By Zhukavets N.M., Shestakov I.P.

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Algebra, as we all know it this day, involves many alternative rules, techniques and effects. an inexpensive estimate of the variety of those diverse goods will be someplace among 50,000 and 200,000. lots of those were named and lots of extra may perhaps (and possibly should still) have a reputation or a handy designation.

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For each n ∈ N, we recursively construct a language Ln in the following way: let L0 be the language L. For n ≥ 1, if Ln−1 has already been constructed, then we obtain Ln by setting In = In−1 , Jn = Jn−1 and Kn = Kn−1 ∪ Mn . 1) from Mn to the set of all existence sentences in the language Ln−1 . 4 Completeness of First-Order Logic 29 indices in a one-to-one and onto manner. Sets Mn and bijections gn of the required kind always exist. 2) of languages. e. we set I = I, J = J and K = n∈N Kn . From this one sees immediately that Sent(L ) = Sent(Ln ) n∈N also holds.

Case 2: y is not x, and x ∈ Fr(ψ ). In this case y cannot occur in t, since, by hypothesis, t is free for x in ∀y ψ (p. 12). 4) a = t A [h] = t A h ay . 4) A |= ψ (x/t) [h ] for all a ∈ |A| ind. hyp. 1) for h y A |= ψ (x/t) h a for all a ∈ |A| def. 9) A |= (∀y ψ ) (x/t) [h] since y is not x. 3. It asserts that everything that can be proved from an axiom system Σ also holds in every model of Σ . Thus, if we have a model of Σ ∪ {¬ϕ }, then obviously ϕ cannot be proved from Σ . 2). 4 (Soundness Theorem).

6. (c) (⇒) If ∀x ϕ ∈ Σ ∗ , then Σ ∗ ∀x ϕ . Then for any t ∈ CT, Σ ∗ ϕ (x/t), by Rule (∀ B) (p. 18), which applies here since every constant term t is, vacuously, free for x in ϕ . 6), we get ϕ (x/t) ∈ Σ ∗ . / Σ ∗ . Then ¬∀x ϕ ∈ Σ ∗ , by (a). From this we would like to (⇐) Assume ∀x ϕ ∈ ∗ deduce ∃x ¬ϕ ∈ Σ . Σ ∗ (¬¬ϕ → ϕ ), since ¬¬ϕ → ϕ is a tautology. 1) Σ ∗ ∪ {∀x ¬¬ϕ } ∀x ϕ . Since ∀x ϕ is a sentence by hypothesis, so is ∀x ¬¬ϕ . 2 to obtain Σ∗ (∀x ¬¬ϕ → ∀x ϕ ). From this we obtain, using (CP) (p.