# Is there any forcing free proof for hard independence results?

We are forced to use forcing for almost all "hard" independence results such as: $Con(ZFC)\longrightarrow Con (ZFC+\neg CH)$. The question simply is:

Primary Question: Is there any "forcing free" proof for $Con(ZFC)\longrightarrow Con (ZFC+\neg CH)$ or $Con(ZF)\longrightarrow Con (ZF+\neg AC)$ or any other "hard" independence results?

Secondary Question: Please list all "non simple" consistency results which have two proofs one by forcing and another without using it.

• – Andrés E. Caicedo Aug 7 '13 at 13:55
• No, there are no forcing-free proofs of any of these results. It would be great to come upon such a new technique. There are, of course, a few independent results that do not rely on forcing, but I do not know i they are of the kind you are seeking. There is the consistency of Aczel's anti-doundation axiom, for example. See also here. – Andrés E. Caicedo Aug 7 '13 at 14:01
• I thought I heard that Moschovakis has a proof of independence of V=L by omitting ordinals, rather than by forcing. – Colin McLarty Aug 7 '13 at 14:10
• In Fraenkel atoms model AC fails. – Eran Aug 7 '13 at 19:22
• @Eran And we have atoms, so we do not have extensionality, and the underlying theory is not $\mathsf{ZF}$. – Andrés E. Caicedo Aug 7 '13 at 23:55

Krivine realizability can be used to obtain independence results over $\mathsf{ZF}$. For instance, in Krivine's paper Realizability algebras II : new models of ZF + DC, Logical Methods in Computer Science 8 (1:10) p. 1-28 (2012), he constructs a realizability model in which $\mathsf{AC}$ fails in a pretty strong way: there is an infinite sequence of infinite subsets of $\mathbb{R}$ strictly decreasing in cardinality.