Principle of explosion: Other arguments?

Question

I've come across a proof-theoretic argument for explosion on Wikipedia, which is as follows:

$A \ \ \wedge\sim A$
$A$
$ \sim A$
$ A \lor B$
$B$
$(A \ \ \wedge \sim A) \implies B$

I've thought of another argument, which isn't on the same Wikipedia page as the above. As far as I can see, it is valid but I would like to see your opinions. Perhaps you could provide me with some more (relatively simple) arguments for explosion within classical logic?

$A \ \ \wedge\sim A$
$A$
$A \lor B$
$ \sim A$
$B$
$(A \ \ \wedge \sim A) \implies B$

You've just swapped the order of lines 3 and 4. Wouldn't that make it essentially the same argument? — Arthur, Jan 26 '15 at 11:56
Yes, you're correct. I have now changed the title of the question. — Five σ, Jan 26 '15 at 11:59
Wikipedia itself, on the same page you apparently reference, has other arguments. — Rory Daulton, Jan 26 '15 at 12:52
Relevant: Can the principle of explosion be removed from constructive logic? — MJD, Jan 26 '15 at 14:36
As a historical complement to the comment by @RoryDaulton, the argument quoted from Wikipedia, using disjunction introduction and disjunctive syllogism, is essentially the one presented by Lewis & Langford in their Symbolic Logic [1932]. The second (quite distinct) argument for explosion to be found at the same Wikipedia entry, using contraposition (and double negation elimination), was offered in the 60s by Popper, in his Conjectures and Refutations. — J Marcos, Jan 26 '15 at 21:35
If you look at classes axiom sets for classical propositional calculus, say in the appendix of A. N. Prior's book Formal Logic, you can find a bunch of them where it is possible to prove the non-organic theorem CNCpNNpq. Since Kpq is defined as NCpNq, this means that such axiomatic systems can argue for CKpNpq, by proving CNCpNNpq. — Doug Spoonwood, Jan 27 '15 at 00:48

score 0 · Answer 1 · answered Jan 27 '15 at 00:44

I use Polish/Lukasiewicz notation. The rule of Negation elimination that I use says that from N$\beta$ having the same scope as an instance of K$\alpha$N$\alpha$ we can infer $\beta$.

assumption                   1 | KaNa
assumption                   2 || Nb
2, 1 Negation Elimination    3 | b
1-3 Conditional Introduction 4 CKaNab.

As a more axiomatic proof, I'll assume that we have the following three axioms:

A1 CpCqp.
A2 CCpCqrCCpqCpr.
A3 CCNpKqNqp.

Then, with Dx.y indicating the condensed detachment with x as the major premise and as the minor premise we can proceed as follows:

assumption                   1 | KaNa
assumption                   2 || Nb
D[A1].1                      3 || CpKaNa
D3.3                         4 || KaNa
2-4 Conditional Introduction 5 | CNbKaNa
D[A3].5                      6 | b

Or as follows:

assumption                   1 | KaNa
D[A1].1                      2 | CpKaNa
D[A3].2                      3 | p

Thus, a fully axiomatic proof can proceed as follows:

axiom 1 CpCqp.
axiom 2 CCpCqrCCpqCpr.
axiom 3 CCNpKqNqp.
D1.3  4 CpCCNqKrNrq.
D2.4  5 CCpCNqKrNrCpq.
D5.1  6 CKpNpq.

score 0 · Answer 2 · answered Jan 27 '15 at 00:58

Here's a 23 step OTTER proof from the 21 letter single axiom of Meredith CCCCCpqCNrNsrtCCtpCsp.

-----> EMPTY CLAUSE at  11.77 sec ----> 7420 [hyper,2,7233] $F.

Length of proof is 23.  Level of proof is 17.

---------------- PROOF ----------------

1 [] -P(C(x,y))| -P(x)|P(y).
2 [] -P(C(N(C(p,N(N(p)))),q)).
3 [] P(C(C(C(C(C(x,y),C(N(z),N(u))),z),v),C(C(v,x),C(u,x)))).
4 [hyper,1,3,3] P(C(C(C(C(x,y),C(z,y)),C(y,u)),C(v,C(y,u)))).
6 [hyper,1,3,4] P(C(C(C(x,C(N(y),z)),u),C(y,u))).
14 [hyper,1,3,6] P(C(C(C(x,x),y),C(z,y))).
16 [hyper,1,14,14] P(C(x,C(y,C(z,z)))).
19 [hyper,1,3,14] P(C(C(C(x,y),N(y)),C(z,N(y)))).
23 [hyper,1,3,16] P(C(C(C(x,C(y,y)),z),C(u,z))).
96 [hyper,1,3,23] P(C(C(C(x,y),z),C(y,z))).
106 [hyper,1,3,96] P(C(C(C(C(N(x),N(y)),x),z),C(y,z))).
111 [hyper,1,96,3] P(C(x,C(C(x,y),C(z,y)))).
113 [hyper,1,96,111] P(C(x,C(C(C(y,x),z),C(u,z)))).
247 [hyper,1,106,96] P(C(x,C(N(x),y))).
250 [hyper,1,106,19] P(C(x,C(y,N(N(x))))).
340 [hyper,1,96,247] P(C(x,C(N(C(y,x)),z))).
405 [hyper,1,96,340] P(C(x,C(N(C(y,C(z,x))),u))).
493 [hyper,1,3,113] P(C(C(C(C(C(x,C(C(C(y,z),C(N(u),N(v))),u)),w),C(v6,w)),y),C(v,y))).
1660 [hyper,1,3,493]     P(C(C(C(x,y),C(z,C(C(C(y,u),C(N(v),N(x))),v))),C(w,C(z,C(C(C(y,u),C(N(v),N(x))),v))))).
5524 [hyper,1,1660,111] P(C(x,C(C(C(y,z),u),C(C(C(z,v),C(N(u),N(y))),u)))).
5540 [hyper,1,5524,5524] P(C(C(C(x,y),z),C(C(C(y,u),C(N(z),N(x))),z))).
5609 [hyper,1,5540,3] P(C(C(C(x,y),C(N(C(C(x,z),C(u,z))),N(C(C(C(z,v),C(N(w),N(u))),w)))),C(C(x,z),C(u,z)))).
6115 [hyper,1,5609,405] P(C(C(x,C(x,y)),C(z,C(x,y)))).
7172 [hyper,1,6115,250] P(C(x,C(y,N(N(y))))).
7189 [hyper,1,7172,7172] P(C(x,N(N(x)))).
7233 [hyper,1,247,7189] P(C(N(C(x,N(N(x)))),y)).
7420 [hyper,2,7233] $F.

------------ end of proof -------------

Principle of explosion: Other arguments?

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