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I found a theorem that is like this:

If A and B are true, then C is true if and only if D is true. In other words:

Can we deduce,

If D and ( not (A) or not (B) ) then not (C)?

Of course "and", "or", and "not" are logical operators.

Thanks a lot.

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    No, that doesn't work. You get counterexamples when C and D are both true and either A or B is false. I'm assuming this is an elementary logic question and your 'if' is a material conditional.
    – Bumble
    Dec 29, 2023 at 14:29

2 Answers 2

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No. If A and B are not both true then no statement is made about C whatsoever.

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I found a theorem that is like this:

If A and B are true, then C is true if and only if D is true. In other words:

Can we deduce,

If D and ( not (A) or not (B) ) then not (C)?

Of course "and", "or", and "not" are logical operators.

Your "theorem":

If A and B are true, then C is true if and only if D is true

translate formally as follows:

A ∧ B ⊢ C ⇔ D

This just means that A and B together imply that C is equivalent to D.

This is of course self-evidently false, therefore not a theorem.

I think you know that already.

However, if the "theorem" you found is:

(A ∧ B) → ((C ⇔ D)) ⊢ (D ∧ (¬A ∨ ¬B)) → ¬C

Then it is not a theorem!

There are cases where (A ∧ B) → ((C ⇔ D)) is true and (D ∧ (¬A ∨ ¬B)) → ¬C is false.

So, it is not true that if (A ∧ B) → ((C ⇔ D)) is true, then (D ∧ (¬A ∨ ¬B)) → ¬C is true.

In fact, if (A ∧ B) → ((C ⇔ D)) is true, then (D ∧ (¬A ∨ ¬B)) → ¬C is false.

This means that we do have a theorem, namely:

(A ∧ B) → ((C ⇔ D)) ⊢ ¬((D ∧ (¬A ∨ ¬B)) → ¬C)

And now it is really a theorem.

Happy New Year to all!

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    Clearly that’s not what OP meant. They’re asking if A&B⊢C<->D is enough to conclude D&(~Av~B)⊢~C.
    – PW_246
    Dec 29, 2023 at 16:56
  • @PW_246 Yes, I understand that, but this is not what it says. Dec 29, 2023 at 17:22
  • Why is "A and B together imply that C is equivalent to D" "of course self-evidently false"?
    – Stef
    Dec 30, 2023 at 22:47
  • @Stef Why do you ask? Dec 31, 2023 at 10:21
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    “In fact, if (A ∧ B) → ((C ⇔ D)) is true, then (D ∧ (¬A ∨ ¬B)) → ¬C is false.“ Huh? Both the original formula and the formula I quoted are satisfiable when A, B, C, and D are all true of all false; in most logics, it follows that both are in fact true.
    – PW_246
    Jan 3 at 23:52

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