I'm definitely coming at the question from the point of view of propositional logic -- something basic because it's already confusing enough as it is :D

Im not versed in logic at all but I'll give my thoughts.

If A is defined specifically (ie. no coin flipping or randomness involved in its definition) and we're assuming "implies" means "neccesitates" and not "can lead to", then it is contradictory.

If B and not B are different, then if A neccesary leads to one, it cannot lead to the other definitionally.

Ofc, if not B and B are the same, then theres no contradiction; and if A can be unspecific then it can lead to both (flipping a coin can get heads or not heads).

That's not the same as "(A implies B) and (A implies not-B)" -- that'd be "(A implies (B and not-B)).

Are they not? First and last rows of the truth tables are all the same. Seems logically equivalent to me.

hrmmm checking now on paper.

I'm using https://en.wikipedia.org/wiki/De_Morgan%27s_laws though I've made many mistakes so I could be wrong...

Eh, that's a bad reference, just the name that came to mind. Material implication is what I mean.

Is what I get. (EDIT: I made a mistake, as pointed out by )

Since the last column is not all "F" it's not a contradiction, I believe. (though I see I confused myself earlier, looking at the T-table)

Yeah, it's not a contradiction, but the final row of the truth table for p→q ∧ p → ¬q is the same p → q ∧ ¬q for all values of p and q.

True.

Maybe that's why it's confusing? It's an implication of two implications, and material implication is already confusing . .. lol see I confused it even in response.

Awwww ... there are minds out there who think like this, and if I'm in the habit I can -- but it's not my normal way.

A related example is Godel's trick in his ontological proof of God as discussed in the other thread, which was to define a property P so as to enforce the condition

¬(g → P(g) ∧ g → ¬P(g))

i.e. ¬¬g, which is a classically acceptable proof of existence.

↪Philosophim What do you think of this?

https://en.wikipedia.org/wiki/Barbershop_paradox — flannel jesus

It was a bit of headox with a sentence like this:

"If Carr is out, then we know this: "If Allen is out, then Brown is in", because there has to be someone in "to mind the shop." Ugh.

So cleaning this up we get this:

They explain that there are three barbers who live and work in the shop—Allen, Brown, and Carr—and some or all of them may be in. We are given two pieces of information from which to draw conclusions. Firstly, the shop is definitely open, so at least one of the barbers must be in. Secondly, Allen is said to be very nervous, so that he never leaves the shop unless Brown goes with him.

So, we have A, B, and C

One must always be in.

If A is out, B is out
Therefore C is in

If B is out, A can be in, as B can leave the shop.
If A is out, B is also out, so C is in

So our combos are as follows:

1. C !A !B
2. !C A B
3. C A !B
4. !C A !B
5. C A B

"Suppose that Carr is out. We will show that this assumption produces a contradiction. If Carr is out, then we know this: "If Allen is out, then Brown is in", because there has to be someone in "to mind the shop." But, we also know that whenever Allen goes out he takes Brown with him, so as a general rule, "If Allen is out, then Brown is out". The two statements we have arrived at are incompatible, because if Allen is out then Brown cannot be both In (according to one) and Out (according to the other). There is a contradiction. So we must abandon our hypothesis that Carr is Out, and conclude that Carr must be in."

This doesn't make any sense. Clearly B can be out and A be in. Clearly C could be out and A and B, or A be in. I don't get it.

I think, at least in philosophy though maybe there's some other argument this stems from that I'm not aware of, that we should separate out implication from modality -- so when you introduce "possibility" and "necessity" those are entirely different operators from implication. — Moliere

This is simply a language issue. What does 'impliciation' mean? That's why I went over all the different possibilities. In the end implication must mean necessary or not necessary, in which case the answer will be different.

What does "simply a language issue" mean?

If implications were horses, then logicians would ride.
If implications were coaches, then logicians would still ride.

