# Quick basic logic lesson…as I only know basic logic!

Logic lesson (basic..it’s all I know anyways!)

A dichotomy is when you have A V ~A such that it is either the case or it is not the case.

So if you instantiate that with “God exist” you have:

It is either the case “God exists” or it is not the case that “God exists”. This can of course easily be re-written as:

God exists or God does not exist

If you predicate with belief you either:

Bp V ~Bp (because of the Law of Excluded Middle(LEM))

Which means “Believes p” or “Do not believe p”

If you believe p for a “normal reasoner” in doxastic logic:

∀x: Bp → BBp

𝘍𝘰𝘳 𝘢𝘭𝘭 𝘰𝘧 x if x 𝘣𝘦𝘭𝘪𝘦𝘷𝘦𝘴 𝘱 implies x 𝘣𝘦𝘭𝘪𝘦𝘷𝘦𝘴 that x 𝘣𝘦𝘭𝘪𝘦𝘷𝘦𝘴 p

One can extrapolate from that I believe for a “normal reasoner”:

∀x: ~Bp → B¬Bp

𝘍𝘰𝘳 𝘢𝘭𝘭 𝘰𝘧 𝘹 if x 𝘣𝘦𝘭𝘪𝘦𝘷𝘦𝘴 p implies x 𝘣𝘦𝘭𝘪𝘦𝘷𝘦𝘴 that x 𝘥𝘰𝘦𝘴 𝘯𝘰𝘵 𝘣𝘦𝘭𝘪𝘦𝘷𝘦 p

Therefore to be a “normal reasoner” one must believe that they believe p or believe that they do not believe p.

This to me seems to indicate there are only two possible states. One normal and one not.

1) x believes that they do not believe p. (normal reasoner)
2) x does not believe that they do not believe p. (not normal reasoner)

1 is clearly defensible. 2 however seems to present to me more of an epistemic challenge. A “peculiar reasoner” is ∃p: Bp ^ B¬Bp (believes p, but does not believe that they believe p) so can we infer from that then ∃p: ¬Bp ^ BBp (does not believe p, but believes that they do believe p)?

Either way there can it seems to me 3 possible epistemic states applied here:

A) S affirms 1)
B) S affirms 2)
C) S affirms neither.

I see no other epistemically logical possibilities assuming one has tried to evaluate p…but if one does not affirm either here because it is a dichotomy they ontologically still either believe or do not believe. You can’t avoid that in classical logic as you would be violating LEM.

So if person say holds to neither 1 nor 2 (not merely not affirming 1) or 2) for whatever reasons like they don’t want to say…but actually holds to neither 1 nor 2 in classical that would be a contradiction for obvious reasons as you can not believe and not believe p at the same time.

Author: Steve McRae

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