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### Disjunction confusion!? Posted: September 29th, 2017, 4:09 am
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)

I can see that premise one can be easily decomposed into p -> (m v s)

I have also come across the rule that I CANNOT decompose a disjunction ... so what direction/s can I take?

I was thinking of using De Morgan's Law on premise (3) by applying negation, but reading that I CANNOT decompose a disjunction has confused me.

The rule I read says I CANNOT do anything with (A v B), meaning I cannot derive either A or B from (A v B). I was thinking of this for premise (3):

negation of (q v r)

therefore ~(q v r) De Morgan's to (~q ^ ~r) then conjunction elimination and double negation to derive either q or r.

Other than that maybe (q v r) <=> ~r -> q, but that doesn't seem to help much :(

I understand that I have to get r to prove p in premise (2) then use that in premise (1).

NOTE: I think it's due to m <=> p, so if I prove (s ^ p) then it's complete?

### Re: Disjunction confusion!? Posted: October 1st, 2017, 4:33 am
BadgerJelly » September 29th, 2017, 2:09 am wrote:I have written 3 random premises and a random thing to prove.

If the premises and conclusion are random, what makes you think the conclusion follows from the premises?

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)

But the conclusion does not follow from those premises. To show that, we only need to demonstrate a truth assigment that makes the premises true and the conclusion false.

Let m be False and r True. I claim this makes each premise true yet the conclusion false.

* Since m is False, the conclusion (s ^ m) is False. Now we need only show that the premises are all true.

1: m is False so ~m is True so (~m v s) is True. Therefore (m ^ p) v (~m v s) is True, and p -> (m ^ p) v (~m v s) is True. So 1 is satisfied.

2: r is true so (p ^ m) v r is True so 2 is satisfied.

3. r is True so q v r is True and 3 is satisfied.

Under the truth assignment m = False and r = True, premises 1, 2, and 3 are True but the conclusion is False. No algebraic manipulations could possibly result in a valid proof of the conclusion from these premises, since this conclusion does not follow from these premises.

ps -- I see that I assumed that p -> (m ^ p) v (~m v s) means p -> ((m ^ p) v (~m v s)). Is that what you intended? If you meant (p -> (m ^ p)) v (~m v s) this might be different. Or maybe not. m False means (~m v s) is True so premise 1 is still satisfied and the rest of it goes through without change. But you should be careful to parenthesize so as to avoid ambiguity.

Also there's some confusion here:

"The rule I read says I CANNOT do anything with (A v B), meaning I cannot derive either A or B from (A v B)."

But of course from (A v B) we can ALWAYS derive either A or B. That's exactly what (A v B) means. You may have misunderstood something you read. From "It's Tuesday or it's raining" you can conclude that either it's Tuesday or it's raining (or both, this being the inclusive or).

### Re: Disjunction confusion!? Posted: October 1st, 2017, 10:06 am
I have no idea what why you are calling (s ^ m) a "conclusion"?

I would have been better with double brackets, but it is not essential because syntax dictates that conjunctions and disjunctions are resolved prior to implications.

For clarities sake :

(p -> ((m ^ p) v (~m v s)))

It is clear enough to me now I must suppose p

so I am left with ...

(m ^ p) v (~m v s)

This seems to me to break down into p -> s, then I place into premise (2) and get (s ^ m) v r

### Re: Disjunction confusion!? Posted: October 1st, 2017, 11:57 am
BadgerJelly » October 1st, 2017, 8:06 am wrote:I have no idea what why you are calling (s ^ m) a "conclusion"?

Because you called it the "thing to prove." That's the conclusion. Isn't it what you say you're trying to prove? But how can you prove it when it doesn't follow from the premises?

You wrote this:

"I want to prove (s ^ m)"

The thing you "want to prove" is called the conclusion. That's why I called it the conclusion. Because it's the thing you say you want to prove. That's what the word means.

BadgerJelly » October 1st, 2017, 8:06 am wrote:I would have been better with double brackets,

What do double brackets do?

Are you trying to prove the "thing to prove" from the premises? Or if not, what are you trying to do?

### Re: Disjunction confusion!? Posted: October 1st, 2017, 3:13 pm
I meant there is no need for the use of parenthesis you mentiones if you understand the hierarchy of the symbols.

