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3 AM Philosophy

Breaking logic: Playing with Modal logic. (False conclusions from valid and sound arguments?)

For me modal logic is extremely complicated and I think far more recondite than most other forms of logic. However, even with my exceptionally little understanding of the basics of modal logic there are some very interesting things which seemingly break logic. I kinda wanted to share how it can be used to challenge our much more mundane way of looking at things using classical logic.

Let’s start with a classical rule of inference Modus Ponens (MP) which can be written a number of different ways such as:

p>q,p,:.q
p->q, p, ∴q
p⊂q, p therefore q
[p ∧ (p → q)] → q

But they all are simply read as something like “If p implies q, p therefore q” or “If p then q, p therefore q”. Meaning if p is true and that implies q is true, then if p is true then q must be true. However, is it possible to have a valid argument with all true premises (sound) with a conclusion that is false? Well no, but if you’re not paying attention carefully enough it can at least appear so. This is called the Modal Scope Fallacy or a fallacy of misconditionalization.

Let’s say p=”I have a dog and 2 cats” and q=”I have to have at least one cat”

p->q, p, ∴q

p>q If I have a dog and 2 cats, then I have to have at least one cat.
p I have a dog and 2 cats.
:. q Therefore I have to have at least one cat.

To be logically valid the only requirement is that it is a valid rule of inference such as Modus Ponens. It makes absolutely no difference to what p and q represent or are instantiated with if it is a valid rule of inference. To be logically sound the argument has to be both valid and the premise be true. If any premise is false or the logic is not valid such as the conclusion doesn’t follow from the premise then the argument is unsound.

So here we have a conclusion of “Therefore I have at least one cat.” which of course has to logically follow by MP and let’s assume that I do in fact have 1 dog and 2 cats, then q has to be true by necessity. But is that the same as it is necessary that q is true? Well no, because it isn’t logically necessary that I have to have at least one cat. It is merely relatively necessary, which means the conclusion is actually false even though it is a valid argument, but not sound…depending upon how the argument is interpreted.

Here is why:

Seemingly the form of the argument using modal logic would be:

p>q H → □C
p H
∴q □C

Read as If I have a dog and 2 cats, then it is necessary (□ is read as necessary) I have to have at least one cat (H is have and C is for cat), I have a dog and 2 cats, therefore it is necessary I have to have at least one cat. But this clearly isn’t something which is logically necessary. I do not logically have to have at one cat as to be logically necessary it would mean I would have to have at least cat in all possible worlds, but of course there are possible worlds where I do not have at least one cat. My having at least one cat is contingent, and not something which must be the case in all possible words. So the conclusion “:.q Therefore I have to have at least one cat.” is false! Even though it is a valid rule of inference and the premise was true, how can it be false as it should be relatively obvious that there is nothing that makes it such that I have to have at least one cat. The first premise is false given this interpretation in that I do not have to have at least one cat as necesary as having cats is conditional thus it is unsound and the conclusion is false for the same reasons as it is not necessily true that I have to have at least one cat.

So how is this resolved? By changing the scope of the modal operator □ from the conclusion to the implication (conditional premise).

p>q □(H → C)
p H
∴q C

Now it the conclusion is true, but not as a logical necessity but a conclusion given the implication that if I have a dog and two cats, then I have to have at least one cat. I have a dog and two cats, therefore I have to have a cat.

By changing the scope of the modal operator we can restrict the modal operator to just the conditional premise, and therefore not have a conclusion (consequent) that appears to be necessarily true, when it isn’t. It isn’t necessarily true, meaning must be the case in all possible worlds, but true by necessity given it being logically deductively inferred and the premises being true.

Another example would be something relating to epistemology something like “To know that p requires p must be true” (“To know that p” essentially for simplicity here is a merely a schema that means “to know some proposition p is true”). Given that you can not have false knowledge as knowledge requires p to be true then if a subject (S) knows that p then it would follow that ‘If S knows that p, then p must be true’.

Using modal logic this would apparently be written as:

Ksp → □p

This would be read as If S knows p then it is necessary that p is true. However, this may not actually be the case. If p=”1+1=2″ then Ksp → □p would seem to hold as “1+1=2″ is a necessary truth in all possible words, so it would be necessary that p is true (□p). But what if it was p=”Steve owns a dog”. In some possible worlds I own a dog, and in other possible words I do not own a dog. It is contingency (true in at least one possible world, but not in all possible worlds), not a necessary truth (true in all possible worlds). Since there are possible words that exist where I do not own a dog then Ksp → □p is not the case as me owning a dog is obviously not a metaphysical nor a logical necessity.

Take for example the argument:

If S knows that p, then p must be true. (Ksp → □p)
If p is true then p cannot be false. (LNC)
If p cannot be false, then p must be necessary (¬◊¬p → □p, ‘if it is not possible (¬◊) p is false, then p is necessary’
If S knows that p, then p must be necessarily true. (Ksp → □p)
If ¬□p then ¬(Ksp → □p) (If p is not necessarily true then “If S knows that p, then p must be true” is false)
:. All possible knowable p to S are necessary truths. (∀p(◊Ksp → □p))

The conclusion seems to imply that if we know p then the truth of is necessary. This seems of course as quite an untenable conclusion as clearly there are truths that are contingent, such as me owning a dog.

So how do you resolve this fallacy of misconditionalization?

Once again, by changing the scope of the modal operator:

□(Ksp → p)

Now the statement “If S knows that p, then p must be true” is necessary given it is by necessity that to know p then p must be true.

If S knows that p, then p must be true. (□(Ksp → p))
If p is true then p cannot be false. (LNC)
If p cannot be false, then p must be necessary (¬◊¬p → □p, ‘if it is not possible (¬◊) p is false, then p is necessary’
If S knows that p, then p must be necessarily true. (□(Ksp → □p))
If ¬□p then ¬(Ksp → □p) (If p is not necessarily true then “If S knows that p, then p must be true” is false)
:. All possible knowable p to S must be true. (∀p□(◊Ksp → p))

We can see the modal error in that “If S knows that p, then p must be necessarily true” (Ksp → □p) is not true as S can know that P, and p must be true, but not necessarily be true in all possible worlds…the conclusion is merely just relatively necessary (□(Ksp → p)) given the scope of the argument.

Source references:
http://www.sfu.ca/~swartz/modal_fallacy.htm
SEP: https://plato.stanford.edu/entries/logic-modal/#WhaModLog

(Special thanks to my friend Zeemo, who is extremely good at logic, who took the time to look this over, and had me fixed a number of things. He was still unsure about the second argument, as am I but I will leave it up to the reader to evaluate it. If there are critical mistakes in the logic, please message me so I can correct them ASAP).

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