Modus Ponens
Valid!
Modus Tollens
Valid!
Converse Fallacy
Invalid!
Inverse Fallacy
Invalid.
However ...
As regards the converse fallacy
Valid!
As regards the inverse fallacy
Valid!
Correct/Incorrect/Both/Neither?
Do nfr things
As alluded in the comments, we need to go to modal logic K (for Kripke) to get to something like "maybe":
We have two new concepts in K-modal logic:
◻ =def "it is necessary that" (i.e., True in all possible worlds)
◊ =def "is is possible that" (i.e., True in at least one possible world)
NOTE: ◊A ≡ ¬◻¬A.
And two new principles of inference:
Necessitation Rule: If A is a theorem, so is ◻A
Distribution Axiom: ◻(A →B) → (◻A →◻B)
So, we could translate your first argument:
p→q
q
∴ "maybe" p
as:
(1) p→q
(2) ◻ (p →q) [by Necessitation Rule]
(3) ◻p → ◻q [by Distribution Axiom]
(4) ◻a ⇒ ¬◻¬a ≡ ◊a [by definition of necessity]
(5) ¬◻q → ¬◻p [Modus Tollens]
(6) q
(7) ◻q [Necessitation Rule]
As you can see, there is no requirement that ◊p. The only requirement of the material conditional is that q not be false and p true. But q is true in all possible worlds, so p can be false in all possible worlds and still this holds; therefore, you cannot conclude ◊p.
I'm assuming (1) that you are using classical logic, (2) you are using valid in the usual conventional way to mean an argument that instantiates a form that has no possible counterexample, and (3) that your p, q are metavariables, not atomic propositions.
Your first four are correct, the last two are not. That is, unless you specify the additional condition that your p, q are logically contingent. In which case, "maybe p" means nothing more than p is logically contingent, and "maybe ¬q" also means nothing more than q is logically contingent, so they would be correct but trivially so, since you have assumed it.
If you don't assume that p, q are logically contingent, then there are counterexamples.
has a counterexample when p is a contradiction and q is true.
has a counterexample when p is false and q is a tautology.
If you actually want to formalise 'maybe' you could use the ◇ operator in modal logic. But those two forms remain invalid.
Maybe ~q Valid! Correct/Incorrect/Both/Neither?
Yes, correct, but trivial!
And maybe q, too.
Because, if we assume (p → q) ∧ ¬p, then we don't assume anything about q or about ¬q.
So maybe ¬q, but also maybe q.
Sorry I couldn't find any scholarly reference here.
