I am trying to show that forking is transitive to the left. i.e if $b\overset{\vert}{\smile}_{A}c $ and $a\overset{\vert}{\smile}_{Ab}c $ than $ab\overset{\vert}{\smile}_{A}c $
Edit: So I think I may got it but not so sure about it: If $p(x,y)=tp(ab/Ac)$ forks over $A$, than there is a forking formula $\varphi (x,y,c)\in p$ i.e $\varphi(x,y,c)$ implies a finite disjunction of formulas over $A$, $$\bigvee _i\varphi _i (x,y,c_i)$$ s.t $\varphi _i (x,y,c_i)$ divides over $A$. So there exist a $A$-indiscernible sequences $I_i=(d_j)_{j<\omega}$ s.t $I_i$ starts with $c_i$ and $$\pi(x,y,I_i)=\{\varphi _i (x,y,d_j)\mid j<\omega\}$$ is inconsistent. But then $\pi(x,b,I_i)$ is also inconsistent, and since $\varphi (x,y,c)\in tp(ab/Ac)$ than $\varphi (x,b,c)\in tp(a/Abc)$ and $\varphi (x,b,c)$ implies $\bigvee _i\varphi _i (x,b,c_i)$. But we know that $\pi(x,b,I_i)$ is inconsistent, therefor $\varphi _i (x,b,c_i)$ divides over $A$. Hence $\varphi (x,b,c)$ forks over $A$. But we want it to fork over $Ab$ so we need to make an adjustment to $I_i$. There is an equivalence for dividing that say's $tp(a/Ab)$ does not divide over $A$ iff for any $A$-indiscernible sequence, there is a $Aa$-indiscernible sequence $I'$ s.t $I'\equiv_{Ab}I$. So since $b\overset{\vert}{\smile}_{A}c $, than in particular $tp(b/Ac)$ does not divide over $A$, Hence for every $i$ there is a $Ab$-indiscernible sequence $J_i$ s.t $J_i \equiv_{Ac} I_i$ and with the right automorphism we can make sure that $J_i$ starts with $c_i$. Thus $\pi(x,b,J_i)$ is inconsistent and therefor $\varphi _i (x,b,c_i)$ divides over $Ab$ and so $\varphi (x,b,c)$ forks over $Ab$ in contradiction to $a\overset{\vert}{\smile}_{Ab}c $. And we are done.
is it right?