So even if I throw (P->Q) → (Q->P) as an additional axiom in I can’t get a general explosion in first order logic.
Sure you can.
R v ~R (axiom)
R → (R v ~R) (by 1)
(R v ~R) → R (by the new axiom)
R (by 1 and 3)
Edit:
(P->Q) → (Q->P) is not a fallacy. ∀P,Q: (P->Q) → (Q->P) is a fallacy, and its negation is ∃P,Q: (P->Q)^~(Q->P) which is indeed a theorem in first order logic.
Huh?? If you allow quantification over propositions, you are no longer using first order logic.
I think you were closer to being on track before your edit. The first thing to realize is that a fallacy is not a false statement. It is an invalid inference scheme or rule of inference.
So, with P and Q taken to be schematic variables (to be instantiated as propositions), the following is a fallacy (affirming the consequent):
P → Q |- Q → P
Or, you could have simply corrected the words “additional axiom” in the quoted claim to “additional axiom scheme”.
Sure you can.
R v ~R (axiom)
R → (R v ~R) (by 1)
(R v ~R) → R (by the new axiom)
R (by 1 and 3)
Edit:
(P->Q) → (Q->P) is not a fallacy. ∀P,Q: (P->Q) → (Q->P) is a fallacy, and its negation is ∃P,Q: (P->Q)^~(Q->P) which is indeed a theorem in first order logic.
Huh?? If you allow quantification over propositions, you are no longer using first order logic.
I think you were closer to being on track before your edit. The first thing to realize is that a fallacy is not a false statement. It is an invalid inference scheme or rule of inference.
So, with P and Q taken to be schematic variables (to be instantiated as propositions), the following is a fallacy (affirming the consequent):
P → Q |- Q → P
Or, you could have simply corrected the words “additional axiom” in the quoted claim to “additional axiom scheme”.
Er, sorry. Meant propositional calculus not first order logic. I think my statement works in that context.
What’s specifically going on here is that (P=>Q) ⇒ (Q=> P) is false whenever P is false and Q is true.
Adding it as an axiom schema to propositional calculus results in a contradiction. It cannot be added as a single axiom to first-order logic.
Yes, you are correct. I was confused in a very stupid way.