Predicate Logic: Soundness and Completeness of Formal Deduction

Predicate Logic: Soundness and Completeness of Formal Deduction

Predicate Logic: Soundness and Completeness of Formal Deduction Alice Gao Lecture 17 CS 245 Logic and Computation Fall 2019 1 / 9 Outline The Learning Goals The Soundness of Formal Deduction Revisiting the Learning Goals CS 245 Logic and Computation Fall 2019 2 / 9 Learning goals By the end of this lecture, you should be able to: ▶ Define soundness and completeness. ▶ Prove that an inference rule is sound or not sound. ▶ Prove that a logical consequence holds using the soundness and completeness theorems. ▶ Show that no natural deduction proof exists for a logical consequence using the soundness and completeness theorems. CS 245 Logic and Computation Fall 2019 3 / 9 The Soundness of Formal Deduction Theorem Formal Deduction for Predicate Logic is sound. Proof Sketch. Since Formal Deduction for Propositional Logic is sound, it suffices to prove the soundness of the ∀−, ∃+, ∀+ and ∃− rules. CS 245 Logic and Computation Fall 2019 4 / 9 The soundness of ∀− Theorem The ∀− inference rule is sound. That is, if Σ ⊨ ∀푥 퐴(푥), then Σ ⊨ 퐴(푡) where 푡 is a term. Proof. Consider a valuation 푣 such that Σ푣 = 1. Since Σ ⊨ ∀푥 퐴(푥), we have that (∀푥 퐴(푥))푣 = 1. By the definition of ∀, 퐴(푢)푣(푢/훼) = 1 for any 푢 ∈ 퐷. Since 푡 is a term, 푡푣 ∈ 퐷. Therefore, we have that 퐴(푡)푣 = 퐴(푢)푣(푢/푡푣) = 1. Therefore, Σ ⊨ 퐴(푡) holds. CS 245 Logic and Computation Fall 2019 5 / 9 The soundness of ∃+ Theorem The ∃+ inference rule is sound. That is, if Σ ⊨ 퐴(푡), then Σ ⊨ ∃푥 퐴(푥) where 푡 is a term. Proof. Consider a valuation 푣 such that Σ푣 = 1. Since Σ ⊨ 퐴(푡), 퐴(푡)푣 = 1. Thus, 퐴(푢)푣(푢/푡푣) = 1. Since 푡 is a term, 푡푣 ∈ 퐷. By the definition of ∃, (∃푥 퐴(푥))=1. Therefore, Σ ⊨ (∃푥 퐴(푥)) holds. CS 245 Logic and Computation Fall 2019 6 / 9 The soundness of ∃− Theorem The ∃− inference rule is sound. That is, if Σ, 퐴(푢) ⊨ 퐵, 푢 not occurring in Σ or 퐵, then Σ, ∃푥 퐴(푥) ⊨ 퐵. Proof. Let 푣 be a valuation such that Σ푣 = 1 and (∃푥 퐴(푥))푣 = 1. There is some 훼 ∈ 퐷 such that 퐴(푢)푣(푢/훼) = 1. Since 푢 does not occur in Σ, we have Σ푣(푢/훼) = Σ푣 = 1. By Σ, 퐴(푢) ⊨ 퐵, we have 퐵푣(푢/훼) = 1. Since 푢 does not occur in 퐵, we have 퐵푣 = 퐵푣(푢/훼) = 1. Hence Σ, ∃푥 퐴(푥) ⊨ 퐵 holds. CS 245 Logic and Computation Fall 2019 7 / 9 The soundness of ∀+ Theorem The ∀+ inference rule is sound. That is, if Σ ⊨ 퐴(푢), 푢 not occurring in Σ, then Σ ⊨ ∀푥 퐴(푥). Proof. Consider a valuation 푣 such that Σ푣 = 1. Since 푢 does not occur in Σ, we have that Σ푣(푢/훼) = Σ푣 = 1 for every 훼 ∈ 퐷. Since Σ ⊨ 퐴(푢), 퐴(푢)푣(푢/훼) = 1 for every 훼 ∈ 퐷. By the definition of ∀, (∀푥 퐴(푥))푣 = 1. CS 245 Logic and Computation Fall 2019 8 / 9 Revisiting the learning goals By the end of this lecture, you should be able to: ▶ Define soundness and completeness. ▶ Prove that an inference rule is sound or not sound. ▶ Prove that a semantic entailment holds using the soundness and completeness theorems. ▶ Show that no natural deduction proof exists for a semantic entailment using the soundness and completeness theorems. CS 245 Logic and Computation Fall 2019 9 / 9.

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