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I am trying to prove to myself weak completeness of natural deduction for FOL.

My progress so far: (incorrect)

let us assume that there exists $\alpha$ so that it is valid and not derivable in natural deduction. if it is not derivable in natural deduction than $\neg\alpha$ is derivable (by consistency of the system) . if I can prove weak soundness of natural deduction I can prove that $\neg\alpha$ is valid and this is a contradiction.

Is there an different way to do that, or do I must first prove weak soundness? If I do, I am not sure I know how to do that. should I prove it by induction on the length of the proof series ? I couldn't find this proof in the literature or online. If anybody knows where I can find it, it will help a lot too.

Thank you

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    Consistency does not mean that if $\alpha$ is not derivable, them $\lnot \alpha$ is. – 2017-01-28
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    You are right... it is absolutely not. so I am truly lost then. Will you help me? – 2017-01-28
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    I am not familiar with 'weak completeness' ... but given your proof attempt, I assume you mean that we can find a proof of every valid statement in FOL. Well, that turns out to be a fairly hard proof .. it was the brilliant logician Kurt Godel who first proved completeness for FOL! So look for that: Godel's Completeness Theorem. – 2017-01-28
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    Every textbook has it; for Natural Deduction, see: Ian Chiswell & Wilfrid Hodges, [Mathematical Logic](https://books.google.it/books?id=JeUDUWYD5eQC&pg=PA89) (2007) **Ch.3.10 Completeness for propositional logic**, page 89. – 2017-01-29
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    @MauroALLEGRANZA Mauro, do you know what is meant by 'weak completeness'? I have never heard of the term. I suppose there is 'strong completeness' as well ... what is the difference? – 2017-01-29
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    @Bram28 - not exactly, but... if [**strong completenss**](https://en.wikipedia.org/wiki/Completeness_(logic)#Strong_completeness) is: "if $\Gamma \vDash \varphi$, then $\Gamma \vdash \varphi$, the "weak" completeness is the (original Godel's) **completeness** : "if $\vDash \varphi$, then $\vdash \varphi$". – 2017-01-29
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    @MauroALLEGRANZA Thanks! If that's the difference, then in terms of 'strength' there is no meaningful difference, right? I mean, if 'strong' complete, then it's automatically 'weak' complete (use $\Gamma = \{P \}$), and while for a 'weak' complete system we may not have $\Gamma \vdash \phi$ (because it may not be a system that is set up to even work with premises), but if $\{ \phi_1, ... \phi_n \} \vDash \phi$ then $\vDash (\phi_1 \land ... \land \phi_n) \rightarrow \phi$, and so at least for the weak system we have $\vdash (\phi_1 \land ... \land \phi_n) \rightarrow \phi$. Right? – 2017-01-29
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    "Weak" complete is : $\Gamma = \emptyset$ and thus: yes, if "strong", then "weak". A proof system (like ND) with derivation from assumptions needs "strong". – 2017-01-29
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    But even in this book they prove 'if Γ⊨φ then Γ⊢φ ' rather than'if ⊨φ than ⊢φ' isn't the latter simpler? I am also correcting myself, I am looking for a proof for propositional logic only. – 2017-01-30

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