New PDF release: A problem course in mathematical logic : is a freeware

By Stefan Bilaniuk

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9. Suppose Σ is a maximally consistent set of sentences and ϕ and ψ are any sentences. Then ϕ → ψ ∈ Σ if and only if ϕ∈ / Σ or ψ ∈ Σ. 10. Suppose Γ is a consistent set of sentences. Then there is a maximally consistent set of sentences Σ with Γ ⊆ Σ. The counterparts of these notions and facts for propositional logic sufficed to prove the Completeness Theorem, but here we will need some additional tools. The basic problem is that instead of defining a suitable truth assignment from a maximally consistent set of formulas, we need to construct a suitable structure from a maximally consistent set of sentences.

Note. It is possible to define first-order languages without =, so = is considered a non-logical symbol by many authors. While such languages have some uses, they are uncommon in ordinary mathematics. Observe that any first-order language L has countably many logical symbols. It may have uncountably many symbols if it has uncountably many non-logical symbols. Unless explicitly stated otherwise, we will 1It is possible to formalize almost all of mathematics in a single first-order language, like that of set theory or category theory.

We will often write M ψ if it is not the case that M |= ψ. Similarly, if Γ is a set of formulas, we will write M |= Γ if M |= γ for every formula γ ∈ Γ, and say that M is a model of Γ or that M satisfies Γ. A formula or set of formulas is satisfiable if there is some structure M which satisfies it. We will often write M Γ if it is not the case that M |= Γ. Note. M ϕ does not mean that for every assignment s : V → |M|, it is not the case that M |= ϕ[s]. It only means that that there is some assignment r : V → |M| for which M |= ϕ[r] is not true.

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A problem course in mathematical logic : is a freeware mathematics text by Stefan Bilaniuk


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