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
 A: 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.
