Can someone clear up this semantic proof of quantification logic?

Question

The idea is to proof validity of

∃x(Fx ^ Gx) / ∃xFx

To do this I understand you assume invalid and get a contradiction. I have the answer but I don't understand the wording

∃x(Fx ^ Gx) is true-in-I, which means some object o∈D satisfies Fx ^ Gx, which means o is such that both o∈ext(F) and o∈ext(G).

∃xFx is false-in-I, which means every object o∈D is such that o∉ext(F).

Since what holds of o` holds of everything in the domain of I, and since o∉D, o∉ext(F). Contradiction, so there is no such interpretation. Thus valid.

what I don't get is the last statement "since what holds of o' holds of everything in the domain of I" how does that go to "since o∉D" when we are talking about o' and not o and we had established o is contained in D already (I get we need the contradiction, but from where was this generated?)

Answer 1

Start with "intuition" :

if there is a black cat, then there is a cat

is an argument that "sound" quite correct.

To prove it "semantically" we have to assume an interpretation I in which the premise is true.

This means that there is an object o of the domain of the interpretation I for which both F and G holds.

But this implies that for that object o the "property" F holds, and this boils down to say that in I there exists an x such that F(x) is true.

Thus, being I an interpretation whatever, we have proved it for all interpretations; i.e. in every interpretation in which the premise is true, also the conclusion is, and this is the definition of logical consequence :

A ⊨ B.

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Note

In proving it by contradiction, we have to assume that the conclusion ∃xFx is false; this amount to saying that for every object x in the domain D of the interpretation, we have that F does not hold for it.

But this is in contradiction with the fact implied by the assumption that there is o ∈ D such that F holds for it.

Answer 2

'o∉D' is a typo, which should be corrected to 'o∈D'. The point simply is this: Since by assumption ∃xFx is false in I we have o*∉ext(F) for all o*∈D. We already know that there is some o∈D. So, o∉ext(F). But wait! Since o satisfies Fx&Gx we have o∈ext(F). So, o∈ext(F) and o∉ext(F). Contradiction.

Actually I think it more elegant to prove the contrapositive. If no o'∈D is such that o'∈ext(F), then it's not the case that both o'∈ext(F) and o'∈ext(G). So, nothing from D satisfies Fx&Gx. So,∃x(Fx & Gx) is false in I.