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let $Asubseteq BsubseteqmathcalM$ (where $subseteq$ is just the subset symbol). $q=mathrmtp(a/B)$ is an heir of $p=mathrmtp(a/A)$ iff for any $varphi(x;y)inmathcalL_A$, $varphi(x;b)in q$ for some $bin B$ implies $varphi(x;a)in p$ for some $ain A$.

I'm trying to prove when $mathcalMsubseteq B$ (possibly $B$ subset of some $mathcalN>mathcalM$ where $mathcalN$ is $|B|^+$-saturated) and $pin S(mathcalM)$, there exists $qin S(B)$ an heir of $p$.

I wanted to find $q$ contradicting "there exists $varphi(x;b)in q$ with $forall ainmathcalM, varphi(x;a)notin p$." So letting $p=mathrmtp(c/ mathcalM)$, I set $Sigma:=varphi(x;b)inmathcalL_Bmid mathcalNmodelsvarphi(c;b)wedge forall ainmathcalM,mathcalNmodelsvarphi(c;a) $. I could prove $Sigma cup p$ is consistent so I took a completion $q$ of this set in $mathcalL_B$. I assumed $q$ is not an heir of $p$ but failed to derive a contradiction. I tried to construct another set but I couldn't find an appropriate one.

In the definition of heir, you forgot the condition that $psubseteq q$.

Hint: You almost chose the right set $Sigma$, but not quite. Instead, look a: $$Sigma = pcup varphi(x;b)mid varphi(x;y)in mathcalL_A, bin B,textand forall ain mathcalM, varphi(x;a)in p.$$

Showing this $Sigma$ is consistent will allow you to conclude, since if $q$ is any consistent completion of $Sigma$, and if we assume for contradiction that $varphi(x;b)in q$ but $varphi(x;a)notin p$ for all $ain mathcalM$, then $lnot varphi(x;a)in p$ for all $ain mathcalM$, so $lnot varphi(x;b)in Sigmasubseteq q$, contradicting consistency of $q$.

This argument won't work with your choice of $Sigma$, since we don't necessarily have $mathcalNmodels lnot varphi(c;b)$.