Counterexample: Γ = {} φ = p(x) ψ = ∀x p(x) Now, applying Deduction Theorem, from prop. logic, substituting above Γ |= (φ ⇒ ψ) iff Γ ∪ {φ} |= ψ yields: ({} |= (p(x) ⇒ ∀x p(x)) ⇔ ({p(x)} |= ∀x p(x)) The LHS of this equivalence is the logical entailment {} |= (p(x) [...]
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