MA plus not CH implies SH

Since CH fails, negating F2 gives a set $A$ with $\mathbb{N}\prec A\prec \mathcal{P}(\mathbb{N})$. By A1 all three sets are well-orderable. F3 and F4 turn the two injections into ${\mathrm{\aleph }}_0\le \mathrm{\mid }A\mathrm{\mid }\le \mathrm{\mid }\mathcal{P}(\mathbb{N})\mathrm{\mid }$. Both inequalities are strict: equality on the left would give $A\approx \mathbb{N}$ by F3, and equality on the right would give $A\approx \mathcal{P}(\mathbb{N})$, contradicting the two strict comparisons that define the witness. With F5 this is ${\mathrm{\aleph }}_0<\mathrm{\mid }A\mathrm{\mid }<2^{{\mathrm{\aleph }}_0}$.

The middle term $\mathrm{\mid }A\mathrm{\mid }$ is a cardinal by F3. Since F6 makes ${\mathrm{\aleph }}_1$ the least cardinal strictly above ${\mathrm{\aleph }}_0$, step 1.1 gives ${\mathrm{\aleph }}_1\le \mathrm{\mid }A\mathrm{\mid }<2^{{\mathrm{\aleph }}_0}$, and hence ${\mathrm{\aleph }}_1<2^{{\mathrm{\aleph }}_0}$.

The cardinal ${\mathrm{\aleph }}_1$ is infinite, so F1 and step 2.1 instantiate the MA scheme at ${\mathrm{\aleph }}_1$. Thus $\mathrm{M}\mathrm{A}({\mathrm{\aleph }}_1)$ holds, and F7 implies that no Suslin tree exists.

Apply the direction “no Suslin tree implies SH” of F8. This yields SH, as required. AC was used in step 1.1 to cardinalize the witness and is also propagated through F7 and F8; no choice-free conclusion is asserted.