# indecidability, independency and company

It might be a stupid question, but I am having a look at independence and this question came to my mind :

Let say you have a proposition P1 independent of ZFC. If you find, in this same axiomatic system, a proposition :

• P2 that imply P1

• P3 that imply (not P1)

• P4 equivalent to P1

Does it mean that P2 is false, that P2 is undecidable, or just nothing ?

Does it mean that P3 is false, that P3 is undecidable, or just nothing ?

Does it mean that P4 is false, that P4 is undecidable, or just nothing ?

If you can advise me good reading about independence, and undecidability it would be nice. If you want to change the tags or the title it could be nice too... I don't know what to write.

Thanks in advance

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When you say "If you find, in this same axiomatic system ..." do you mean ZFC or ZFC+P1? – Asaf Karagila Jul 18 '11 at 19:27
I mean just ZFC – Ricky Bobby Jul 18 '11 at 19:29
Is the independency in the title intentional for having three words ending in -y? I think independence is much more common... Could you expand a bit on your background? I mean, on what level do you want good reading? Do you want a popular account (like Nagel-Newman) or an introductory book like Ciesielski or a more advanced text like Kunen? – t.b. Jul 18 '11 at 19:30
:D yes.. the y was for the rhyme. I kind of feel pretty frustrated when I cannot go as "far" as I want in the reading of a proof, so I would say that I'm looking for a pretty "complete" book starting from some basics. Something like the introduction book may be. thanks for your advices. – Ricky Bobby Jul 18 '11 at 19:39
I like the title :). I found Kunen quite tough reading but I'm an amateur when it comes to set theory. Ciesielski is really nice (if you've got your set theory right, you can start right away with the independencY-results in the second part). The bible is of course Jech. – t.b. Jul 18 '11 at 19:46

## 1 Answer

If $P_1$ is independent of the theory $T$ it means that $T\cup\{P_1\}$ has no new contradictions, as well $T\cup\{\lnot P_1\}$ has no new contradictions (that is if $T$ was consistent then so are the new theories).

If $T$ proves that $P_2\rightarrow P_1$ then either $T$ proves $\lnot P_2$ and then $P_2\rightarrow P_1$ is always true in models of $T$; or $P_2$ is also independent.

For example $GCH$ (the Generalized Continuum Hypothesis) implies that Axiom of Choice. Both of these claims are independent of $ZF$. However $0=1$ also implies $GCH$ (simply because contradiction implies everything), but $0=1$ is provabily false from $ZF$ (since $0=\varnothing$ and $1={\varnothing}$).

The same deal goes for $P_3$ such that $T$ proves $P_3\rightarrow\lnot P_1$, simply because $P_1$ is independent if and only if $\lnot P_1$ is independent of $T$.

For $P_4$ note that equivalency means that it implies $P_1$ and therefore independent as well, but also that $P_1$ implies it so if $T\cup\{P_1\}$ is consistent so is $T\cup\{P_4\}$, and therefore in this case $T$ cannot prove $P_4$ to be false.

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Thanks a lot! that's exactly what I was looking for. – Ricky Bobby Jul 18 '11 at 19:45