Let A = "Unenlightened's testimony is unreliable"
Let B = "Unenlightened tells the truth"
not B ="Unenlightened does not tell the truth"

Would that make a difference? 0/1=F/T as I understand it. — Moliere

No, 0/1 is exactly the same as True/False for the purpose of logic tables. What I meant to say is that some people seem to think that A here implies not a variable (that may take 0 or 1) but a proposition that is being asserted as True (which is to say that it is 1). When A is 1, A→B and A→notB give opposite results.

In the end implication must mean necessary or not necessary, in which case the answer will be different. — Philosophim

We are not talking about modal logic.

((p→q)∧(p→¬q)) and (p→(q∧¬q)) are the same formula

Both are contradictory (thus False) if A is True. But as by the definition of material implication, both are True if A is False.

↪Philosophim What does "simply a language issue" mean? — Moliere

Often times in logic or arguments, the solution is clear. What often happens is we use language with poor definitions. Thus people make assumptions or conclusions that the user of the unclear language didn't intend, or the user themself is doing the same.

Notice how he used the word 'imply' and people immediately thought, Oh A -> B. But that's not imply. that's "Necessarily leads to." And its obvious that if A -> B, that A -> !B is a contradiction if you use the correct definition. "Imply" is vague enough that some thought 'necessarily', while others thought 'maybe'. And because you can get halfway there with the word, some might even thing A ->B means only sometimes.

If this is about material implication then the answer is utterly simple:

A -> B
A -> ~B

are not together inconsistent, since they are both true when A is false, and the propositional calculus doesn't permit an inference of a contradiction from two formulas such that there is a model in which they are both true.

The digressions in this thread don't affect that very simple fact.

It is troublesome to talk about these things if one is not using very specific terminology. What does "contradictory" really mean? — Lionino

Yours is the best post in the thread imo. It is especially interesting that what I called fallacious your source explicitly calls a contradiction (). There are two different notions of contradiction occurring in the thread.

But OP is asking are the two contradictory with each other?

I think what is being asked here is whether one is the denial of the other. And the answer is no. Putting it in logical tables, denial would be whenever (A → B) yields True (A → ¬B) yields False. — Lionino

A classical definition says that two propositions are contradictory if the denial of either entails the affirmation of the other, and vice versa. So there are materially four different relations, given that each of the two propositions can be denied or affirmed. This is what I was trying to get at in my first post.

What's interesting is that it is not possible to contradict a material implication even on the classical understanding of contradiction. This is why I think the more interesting question prescinds from material implication:

NB: Given the way that common speech differs from material implication, in common speech the two speakers would generally be contradicting one another. — Leontiskos

The two statements are not contradictory. They simply imply ~A.

The two statements are not contradictory. They simply imply ~A. — hypericin

You think the two propositions logically imply ~A? It seems rather that what they imply is that A cannot be asserted. When we talk about contradiction there is a cleavage, insofar as it cannot strictly speaking be captured by logic. It is a violation of logic.

A contradiction is of the form "P ^ ~P" — Moliere

Your presupposition here straddles the two definitions of a contradiction in an interesting way. Using the same example I gave privately:

"The car is wholly green." "No, the car is not wholly green."

This is a contradiction on all definitions.

"The car is wholly green." "No, the car is wholly red."

This is a contradiction classically but not according to symbolic logic, and your method would not find it to be a contradiction.

I think the third way to give a contradiction, besides the two already noted, is to use symbolic logic and say, "Assume P and suppose Q. If an absurdity results on Q, then P and Q are contradictory." This gets closer to the classical definition. It shows that they cannot both be true, but it does not show that they cannot both be false, and it does not show that the trueness or falseness of one results from the falseness or trueness of the other.

A classical definition says that two propositions are contradictory if the denial of either entails the affirmation of the other, and vice versa. So there are materially four different relations, given that each of the two propositions can be denied or affirmed. — Leontiskos

A contradiction is a formula of the form P & ~P, or in other contexts the pair {P ~P}.