If not I am trying to disprove them.

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?

### Re: Disjunction confusion!? Posted: October 1st, 2017, 4:05 pm
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.

You're right, v precedes -> and no parens were needed.

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?

Let's strip this down to the simplest case. Let P and Q be propositions. (I prefer upper case variables for propositions, I hope that's ok with you).

Now, is there a syntactic derivation from P to Q? No there isn't.

Is there a syntactic derivation from P to ~Q? No there isn't.

That's not strange at all. It corresponds to the fact that some truth value assignments make P true and Q false, and other truth value assignments make P true and Q true.

It's simply not the case that given any set of premises and some claimed conclusion, that there must be a syntactic derivation from the premises to either the conclusion or its negation.

By the way "conclusion" just means "the thing we're trying to prove" I'm not sure why there's confusion about this point but I hope it's clear. You have a bunch of premises and some statement you're trying to prove from the premises. The statement you're trying to prove is called the conclusion.

### Re: Disjunction confusion!? Posted: October 1st, 2017, 10:08 pm
I am playing with this. That is how I learn.

The only premise that matters is (1) as far as I can see.

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! haha! It looks to me to at least show something of a loop?

If I have a premise with both m and ~m what does this tell me straight of the bat?

note I can just as easily use this ... (~s -> ~m) instead of ... (m -> s)

Like I said I am just playing around to see what I can do and what I can find out by juggling things around. 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.

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.

I can derive A or B if I have the premise of (A -> C) and (B -> C). Then I can use disjunction elimination on (A v B) to prove C and 'free up' the A and the B.

note: I thought there was a difference between a proof and a conclusion. I thought conclusion was the same as a conditional proof? Anyway, I sort through the semantics later. Still struggling to break out of algebraic mindset.

I think by what I've done I've shown that If p then s, if s then m, if m then p etc ...

I have shown it is a tautology. So (s ^ m) is true because s is the same as saying m. Premises 2 and 3 do nothing as far as I can tell?

### Re: Disjunction confusion!? Posted: October 2nd, 2017, 12:30 am
BadgerJelly » October 1st, 2017, 8:08 pm wrote:I am playing with this. That is how I learn.

Whatever works for you is fine with me. I'll just make my comments and perhaps you'll find something of value.

BadgerJelly » October 1st, 2017, 8:08 pm wrote:The only premise that matters is (1) as far as I can see.

Since I already showed that the three premises can't prove the conclusion, throwing out two of the premises can't possibly improve things.

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!

Well we agree on that! I was going to write, "What are you doing?" I can't tell what you're doing.

BadgerJelly » October 1st, 2017, 8:08 pm wrote:haha! It looks to me to at least show something of a loop?

There are no loops involved in sentential logic. I understand that we all have our own learning styles, but you have some fundamental misconceptions going on. There are no loops involved in any of this.

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?

Well you don't. Not by themselves. They're buried in different clauses. You can't draw conclusions without considering the rest of the context.

BadgerJelly » October 1st, 2017, 8:08 pm wrote:note I can just as easily use this ... (~s -> ~m) instead of ... (m -> s)

(m -> s) does not appear in your problem. I don't understand where this came from.

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.

Well what are you trying to do? Perhaps that's the part I'm not understanding.

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.

I already gave the proof, which is very simple. The truth value assignment m = False and r = True makes each of your three premises true and your conclusion false. So there can be no syntactic derivation from these premises to the conclusion.

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.

If you have A v B you can derive either A or B.

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).

But that isn't true. Suppose A and B are both false. Then (A -> C) is true and (B -> C) is true regardless of what C is. Therefore "(A -> C) and (B -> C)" is true, yet neither A nor B are true. That refutes what you just claimed.

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.

But you can't derive A v B from (A -> C) and (B -> C). What happens if A and B are false? Then both implications are true hence the conjunction of the implications is true. Yet A and B are both false.

BadgerJelly » October 1st, 2017, 8:08 pm wrote:note: I thought there was a difference between a proof and a conclusion.

Yes of course. The proof is the chain of reasoning that gets from the premises to the conclusion. In the present case your premises 1, 2, and 3 are the premises and (s ^ m) is the conclusion. As it happens there is no proof from these premises to this conclusion. But if there were, the premises would be the premises, the conclusion would be the conclusion, and the proof would be the chain of reasoning (in whatever form) that leads from the premises to the conclusion.

