The L deductive System in Propositional Logic I am trying to proof a few statements in the deductive system L, in propositional logic. The system contains 3 axioms (I, II, III below) and a few proven statements (1,2,3,4). In addition, the only inference rule is the modus ponens. In a book I found, the system differs from mine in the 3rd axiom. In the book, the 3rd axiom is (~b->~a)->((~b->a)->b) while my 3rd axiom is  (~b->~a)->(a->b). I wanted to ask, is it possible, given my axioms, statements the MP rule and the deduction theorem, to prove the 3rd axiom from the book? Thank you.


 A: You will need Law of Clavius. For how to derive it using your System L, see help with some Hilbert style proofs in a propositional logic axiom system. Here's an answer to your question.
Theorem $\vdash(\neg\beta\rightarrow\neg\alpha)\rightarrow((\neg\beta\rightarrow\alpha)\rightarrow\beta)$
Proof:


*

*$(\neg\beta\rightarrow\neg\alpha)$ (premise)

*$(\neg\beta\rightarrow\alpha)$ (premise)

*$(\neg\beta\rightarrow\neg\alpha)\rightarrow(\alpha\rightarrow\beta)$ (Axiom 3 in your System L)

*$(\alpha\rightarrow\beta)$ (m.p. 1,3)

*$(\neg\beta\rightarrow\beta)$ (Theorem/Statement 4 in your System L, combining 2,4)

*$(\neg\beta\rightarrow\beta)\rightarrow\beta$ (Law of Clavius)

*$\beta$ (m.p. 5,6) (conclusion)

*Therefore $(\neg\beta\rightarrow\neg\alpha),(\neg\beta\rightarrow\alpha)\vdash\beta$ (1,2,7,)

*$(\neg\beta\rightarrow\neg\alpha)\vdash(\neg\beta\rightarrow\alpha)\rightarrow\beta$ (deduction theorem)

*$\vdash(\neg\beta\rightarrow\neg\alpha)\rightarrow((\neg\beta\rightarrow\alpha)\rightarrow\beta)$ (deduction theorem)
A: You would need some sort of rule of inference for you to change deductions from I., II., or III. to a form like p -> (q -> (r -> p).  Let '{' and '}' stand in place of the third parenthesis form types, considered by their size. Let '[' and ']' stand in place of the second form types, also considered by their size.  
That might require a rule of inference to change 
{p -> [q -> (r -> p)]} 
into 
p -> [q -> (r -> p)].  
That last step does not accord with the rules of formation.
That all said, something similar does appear to seem possible:

