Strictly speaking, it is not clear how your described interpretation relates to the fact that F2 does not logically imply F3. The latter fact says thatLet the domain be {0,1}

Let P(x,y,z): x+y>z

Set x=0 y=1 u=0 z=1 w=0 v=0 t=0

Then F2 does not logically imply F3there existsan interpretation where F2 is true and F3 is false. Did you describe that alleged interpretation? Then say directly that it is in the interpretation just describes that F2 is true and F3 is false.

First, you may need to clarify what is 1 + 1 since the domain is {0,1}. Since + is not in the formula vocabulary, it does not have to be interpreted by a function from {0,1} to {0,1}; so I assumed that you mean regular addition. I agree that F2 is true because its premise, namely P(x,v,t) is false. However, both sides of F3 are true: take u = 0 and v = 1.

Assuming F1, it is easy to show -> implication in F3. Indeed, assume P(x,y,u) and P(u,z,w) for some u. By F1, there exist v and y such that P(y,z,v) and P(x,v,t). Then by F2, w = t, so P(x,v,w) and thus ∃v(P(y,z,v) ∧ P(x,v,w)). At the moment I am not sure about the converse implication. Are you sure it is an equivalence?