BadgerJelly » October 1st, 2017, 8:08 pm wrote:I thought conclusion was the same as a conditional proof?

No not at all. In your problem, (s ^ m) is the conclusion. It's the thing you're trying to prove.

BadgerJelly » October 1st, 2017, 8:08 pm wrote:Anyway, I sort through the semantics later. Still struggling to break out of algebraic mindset.

You have some fundamental misunderstandings. Are you still planning to take that MOOC? That will probably help a lot. Self-study can be inefficient.

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 ...

Lost me here I'm afraid.

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.

No that's just not true. This is not a matter of learning style. What you said is not true.

* Suppose m is false.

* Then (s ^ m) is false, since it's an "and" with one of the clauses false.

* Then (m ^ p) is false for the same reason.

* (~m v s) is true, since ~m is true and an "or" only needs one of the clauses to be true in order for the "or" to be true.

* Then (m ^ p) v (~m v s) is true for the same reason. The clause (~m v s) is true.

* Then p -> (m ^ p) v (~m v s) is true because it's a conditional with the consequent (the part to the right of the arrow) true.

So we've just shown that the premise (premise 1) is true yet the conclusion (s ^ m) is false, given this particular assignment of truth values. Therefore there can be no syntactic derivation from premise 1 to the conclusion (s ^ m).

BadgerJelly » October 1st, 2017, 8:08 pm wrote:Premises 2 and 3 do nothing as far as I can tell?

Yes but premise 1 does't do anything either in this case.

Well like I say if anything I said is helpful that's all to the good. I really think having a formal class with an instructor will make a big difference.

### Re: Disjunction confusion!? Posted: October 2nd, 2017, 3:19 am
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))

If you don't follow I was just using this:

(~A -> B) <=> (A v B) ... (I assume you understand this rule of equivalence?)

SO if p is true or false then I get m, because it is a tautology, and if m then s, if s then p, if ~p then ~s, if ~p then s, EVERYTHING is provable because it is a tautology. This is what I meant by "loop".

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!)

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.

IF we assume m is false, OR any other combination for m, p or s then premise (1) is true. Premises (2) and (3) do what then? 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. 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)

### Re: Disjunction confusion!? Posted: October 2nd, 2017, 2:33 pm
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!)

That's right. Assuming m is false, premise 1 is true while conclusion (s ^ m) is false. This shows that there can be no possible formal derivation from the premise to the conclusion.

But how do I know this? It's been on my mind that I owe you an explanation of this point. I'm glad you brought it up. I will discuss that in this post, and I'll comment on the rest of your post a bit later.

I have been using the following fact.

(*) Suppose P1, P2, ..., Pn are premises; and C is a conclusion. Then there is a syntactic derivation from the premises to the conclusion if and only if every possible assignment of truth values to the variables makes the conditional P1 ^ P2 ^ ... ^ Pn -> C true.

If you believe that, then in order to show that there's no syntactic derivation from your premises 1, 2, and 3 to the conclusion (s ^ m), all I have to do is find some assignment of truth values that makes (s ^ m) false while making each of 1, 2, and 3 true. This can be done by the assignment m = False as I've demonstrated a couple of times.

But what is my justification for (*)? This turns out to be something called the completeness theorem for propositional logic.

Suppose we want to derive A from A. In other words we want to show that the conditional A -> A is true under any possible assignment of truth values to the propositions. Well if A is true, we have "true implies true" which is true. And if A is false, we have "false implies false" which is also true.

In anybody's formal system of deduction (Fitch, natural deduction, etc.), given A we can always infer A. This is an application of the completeness theorem. There's a formal deduction if and only if every assignment of truth values makes the associated conditional true.

On the other hand, suppose we're given A and we wish to derive B. Can we do this?

The corresponding conditional is A -> B. Is this true under every possible assignment of truth values?

Well, if A and B are both true, A -> B is true. If A and B are both false, A -> B is true. And if A is false and B is true, then A -> B is true. We're almost there!

BUT!!!! If A is true and B is false, then A -> B is false. So we've found a truth value assignment that makes the conditional A -> B false. And therefore by the completeness theorem there can never be a derivation from A to B, no matter how hard we try.

