Logic Chapter 6 - Natural Deduction

अब Quizwiz के साथ अपने होमवर्क और परीक्षाओं को एस करें!

¬Elim

[¬φ]···ψ, [¬φ]···¬ψ | φ

↔Intro

[φ]···ψ , [ψ]···φ | φ↔ψ

→Intro

[φ]···ψ | φ→ψ

¬Intro

[φ]···ψ, [φ]···¬ψ | ¬φ

∀Intro

···φ [t/v] | ∀vφ {provided the constant t does not occur in φ or in any undischarged assumption in the proof of φ [t/v]}

∨Intro

···φ | φ∨ψ, ···ψ | φ∨ψ

→Elim

···φ, ···φ→ψ | ψ

↔Elim

···φ, ···φ↔ψ | ψ, ···ψ, ···φ↔ψ | φ

∧Intro

···φ, ···ψ | φ∧ ψ

∧Elim

···φ∧ψ | φ, ···φ∧ψ | ψ

∀Elim

···∀vφ | φ [t/v]

∃Elim

···∃vφ, [φ [t/v]]···ψ | ψ {provided the constant t does not occur in ∃vφ, or in ψ, or in any undischarged assumption other than φ [t/v] in the proof of ψ}

∃Intro

φ [t/v] | ∃vφ

Proof in Natural Deduction

φ is provable from Γ (Γ⊢φ) if and only there is a proof of φ with only sentences in Γ as undischarged assumptions. If Γ is empty, we can write ⊢φ. The proof system is defined in purely syntactic terms, on which semantics have no bearing

∨Elim

φ∨ψ, [φ]···χ, [ψ]···χ | χ


संबंधित स्टडी सेट्स

AP Language and Composition Exam Skills Test

View Set

unit 10 - political parties and elections

View Set