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    If Γ ⊨ φ, then Γ ∪ {¬φ} is not satisfiable and has no model — Carmelics
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    Supports→Strong completeness holds for many-sorted logic: if Γ ⊨ φ then Γ ⊢ φ

    If Γ ⊨ φ, then Γ ∪ {¬φ} is not satisfiable and has no model

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    Related propositions within the same area of thought.
    By Henkin's theorem, if Γ ∪ {¬φ} has no model then Γ ∪ {¬φ} is contradictoryStrong completeness holds for many-sorted logic: if Γ ⊨ φ then Γ ⊢ φThe calculus rules allow elimination of ¬φ to infer Γ ⊢ φ

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    If Γ ⊨ φ, then Γ ∪ {¬φ} has no model (is unsatisfiable)96%If Γ ⊨ φ, then Γ ∪ {¬φ} has no model and is therefore, by Henkin's the...91%By Henkin's theorem, if Γ ∪ {¬φ} has no model then Γ ∪ {¬φ} is contrad...89%If Γ ∪ {¬φ} is contradictory, then Γ ∪ {¬φ} ⊢ φ81%

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    , determining validity, or equivalently, testing for satisfiability of given formulas) for many-sorted logic is undecidable. So, we are in the same situation encountered in one-sorted first-order logic. Of course, if a calculus is to be helpful it would never allow erroneous reasonings: it is not going to drive us from true hypotheses to false conclusions. It must be a sound calculus. Further, it is highly desirable that all the consequences of a set \(\Gamma\) of hypotheses could be derived fr

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