I don't know if the completeness theorem is taught in your class or your book or if you've run across it; and I should not have invoked it without explanation. But now that I've explained it, this is the principle I've been using.

Let's look at this another way. A tautology is a statement that evaluates to true no matter how you assign truth values to its individual components. A -> A is a tautology, it's "true in all possible worlds" as some philosophers would put it.

But A -> B is not a tautology because it's true in some worlds and false in others. It's true in the worlds where A and B are both true, A and B are both false, or A is false and B is true. But it's false in the world where A is true and B is false. Therefore A -> B is not a tautology.

A theorem is a statement that can be derived formally (via syntactic derivation) from some premises.

So another way to state the completeness theorem is that:

A statement is a theorem if and only if it's a tautology.

That is, there is a formal derivation if and only if every assignment of truth values makes the associated conditional true.

Now in the case of premises 1, 2, and 3 and the conclusion (s ^ m), if we assign m = False and r = True, then the conjunction 1 ^ 2 ^ 3 is true, while (s ^ m) is false. So the conditional

1 ^ 2 ^ 3 -> (s ^ m)

is false under that particular assignment of truth values, namely m = false and r = true. Therefore by the completeness theorem there can not be a formal derivation of premises 1, 2, and 3 to the conclusion (s ^ m).

Likewise if we just take premise 1 by itself, the assignment m = False makes 1 -> (s ^ m) false, so there can not possibly be any syntactic derivation from 1 to (s ^ m).

### Re: Disjunction confusion!? Posted: October 2nd, 2017, 4:33 pm
Remarks on the rest of your post.

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))

Yes ok but to what end? What are you trying to do?

Also you have an error here:

BadgerJelly » October 2nd, 2017, 1:19 am wrote:(p -> m) v (p -> (m -> s))

Which then goes to

(~p -> m) -> (p -> (m ->s))

No. You can't replace (p -> m) with (~p -> m).

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?)

Yes but what are you trying to do? You haven't established that (s ^ m) must necessarily be true.

BadgerJelly » October 2nd, 2017, 1:19 am wrote:SO if p is true or false then I get m

This I do not see. And it must be false as I've already shown, because if m is false then premise 1 is true yet (s ^ m) is false.

BadgerJelly » October 2nd, 2017, 1:19 am wrote:, because it is a tautology, and if m then s

Premise 1 may be a tautology but so what if m then s? If m is false then "if m then s" is still true.

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.

No that is wrong. Everything is provable from a contradiction. From A ^ ~A we can prove anything. But from A V ~A, which is a tautology, we can't prove anything else.

BadgerJelly » October 2nd, 2017, 1:19 am wrote:This is what I meant by "loop".

I can't make you think otherwise, but there are no loops involved in propositional logic.

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!)

Yes but your conclusion (s ^ m) would be false. So your conclusion would not follow from your premise, showing that (by the completeness theorem) there is no formal derivation from 1 to (s ^ m).

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.

The purpose of examining m = false is to show that there is a truth value assigment that makes 1 true and (s ^ m) false.

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.

Yes. And that's exactly the point. (1) is true yet (s ^ m) is false. Showing that there can be no derivation from (1) to (s ^ m).

BadgerJelly » October 2nd, 2017, 1:19 am wrote:Premises (2) and (3) do what then?

I have no idea. They're your premises. You tell me what they do.

I did note that if in addition to m = false we also take r = true, then not only (1) is true but (2) and (3) are as well, and still (s ^ m) is false.

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.

Correct. From a truth you can't conclude anything. If I tell you A is true, then I ask about B, you can't draw any inference. A -> B is not a tautology nor is A -> ~B. From a tautology you can't prove anything. From a contradiction you can prove everything.

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)

As I've noted, there is no derivation of (s ^ m) from your premises, either 1 by itself or 1, 2, and 3. You are attempting something that can't be done.

### Re: Disjunction confusion!? Posted: October 3rd, 2017, 12:11 am
"BadgerJelly » October 2nd, 2017, 1:19 am wrote:
(p -> m) v (p -> (m -> s))

Which then goes to

(~p -> m) -> (p -> (m ->s))"

you said:

"No. You can't replace (p -> m) with (~p -> m)."

I didn't. I used equivalence to get rid of the disjunction.

I see that I cannot prove (s ^ m) now. There are still some things you've said I did further clarity on. Later ...

### Re: Disjunction confusion!? Posted: October 3rd, 2017, 12:44 pm
Ok. Completeness theorem is philosophically interesting. It connects syntax with semantics. Syntax is just about manipulating the symbols according to rules. I think that's what you mean by the algebraic approach.

Semantics is concerned with meaning. Mapping the symbols to facts about the world (or possible worlds) that may be true or false.

### Re: Disjunction confusion!? Posted: October 3rd, 2017, 1:22 pm
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)

The problem I see is that the premises are vacuous.
2 and 3 don't tell you anything because of the (v r) tacked on the end.
1 reduces to p->m v ~m v s, which is vacuously true, telling you nothing about s.
Thus I agree that s ^ m cannot be proven from these premises.

BadgerJelly » September 29th, 2017, 3:09 am wrote:I have written 3 random premises and a random thing to prove.

Which just goes to show that making up workable problems for students is very much harder than this.

### Re: Disjunction confusion!? Posted: October 5th, 2017, 8:06 pm
I got an email reminding me that I'm actually still enrolled in the Coursera logic class that BadgerJelly is taking. I found a very significant omission in the Week 1 material that bears on this discussion.

In lesson 1.4, they define entailment, also known in logic as semantic entailment, as the condition of a statement or a set of statements being true in all possible worlds.

Here is the relevant passage.

https://s1.postimg.org/6e63yv649r/Scree ... .32_PM.png

This is a semantic notion because it relates to meaning. We are not just talking about the formal manipulations of abstract symbols -- the "algebra" of logic. Rather, we are assigning truth values to our variables; and considering the notion of "possible worlds". What can these things possibly mean in a conversation about formal symbol manipulation? Nothing, in fact. Assignments of truth values and consideration of whether a given statement is "true" in some "possible world" is a question of semantics -- that is, meaning. It's the exact opposite of the concept of the purely formal manipulation of meaningless symbols.

Then in the next section they say this:

https://s1.postimg.org/6l9buas1zj/Scree ... .51_PM.png

In other words they claim, without any explanation, that if a sentence or set of sentences happen to be true in all "possible worlds" -- whatever that may mean -- then there must necessarily be a formal symbolic derivation of those sentences.

But why is this? Why should there be such a mysterious connection between the world of meaning, and the world of meaningless symbols? Why is it the case that there happens to be a formal derivation of a sentence just in case that sentence is true in all possible worlds?

That is actually a very deep point! It is exactly the content of the completeness theorem.

I can't say I agree with the pedagogy. There is a priori no reason at all that a sentence that happens to be true under every possible truth value assignment should necessarily have a formal symbolic proof. This is a theorem, and a rather deep one. The instructor did nobody a service by leaping so casually over this point, and by failing to explain the difference between syntax and semantics.

[Meta -- img tags don't appear to work].

### Re: Disjunction confusion!? Posted: October 5th, 2017, 9:39 pm
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.

### Re: Disjunction confusion!? Posted: October 5th, 2017, 9:55 pm
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.

I thought the format was awful too. You're better off with a different book. What are you reading if I may ask? Afraid I haven't any specific recommendations.

### Re: Disjunction confusion!? Posted: October 5th, 2017, 11:34 pm
Nothing other than what I find on youtube and some random sites, docs. PLUS I am studying linguistics and math, so they all play off each other to some degree.

I am going to order The Logic Book, which is used by a guy on youtube "Trevtutor", he is pretty good and briefly covers the general area I am working in regarding math, logic and linguistics.

Will be delving into Kripke later on ... just getting basics down at the moment. Then I'll get more into neurolinguistics and computer programming.

Sadly I am not in a position where I can just go down to the bookstore and buy a book. Nor can I order on amazon, so I have to use my friends/family back in UK as proxy to deliver my books :(

From what I can tell The Logic Book by Bergmann seems to be the standard university textbook. In the meantime I'll start to work through the book that mitch posted the pdf for. I generally prefer to work from an actual book though and not off a computer screen so would be damn helpful if I could find a reasonable textbook that I can print out.

A friend in Canada has some books on logic from his philosophy course, but I am hardly going to ask him to post them to me! (no way of paying him at the moment.)