pm wrote:Like I said I am just playing around to see what I can do and what I can find out by juggling things around.
BadgerJelly » October 1st, 2017, 8:08 pm wrote:If I cannot prove (s ^ m) then I should at least be able to show my reasoning as to WHY I cannot do this.
I imagine it is likely to be a very long and monotonous proof to show why it cannot be done.
BadgerJelly » October 1st, 2017, 8:08 pm wrote:If I say A v B I cannot derive EITHER without the premise of A or B. The point above about not being able to derive A or B from this what was meant. I cannot simply assume A or B to derive either.
BadgerJelly » October 1st, 2017, 8:08 pm wrote:I can derive A or B if I have the premise of (A -> C) and (B -> C).
BadgerJelly » October 1st, 2017, 8:08 pm wrote:Then I can use disjunction elimination on (A v B) to prove C and 'free up' the A and the B.
BadgerJelly » October 1st, 2017, 8:08 pm wrote:note: I thought there was a difference between a proof and a conclusion.
BadgerJelly » October 1st, 2017, 8:08 pm wrote:I thought conclusion was the same as a conditional proof?
BadgerJelly » October 1st, 2017, 8:08 pm wrote:Anyway, I sort through the semantics later. Still struggling to break out of algebraic mindset.
BadgerJelly » October 1st, 2017, 8:08 pm wrote:I think by what I've done I've shown that If p then s, if s then m, if m then p etc ...
BadgerJelly » October 1st, 2017, 8:08 pm wrote:I have shown it is a tautology. So (s ^ m) is true because s is the same as saying m.
BadgerJelly » October 1st, 2017, 8:08 pm wrote:Premises 2 and 3 do nothing as far as I can tell?
BadgerJelly » October 2nd, 2017, 1:19 am wrote:I don't understand why you would "suppose m is false" this would still show the whole of premise (1) to be true (because it's a tautology!)
BadgerJelly » October 2nd, 2017, 1:19 am wrote:This is what was interesting me the most:
p -> (m ^ p) v (~m v s) <=> (p -> m) v (p -> (~m v s))
then use
(~m v s) <=> (m -> s)
so
(p -> m) v (p -> (m -> s))
Which then goes to
(~p -> m) -> (p -> (m ->s))
BadgerJelly » October 2nd, 2017, 1:19 am wrote:(p -> m) v (p -> (m -> s))
Which then goes to
(~p -> m) -> (p -> (m ->s))
BadgerJelly » October 2nd, 2017, 1:19 am wrote:If you don't follow I was just using this:
(~A -> B) <=> (A v B) ... (I assume you understand this rule of equivalence?)
BadgerJelly » October 2nd, 2017, 1:19 am wrote:SO if p is true or false then I get m
BadgerJelly » October 2nd, 2017, 1:19 am wrote:, because it is a tautology, and if m then s
BadgerJelly » October 2nd, 2017, 1:19 am wrote:, if s then p, if ~p then ~s, if ~p then s, EVERYTHING is provable because it is a tautology.
BadgerJelly » October 2nd, 2017, 1:19 am wrote:This is what I meant by "loop".
BadgerJelly » October 2nd, 2017, 1:19 am wrote:I don't understand why you would "suppose m is false" this would still show the whole of premise (1) to be true (because it's a tautology!)
BadgerJelly » October 2nd, 2017, 1:19 am wrote:It makes no sense to prove the conclusion by using it as a premise? Why would I suppose ~m for the conclusion by using it in the conclusion? That sounds insane.
BadgerJelly » October 2nd, 2017, 1:19 am wrote:IF we assume m is false, OR any other combination for m, p or s then premise (1) is true.
BadgerJelly » October 2nd, 2017, 1:19 am wrote:Premises (2) and (3) do what then?
BadgerJelly » October 2nd, 2017, 1:19 am wrote:If one of the premises is tautological, and that premise happens to be the only one with s in it, then I can go nowhere.
BadgerJelly » October 2nd, 2017, 1:19 am wrote:I can only attempt to derive m and the best I could do from there is disjunction introduction which would get me (m v s)
BadgerJelly » September 29th, 2017, 3:09 am wrote:I have written 3 random premises and a random thing to prove.
Premises:
1- p -> (m ^ p) v (~m v s)
2- (p ^ m) v r
3- q v r
I want to prove (s ^ m)
BadgerJelly » September 29th, 2017, 3:09 am wrote:I have written 3 random premises and a random thing to prove.
BadgerJelly » October 5th, 2017, 7:39 pm wrote:I stopped using that site. The vids constantly freeze and the logic course is too brief. I am using other sources instead. Will buy The Logic Book though.
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