# Is rejecting A not equivalent to accepting ~A?

A fundamental misconception that many laymen hold is that rejecting claim A is equivalent to accepting its inverse, namely ~A.

How can we formally differentiate these claims, which I believe are all different:

• "I believe that Mike is a good dog owner." <- Assertion
• "I do not believe that Mike is a good dog owner." <- Rejection
• "I believe that Mike is not a good dog owner." <- Assertion

It seems to me that the rejection of a claim (i.e. #2) does not, in any way, necessitate asserting claim #3, but I have no way to demonstrate this.

• You're asking for a proof that rejecting A is not equivalent to accepting ~A. You've given an example in which it's clearly possible to reject A without accepting ~A. That example is a proof. What more proof could you need? Oct 3, 2015 at 22:45
• No. A counterexample to X is a proof of not-X. Oct 3, 2015 at 22:51
• X is the statement "If you reject A you must accept ~A". You have given a counterexample example to this statement, and hence proved ~X, which is to say you have proved that "It is not true that if you reject A you must accept ~A". I thought this was exactly what you were trying to prove. Were you trying to prove something different? Oct 3, 2015 at 23:06
• I see. I think we were discussing two different things, but I see what you mean. Thanks. Oct 3, 2015 at 23:09
• IMO this is more a matter of language than philosophy.
– user6559
Oct 4, 2015 at 17:58

"Belief" is a modality; thus, you are right in saying that "do not believe P" is not equivalent to "to believe not-P".

Compare with possibly and necessary :

The operator (for ‘possibly’) can be defined from [‘it is necessary that’] by letting ◊P = ¬□¬P.

This means that e.g. : ¬◊P is not equivalent to ◊¬P.

The same thing happens with quantifiers; ¬∃xP(x) is not equivalent to : ∃x¬P(x) but to ∀x¬P(x).

This does not contradict Excluded Middle.

See Doxastic or Epistemic Logic :

B_c(A) reads "Agent c believes A".

• You are right the "rejection (b)" does not negates the statement but the belief. I think it is a mistake of the OP unless i haven't understood correctly. Of course negation of a belief means nothing unless proven that a belief has a true reference Oct 3, 2015 at 20:10
• One might define □P as "I am certain that P is true", and ◊P as "I can imagine that P is true." One then has only to decide how the accessibility relation should behave. Oct 3, 2015 at 22:27
• This answer proves nothing. You have stated your hypothesis `∃P. ¬◊P ≠ ◊¬P` but have not proved it. Now, I know that `∃P. ¬◊P ≠ ◊¬P` is probably provable from simple and common axioms of modal logic but you have not given a proof that modal logic is a useful representation of the reality around us. Why does modal logic represent belief in useful ways? Oct 4, 2015 at 0:08
• @StevenStewart-Gallus Mauro Allegranza is absolutely correct to say that "not possibly p" does not entail "possibly not p." The reason is that this is the way we've defined the negation operator. The case doesn't turn on anything about modal logic, cf. (i) "~ for all x, P" does not entail (i) "for all x, ~P." (Imagine a box with 2 red cubes and 2 yellow balls. In this case (i) it is false that every item is red, but this doesn't entail that (ii) every item is non-red. ) That's just a fact about the syntax of first-order languages.
– user5172
Oct 4, 2015 at 0:59
• @shane No one has given a proof that `∀P. ¬◊P = ◊¬P` and one would be right in being cautious in assuming that. But that does not necessarily imply that `∃P. ¬◊P ≠ ◊¬P`. One actually has to provide a proof of that and which in fact you did do now with the 2 red cubes and 2 yellow balls example which might in fact be a good example to edit into the answer and make the answer better. Oct 4, 2015 at 18:27

Statement no.1 states

The person believes a certain issue.

Its contradiction negates no.1 and says:

It is wrong that the person believes a certain issue,

which is equivalent to the statement from no. 2. But no.2 is not equivalent to no. 3. Because

A does not believe B

is not equivalent to

A believes not B:

Possibly A has no believe at all concerning issue B.

• Yup. Mike who?? You don't know anything about Mike, thus you have no reason to believe anything about his dog ownership abilities. Oct 5, 2015 at 5:18
• Yes, you got it :-) Oct 5, 2015 at 6:47

It does not always. If Mike is a dog owner then he can be a good one or a bad one or a mediocre, or sometimes good and sometimes bad, or depending of what aspect of "dog ownership" is investigated any other type. But there are specific types of "treatments" that will make Mike a "good owner" and if he deviates from these a bad one.

So i think it is not always a fundamental misconception of laymen thought.

If statement A is true then a true negation of it leads to the opposite statement to become true.

https://en.wikipedia.org/wiki/Law_of_excluded_middle

It depends heavily on how exacting your wording is. It is possible to state "I reject A" in such a way that is equivalent to declaring an acceptance of ~A, because of the Law of the Excluded Middle, which is an accepted axiom of propositional logic (A proposition is always either true or false). However, it is also reasonable to restate the rejection slightly: "I reject this proof of A." This states nothing about the truth or falseness of A, merely that the proof being offered is not sufficient. This is especially important in the handling of axioms. I may believe that Mike is a good dog owner, but not with a sufficient conviction to blindly assume any prepositional logic which may follow from that:

Assume: Mike is a good dog owner
Assume: Good dog owners pick up their dog's poop
Observe: There is dog poop in my back yard
Assume: Mike is the only dog owner with keys to my back yard
Thus: Since Mike is a good dog owner, and good dog owners pick up their dog's poop, Mike would have picked up any poop his dog left in my yard.
Thus: Since no other dog owner has keys to my back yard, my dog must have left the poop
Observe: I did not pick up the dog poop
Thus: I must be a bad dog owner.

You can see why I might like to argue some semantics regarding the validity of these assumptions, but I'd be quick to claim Mike is a good dog owner and good dog owners pick up poop after their dog. Belief is a wiggly thing that way.

• It could be interesting to distinguish the case in which I talk about my own beliefs and the case in which I talk about someone else's belief. The logical reasons given above are perfectily convincing for statements of the form " That person does not believe that P". Is it also the case for " de se" statements, statements obout oneself?
– user37859
Mar 30, 2019 at 19:50
• I mean it seems difficult to accept as consistent the following proposiiton, provided the speaker understands what he is saying ( that is, grasps the concepts involved in the proposition he utters) : " I do not believe that the nazis won WWII, but I do not believe either that the nazis didn't win WWII" ?
– user37859
Mar 30, 2019 at 20:13
• Could one talk about a pragmatic inconsistency?
– user37859
Mar 30, 2019 at 20:16
• @RayLittleRock If we talk pragmatism, the utterances are made within a code which specifies whether the law of the excluded middle applies or not. In layspeak, we almost always assume it (I'm not aware of many cultures that do not). As for your example, the result of WWII has copious evidence to justify statements as to who won, so only the most radical of skeptics would question it. However, other phrasing do work. "I do not believe we were put here by aliens, but I do not believe either that we were not put here by aliens." Mar 31, 2019 at 2:45