Do you want only Corvus to reply?

if you agree with corvus, I wouldn't mind talking to you about why.

If you disagree, it would be appreciated if you expressed briefly why but, ideally, exited the thread after that to avoid too many overlapping debates and confusion.

If corvus has other preferences, I welcome him to express them and hopefully they can be accommodated.

Logic is a science of inference. That is one of my old logic book says. To prove a statement for true or false, you must start with a premise to prove, then make the relevant inference. (Don't make irrelevant nonsensical inferences such as hosing a garden into this argument, and you get told, are you having a laugh mate?)

If it rains, then the ground will be wet. (Premise to prove)
R -> W

If it doesn't rain, then the ground will not be wet. (Inference = WHY NOT??)
Not R -> Not W

At this moment, you check to see if it rains, and if the ground is wet in reality.
It doesn't rain, and the ground is not wet. (TRUE from the inspection and observation)
Not R -> Not W = True

Therefore If it rains, then the ground will be wet. (True)
R -> W

Therefore for proving A->B is true or false, Not A -> Not B premise works fine.

You can use a proof based on the Truth table too.
R | W | R and W | R or W | R-> W | NotR | NotW | NotR -> NotW
T | T | T | T | T | F | F | T
T | F | F | T | F | F | T | T
F | T | F | T | T | T | F | F
F | F | F | F | T | T | T | T

Would you mind explicitly stating if every (a implies b) also leads to (not a implies not b), or can you only do that for specific (a implies b) statements?

Are there any exceptions?

The truth table value is from the rule. But you also check with the reality observations here too, because it is an empirical argument, not analytic.

That sounds like you're confirming that yes, it's always applicable any time you have (a implies b), am I interpreting that correctly?

I think you are correct. But you don't have to always apply the contradiction. It depends on the case, and your inference.

Do you also agree with the contrapositive rule, which states that if (a implies b), then (not b implies not a)?

It would also depend on the case by case.

do you have examples where it doesn't apply?

Examples of a implies b where is not true that not b implies not a

I will get back on that, as I must go out now. cheers.

do you have examples where it doesn't apply? — flannel jesus

If not sure, make up an extended truth table for all the possible scenarios, and see all the cases for the TF values. It gets apparent.

wait you edited this response, we have to go back. We need clarity on this conversation or nothing will work.

It either always applies, or it doesn't always apply. When you say you think I'm right, that means it always applies. But then you say it depends on the case. It either always applies, or it depends on the case, it can't be both.

Does (a implies b) always lead to (not a implies not b), or only sometimes?

It is not "lead to". It is an assumption, which you bring into the argument that you think most relevant and reasonable for the premise that you want to prove. But the truth table contains all the cases and TF values. If it is an empirical case, you must check with the reality for the TF values for the real TF values.

What's the assumption? Specifically.

In the case of If it rains the the ground will be wet. You think and infer, which will be most relevant and good to test, if it is true? Ah... how about if it doesn't rain, and the will not be wet? NotR -> NotW. And check it with the reality.

But you can bring other assumptions as counter reasonings and inferences for the proof process. If not sure, make a truth table, insert all the TF values according to the rules. Ok, Later~

. Ignore this, see next post

I think I misread this before. You're saying you "don't have to apply it", but you always can right? You CAN always apply it, because it's always true, you're just saying there's not always a contextual need to apply it, is that right?

So any time you have (a implies b) , it's always true that (not a implies not b), you just don't always need to bring that up. Is that what you're saying?

And is the same thing true about the contrapositive? For every (a implies b) it's always true that (not b implies not a), correct? Even if it's not always useful to bring it up, it's always true?

Every case in the truth value can be applied for your inference. If it helps to contrapositive for your argument, and it matches the reality event, then yes it can be adopted.

If you look closely the truth table, it only ever contains P and Q cases with various connectives and extensions. You never see in the table H -> W, once it started with P, Q. It tells you that in an argument, you don't bring in some unrelated cases and argue, the logic is incorrect. That is just sheer nonsense.

If you started with H, G in an argument, then it only have cases with H and G in the truth table. That is the rule No.1.

Ok, we are going out for a meal here. I am not sure if it is under determinism or freewill. Have a good day, and talk to you later for any other points.

I need clear, unambiguous answers. Preferably Yes or No.

One thing at a time

So do you think any time you have (a implies b) , it's always true that (not a implies not b), you just don't always need to bring that up. Is that what you're saying?

Yes, it is correct. You don't need to always bring in Not A -> Not B for your inference. It is totally under your discretion of your inference for the case. It could be any case in the truth table you can bring in as your inference.

Yes, it is correct. — Corvus

Awesome, and this one?

And is the same thing true about the contrapositive? For every (a implies b) it's always true that (not b implies not a), correct? Even if it's not always useful to bring it up, it's always true?

That would depend on the case under the proof. I will get back on that later. cheers.

if it depends, I would love to see some examples. I would love to see an example from you where the answer is "yes" and an example from you where the answer is "no".

An example where a implies b, and not b implies not a,

and an example where a implies b, but it's not true that not b implies not a.

I'm especially interested in the second example. The first example is very agreeable. The second is much more tricky, I think.

This is a pretty common example in logic textbooks, but it is not the case that if A -> B then ~A -> ~B. To see why, consider a lawn with a sprinkler system. A person sees it has not rained (~A), but then goes out to find the lawn is wet (B). This is possible because there are many ways for the lawn to get wet (B). If it rains, the lawn will be wet, but the lawn might also be wet for other reasons.

I know you disallowed hoses, sprinklers, etc., but in that case, when the only way for the lawn to get is from the rain (A), you should frame it as an iff/biconditional.

To get ~A -> ~B the starting premise would need to be "if and only if it rains (iff A) then the lawn will be wet (B),or A <->B. In such a case, B also implies A.



This would be "denying the antecedent."



But Corvus seems to be assuming an iff relationship, in which case the inference would be valid.

What the Count said.

For every (a implies b) it's always true that (not b implies not a), correct? Even if it's not always useful to bring it up, it's always true? — flannel jesus

It's always valid (if not true), and that's called modus tollens. You're right about that, but wrong about modus ponens.

EDIT: Sorry I seem to have misunderstood flannel jesus! I thought he was agreeing with Corvus, but after a PM exchange it's clear that he isn't.

A person sees it has not rained (~A), but then goes out to find the lawn is wet (B). This is possible because there are many ways for the lawn to get wet (B). If it rains, the lawn will be wet, but the lawn might also be wet for other reasons. — Count Timothy von Icarus

The argument was meant to prove R -> W, not (R or H) -> W.
If you wanted to prove (R or H) -> W, then yes of course, the premise should have begun with it.

For (P1 ..... Pn)-> W, you would need to go into Higher Order logic, wouldn't you?
We were trying to probe P1 -> W, but if you bringing in P31-> W, then you would be brining in irrelevant inference into the argument.

IFF must be regarded as presumed in the argument looking at the starting premise from the number of variables in the premise i.e. 1.

Do you have an example for this?

An example where a implies b, but it's not true that not b implies not a.