[From "Introduction to Mathmatical logic" by Elliott Mendelson,exercise 1.58]
Axioms of $L_1$:
(A1) $B \rightarrow (C \rightarrow B)$
(A2) $(B \rightarrow (C \rightarrow D)) \rightarrow ((B \rightarrow C) \rightarrow (B \rightarrow D))$
(A3) $(\lnot C \rightarrow \lnot B ) \rightarrow ((\lnot C \rightarrow B) \rightarrow C)$
only rule of inference is $MP$
Axiom system of $L_2$:
(B1) $(\lnot B \rightarrow B) \rightarrow B$
(B2) $B \rightarrow (\lnot B \rightarrow C)$
(B3) $(B \rightarrow C) \rightarrow ((C \rightarrow D) \rightarrow (B \rightarrow D))$
only rule of inference is $MP$
I have to prove that $L_1$ and $L_2$ have the same theorems.For that I have to prove that the axioms of $L_1$ can be proven from axioms of $L_2$ and vice-versa (both systems have the same inferences so I don't have to worry about that).I was able to prove the axioms of $L_2$ from the axioms $L_1$ (in $L_1$) , but failed at proving the axioms of $L_1$ from the axioms of $L_2$ (in $L_2$).
The theorems I was able to derive in $L_2$ so far were:-
(1) $\vdash_{L_2} B \rightarrow B$
(2) $B \rightarrow C, C \rightarrow D \vdash_{L_2} B \rightarrow D$
(3) $B \rightarrow C \vdash_{L_2} ((C \rightarrow D) \rightarrow (B \rightarrow D))$
(4) $\vdash_{L_2}(\lnot B \rightarrow B) \rightarrow (\lnot B \rightarrow C)$
(5) $\vdash_{L_2}B \to (\lnot B \to B)$
(6) $\vdash_{L_2}(B \to C) \to ((\lnot B \to B) \to C)$
(7) $\vdash_{L_2}((\lnot B \to B) \to C) \to (B \to C)$
(8) $\vdash_{L_2}B \to ((B \to D) \to (\lnot B \to D))$
(9) $\vdash_{L_2}\lnot (B \to B) \to (B \to B)$
(10) $\vdash_{L_2} (B\to(\lnot B \to \lnot C)) \to (((\lnot B \to \lnot C) \to (C \to B)) \to (B \to (C \to B)))$
(11) $\vdash_{L_2} ((\lnot B \to \lnot C) \to (C \to B)) \to (B \to (C \to B))$
(12) $\vdash_{L_2} (B\to(\lnot B \to \lnot C)) \to (((\lnot B \to \lnot C) \to D) \to (B \to D)))$
(13) $\vdash_{L_2} ((\lnot B \to \lnot C) \to D) \to (B \to D) $
(14) $\vdash_{L_2} ((\lnot B \to \lnot C) \to C) \to (B \to C) $
But I still am a long way from proving the axioms of $L_2$ from $L_1$ .The problem is that I have ran out of ideas about how I can manipulate the axioms of $L_2$.Can someone give me a hint about this?
Edit:I found another axiom system in another post which I will call $L_3$.
Axioms of $L_3$:
(C1) $B \to (C \to B)$
(C2) $(B \to (C \to D)) \to ((B \to C) \to (B \to D))$
(C3) $(\lnot B \rightarrow B) \rightarrow B$
(C4) $\lnot B \rightarrow (B \rightarrow C)$
idk , maybe I should try to prove (C1) and (C2) from (B3) in $L_2$ as a subgoal/lemma [Not sure if it will help me with my main problem of proving the axioms of $L_1$ from the axioms of $L_2$ (in $L_2$)].
I found a wiki article List of Hilbert systems where, $L_1$ is called "Mendelsons axiom system" and $L_2$ is called "Łukasiewicz First Axiom system".I tried to find a paper,book or article about "Łukasiewicz First Axiom system" instead of the wiki article of Hilbert systems, but the only thing I was able to get that uses "Łukasiewicz First Axiom system" was this MIT OCW lecture slide An example of usage of "Łukasiewicz First Axiom system".There, they derived $\vdash_{L_2} B \to B$ and stopped.I also tried to searching "lukaweicz" to see if someone has a similar question like mine,but the questions I found were actually about "lukaweicz third axiom system", not "lukaweicz first axiom system" an example. I will be glad if someone shows me some other reference that I can read.
Best Answer
Example from Jan Łukasiewicz, Elements of Mathematical Logic (Pergamon Press, 1963), page 42.
Starting from your axiom (B3) $[(p \to q) \to ((q \to r) \to (p \to r))]$ with substitution: $[p \leftarrow (p \to q), q \leftarrow ((q \to r)\to(p \to r)), r \leftarrow s]$ we get:
The antecedent is (B3) again; thus, by Modus Ponens, we get:
In the same way (from (T4) and a suitable substitution) we get:
The basic "proof technique" is to identify a suitable substitution in some axiom or already proven thesis such that the result is a "big" conditional where the antecedent is an axiom or an already proven thesis and the consequent is the thesis to be proved.
We can see it in the derivation of $p \to p$ [page 45: your (1)].
From (B3) again, with substitution $[q \leftarrow (\lnot p \to q)]$ we get: $[p \to (\lnot p \to q)] \to [((\lnot p \to q) \to r) \to (p \to r)]$.
The antecedent is axiom (B2) and thus by MP we get (T9): $((\lnot p \to q) \to r) \to (p \to r)$ [that seems quite strange...].
Now, from (T9) with substitution $[q \leftarrow p, r \leftarrow p]$ we get: $((\lnot p \to p) \to p) \to (p \to p)$.
The antecedent is axiom is (B1); using MP we get: (T16) $p \to p$.
Going on, we have [page 45]: (T18) $q \to (p \to q)$ [which is (A1) of L1], and "the very important law of commutation" [page 47]: (T21) $(p \to (q \to r)) \to (q \to (p \to r))$.
Then [page 49]: (T35) $(p \to (q \to r)) \to [(p \to q) \to (p \to r)]$ [which is (A2) of L1].
The main difficulty in Łukasiewicz's original book is the lack of the $\text { Deduction Theorem}$.
Regarding axiom (A3) of L1, we can use "the very important law of double negation" [page 50]:
as well as "the four laws of transposition" [page 51]:
What we can do now is to use (T46): $(\lnot q \to p) \vdash (\lnot p \to \lnot \lnot q)$ with (T39) and your (2) above to get:
Using it with $(\lnot q \to \lnot p)$ and (2) again we get:
Finally, using axiom (B1) of L2 and MP we get: