I have stumbled on a few different formulations of the Singular Cardinals Hypothesis. The most common are:
SCH1: $\quad 2^{cf(\kappa)}<\kappa \ \Longrightarrow \ \kappa^{cf(\kappa)}=\kappa^+$ for infinite cardinals $\kappa$
SCH2: $\quad\kappa^{cf(\kappa)}=\max\{\kappa^+,2^{cf(\kappa)}\}\quad(=\kappa^++2^{cf(\kappa)}=\kappa^+ \cdot 2^{cf(\kappa)})$ for infinite cardinals $\kappa$
SCH1 and SCH2 are easily seen to be equivalent when you note that $2^{cf(\kappa)}\geq\kappa$ implies $\kappa^{cf(\kappa)}=2^{cf(\kappa)}$ and recall that $\kappa^{cf(\kappa)}>\kappa$.
There are also three other formulations which are found in Wikipedia (F1) and in some notes on set theory (F2 & F3):
F1: $2^\kappa=\kappa^+$ for singular strong limit cardinals $\kappa$
F2: $2^\kappa=\min\{\lambda:\lambda\geq 2^{<\kappa}\text{ and }cf(\lambda)>\kappa\}$ for singular cardinals $\kappa$
F3: $\kappa^{cf(\kappa)}=(2^{<\kappa})^+$ for singular cardinals $\kappa$
SCH1 implies F1 since $2^\kappa=\kappa^{cf(\kappa)}$ for strong limit cardinals $\kappa$.
If the continuum function $\lambda\mapsto 2^\lambda$ is eventually constant below $\kappa$, then $2^\kappa=2^{<\kappa}$. Otherwise $cf(2^{<\kappa})=cf(\kappa)\leq\kappa$ and, assuming SCH1, $2^\kappa=(2^{<\kappa})^+$. This is how one can see why SCH1 implies F2.
Of F3 I'm the most doubtful that it even is equivalent to SCH(1/2): if $2^\kappa=2^{<\kappa}$ (for example when the continuum function is eventually constant below $\kappa$), then $(2^{<\kappa})^+>2^\kappa\geq\kappa^{cf(\kappa)}$.
Question: Are F1, F2 and F3 equivalent to SCH(1/2) and if they are, how can one prove the implications F1 $\Rightarrow$ SCH, F2 $\Rightarrow$ SCH and F3 $\Leftrightarrow$ SCH?