A base of the free alternative superalgebra on one odd by Zhukavets N.M., Shestakov I.P.

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

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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.

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