We don't have to check four different things to see that formula, or in other contexts a pair of formulas, is a contradiction.

A set of formulas is inconsistent if and only if it proves a contradiction.

it is not possible to contradict a material implication — Leontiskos

Wrong. P -> Q is contradicted by ~(P -> Q).

Assume P and suppose Q. If an absurdity results on Q, then P and Q are contradictory. — Leontiskos

Yes, people say "contradictory", but as a terminological preference, I would say they are together inconsistent. A pair of formulas is a contradiction iff they are of the form P and ~P. A set of formulas is inconsistent if and only if it implies a contradiction.

It shows that they cannot both be true, but it does not show that they cannot both be false — Leontiskos

Wrong. If one is true then the other is false. If one is false then the other is true. [EDIT:] Correction: Obviously, it is not correct that it is always the case that there is at lease one true statement in an inconsistent set of statements.

it does not show that they cannot both be false, and it does not show that the trueness or falseness of one results from the falseness or trueness of the other — Leontiskos

Wrong. If the members of a set of sentences that includes P are all true, and that set of sentences along with Q (that is not in the set) proves a contradiction, then Q is false. Put another way, we cannot derive a contradiction from a set of all true sentences. [EDIT:] Correction: Obviously, it is not correct that it is always the case that there is at lease one true statement in an inconsistent set of statements.

"The car is wholly green." "No, the car is wholly red."

This is a contradiction classically but not according to symbolic logic — Leontiskos

Classical logic includes propositional logic such as treated in symbolic logic.

"The car is green" and "The car is red" is not a contradiction. But if we add the premise: "If the car is red then the car is not green," then the three statements together are inconsistent. That's for classical logic and for symbolic rendering for classical logic too.

You think the two propositions logically imply ~A? — Leontiskos

They imply ~A.

When we talk about contradiction there is a cleavage, insofar as it cannot strictly speaking be captured by logic. It is a violation of logic. — Leontiskos

I don't know what you mean by 'cleavage' and 'captured' in this context. But in logic systems we can write contradictions. Indeed, we often intentionally prove contradictions from premises in order to refute the premises.

A -> B. But that's not imply. that's "Necessarily leads to." — Philosophim

Wrong. Material implication does not require necessity.

They imply ~A. — TonesInDeepFreeze

Then give your proof.

((p→q)∧(p→¬q)) and (p→(q∧¬q)) are the same formula — Lionino

If p is False, both propositions are true for any value of q.

Taking a look at (p→(q and ¬q)):

0 = False, 1 = True
(q and ¬q) is always 0 – definition of contradiction
a→b is equivalent to (¬a or b) – definition of material implication
our b in this case is (q and ¬q), thus b is always 0
if p is 0, ¬p is 1, so from (¬p or (q and ¬q)) we have (1 or 0)
The or operator will return 1 if any of the variables is 1 – by definition
So (1 or 0) returns 1, so (¬p or (q and ¬q)) returns 1 if p is 0, so ((p→q)∧(p→¬q)) returns 1 if p is 0

This kind of stuff is better to understand from the POV of electronics. The words "True" and "False" make things confusing. "1 or 0" is a logical gate with two inputs, returning 1, "True or False" is like "Duh, what else could it be?"

So a few conclusions.

(A implies B) and (A implies notB) do not contradict one another.

It would be useful to have a page that generates an image of a given truth table.

Around a third of folk hereabouts who have an interest in logical issues cannot do basic logic.

might note that 's testimony is reliable.

Oh, and , (A→B)∧(A→¬B)↔¬A.

- very clear.

It would be useful to have a page that generates an image of a given truth table. — Banno

https://web.stanford.edu/class/cs103/tools/truth-table-tool/ ?

- Thanks - I concede your point. I keep reading the OP in terms of a dialogue between two people, probably because of some reading I have been doing on non-deductive reasoning.

Thanks - I concede your point. — Leontiskos

I am mostly complementing my own posts. I didn't know you were arguing the opposite, but alright :up: