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As usually happens, a statement (p) and its negation (~p) contradict each other. So, e.g. God does not exist, the negation of, God exists, together form a contradictory pair. A statement (p) and its negation (~p) are such that they're mutually exclusive (if one is true the other has to be false and if one is false the other has to be true) and jointly exhaustive (at least one of the two has to be true).

Take now the statement God may exist and its negation, the statement God may not exist. They do not contradict each other. The compound statement that goes God may exist AND God may not exist is not a contradiction.

  1. What does this have to do with the principle of bivalence?
  2. What are the implications for the law of the excluded middle (LEM)?
  3. What does this say about the law of noncontradiction (LNC)?
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    Let P be "god may exist". Then not P is "not the case that god may exist". under standard modal semantics, this is "impossible that god exist" and not "god may not exist". So maybe you mean something different by negation?
    – emesupap
    Sep 7 at 4:51
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    Not much to do with your 3 classic laws of thought but your modal scope. Even weirder consider the correct claim from elementary arithmetic that 'it's necessary that 8 is less than 10', also we know '8 equals the number of planets of our solar system', but you cannot substitute 8 and confidently claim as before 'it's necessary that the number of planets of our solar system is less than 10'...
    – Double Knot
    Sep 7 at 5:13
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    Modality; the first one is "It is possible that..." Negating it, we get "It is not possible that..." that is equivalent to "It is necessary not..." The second instead, is "It is possible that not..." and this, as you say, is not the contradictory of the first one. Contradictory pairs are produced putting the negation IN FRONT of the statement.
    – Mauro ALLEGRANZA
    Sep 7 at 6:01
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    @MauroALLEGRANZA, Please read my reply to Bumble
    – Agent Smith
    Sep 7 at 6:15
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    See also de dicto vs de re.
    – Mauro ALLEGRANZA
    Sep 7 at 13:42

2 Answers 2

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This is one of those cases where it helps to use some formal logic, because one of the things formal logic is especially good at is exhibiting the scope of operators.

Suppose we use the propositional symbol G to represent the proposition "There is a god". We can overlook the issue of how to make sense of attributing existence to a thing. And let's use ◇ as a modal operator that we will understand as "possibly". We can overlook exactly what kind of possibility for the present purposes.

When you say "God may exist" this is naturally understood de dicto as "It is possible that there is a God". So we might write this as:

◇G

When you say "God may not exist" the negation may sit inside or outside the possibility operator. So,

◇¬G

means "it is possible that there is no God", while

¬◇G

means, "it is not possible that there is a God".

The sentences ◇G and ¬◇G together form a contradiction, so they satisfy your criteria that they cannot both be true and cannot both be false. This is entirely consistent with bivalence, excluded middle and non-contradiction.

The sentences ◇G and ◇¬G are not a contradiction. It is possible there is a God; it is possible there is no God.

We might also add that ◇(G ∧ ¬G) would be a weird sentence stating that it is possible for the contradiction "there is a God and there is no God" to hold. This sentence would never be true in what are called normal modal logics.

I think that answers your question. Things get more complicated if we combine modal logic with quantifiers and write things like:

◇(∃x)Divine(x)

or even

(∃x)◇Divine(x)

This introduces all kinds of additional problems concerning how to understand potential existence and potential properties, and how to quantify into referentially opaque contexts.

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  • In colloquial usage I think the word "possible" means "we cannot at this time rule it out." In other words, X is possible if ~X is not "easily" proved from the current state of knowledge. Is there any logic that tries to model this concept?
    – causative
    Sep 7 at 5:44
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    @Causative That sounds like 'epistemically possible'. On this understanding □P means P is known to be true. □¬P means P is known to be false. ◇P is equivalent to ¬□¬P so it means we don't know P is false, hence P might be true for all we know, or P being true is consistent with what we know. Some accounts of epistemic logic dispense with the □ symbol and use K. There is disagreement about whether S4, S5, or some other modal system provides the best logic. plato.stanford.edu/entries/logic-epistemic
    – Bumble
    Sep 7 at 6:38
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    @AgentSmith You are correct that ◇G and ◇¬G cannot both be false. We would need a little modal logic to prove it, but ¬◇G ∧ ¬◇¬G entails □¬G ∧ □G which entails □(G ∧ ¬G) which entails a contradiction in normal modal logics. As per my response to Causative, we could take 'possibly' to be epistemic.
    – Bumble
    Sep 7 at 6:38
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    @AgentSmith It is not correct to say that God may exist is logically equivalent to God may not exist. Possibility does not exclude necessity. ◇P does not rule out □P. It is analogous to saying that some ravens are black does not rule out that all are.
    – Bumble
    Sep 7 at 6:39
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    @AgentSmith: Your intuition that “might” could imply “might not” comes from human language pragmatics, and particularly implicature, subtly different from implication. If you say “I’m hungry!” and I say “Oh, I’ve got some chocolate here”, the conversational context suggests I’m about to offer you some, but my statement doesn’t logically imply that. Similarly, if I say “It might rain today”, you can normally infer I also think it might not (because if I was sure, I’d have said so); but that’s implicature, not logical implication.
    – Peter LeFanu Lumsdaine
    Sep 9 at 19:59
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  1. Nothing.

  2. There are none.

  3. Nothing.

May already entails perhaps, or perhaps not. If you are uncomfortable with perhaps, perhaps not, that is absolutely hunky-dory, but then best stay away from may (and similar indicators of modality).

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  • Leaks assume ex arguendo that none of us disputes that 1+ 1 = 2. Can one then dispute the proposition that one plus one may equal two? In any case, I, for one, would dispute the proposition that one plus one may not equal two.
    – Paul Tanenbaum
    Sep 8 at 19:26

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