Let $\lambda$ be an infinite cardinal. Recall that Weak diamond $\Phi_S$ on $S\subseteq\lambda^+$ is the following principle:

For every function $F:2^{<\lambda^+}\rightarrow 2$, there exists $g\in 2^{\lambda^+}$ such that for all functions $f:\lambda^+\rightarrow 2$, the set $\{\alpha\in S:~F(f\restriction\alpha)=g(\alpha)\}$ is stationary.

The following theorem is well known:

Theorem: (Devlin-Shelah) For every infinite $\lambda$, $\Phi_{\lambda^+}\Leftrightarrow 2^\lambda<2^{\lambda^+}$.

Unlike the above situation, failure of the weak diamond at a stationary set is not enough to get $2^\lambda=2^{\lambda^+}$. In fact, we have the following.

Theorem: (Shelah) The theory ${\rm ZFC}+\neg\Phi_{S^{\omega_2}_\omega}+2^{\omega_1}<2^{\omega_2}$ is consistent.

where $S^{\omega_2}_{\omega}=\{\alpha<\omega_2: {\rm cof}(\alpha)=\omega\}$, and similarly for $S^{\omega_2}_{\omega_1}=\{\alpha<\omega_2:{\rm cof}(\alpha)=\omega_1\}$ we have:

Theorem:(Shelah) The theory ${\rm ZFC}+\neg\Phi_{S^{\omega_2}_{\omega_1}}+{\rm GCH}$ is consistent.

Now in the light of above theorems, I would like to know what is the answer to the following question.

###Question: Does simultaneous failure of $\Phi_{S^{\omega_2}_{\omega }}$ and $\Phi_{S^{\omega_2}_{\omega_1}}$ imply $2^{\aleph_1}=2^{\aleph_2}$?

##Edit:

For a proof of 2nd theorem look at Shelah's paper "More on the weak diamond"-1985. A reference for the 3rd one was given in the comments by Golshani.