Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
The concept of a deductive system for propositional calculus, focusing on the language l0, its axioms, and rules of inference. It also covers the definition of deducibility and provides examples of deductions. The document concludes with the soundness theorem, stating that the system is sound.
Typology: Study notes
1 / 8
This page cannot be seen from the preview
Don't miss anything!
6.1 Definition Let L 0 := L[{¬, →}] (which is an adequate lan-
guage). Then the system L 0 consists of the
following axioms and rules:
Axioms
An axiom of L 0 is any formula of the following
form (α, β, γ ∈ Form(L 0 )):
A1 (α → (β → α))
A2 (((α → (β → γ)) → ((α → β) → (α → γ)))
A3 ((¬β → ¬α) → (α → β))
Rules of inference
Only one: modus ponens
(for any α, β ∈ Form(L 0 )) MP From α and (α → β) infer β.
6.3 Example For any φ ∈ Form(L 0 )
(φ → φ)
is a theorem of L 0.
Proof:
α 1 (φ → (φ → φ)) [A1 with α = β = φ] α 2 (φ → ((φ → φ) → φ)) [A1 with α = φ, β = (φ → φ)] α 3 ((φ → ((φ → φ) → φ)) → → ((φ → (φ → φ)) → (φ → φ))) [A2 with α = φ, β = (φ → φ), γ = φ] α 4 ((φ → (φ → φ)) → (φ → φ)) [MP α 2 , α 3 ] α 5 (φ → φ) [MP α 1 , α 4 ]
Thus, α 1 , α 2 ,... , α 5 is a deduction of (φ → φ) in L 0.
2
6.4 Example
For any φ, ψ ∈ Form(L 0 ):
{φ, ¬φ} ⊢ ψ
Proof:
α 1 (¬φ → (¬ψ → ¬φ))
[A1 with α = ¬φ, β = ¬ψ]
α 2 ¬φ [∈ Γ]
α 3 (¬ψ → ¬φ) [MP α 1 , α 2 ]
α 4 ((¬ψ → ¬φ) → (φ → ψ))
[A3 with α = φ, β = ψ] α 5 (φ → ψ) [MP α 3 , α 4 ]
α 6 φ [∈ Γ]
α 7 ψ [MP α 5 , α 6 ]
i = 1 either α 1 is an axiom, so v˜(α 1 ) = T or α 1 ∈ Γ, so, by hypothesis, v˜(α 1 ) = T.
Induction step Suppose (⋆) is true for some i < m. Consider αi+1.
Either αi+1 is an axiom or αi+1 ∈ Γ, so v˜(αi+1) = T as above,
or else there are j 6 = k < i + 1 such that αj = (αk → αi+1).
By induction hypothesis
v^ ˜(αk) = v˜(αj) = v˜((αk → αi+1)) = T.
But then, by tt →, v˜(αi+1) = T (since T → F is F ).
2
For the proof of the converse
Completeness Theorem
If Γ |= α then Γ ⊢ α.
we first prove
6.6 The Deduction Theorem for L 0
For any Γ ⊆ Form(L 0 ) and
for any α, β ∈ Form(L 0 ):
if Γ ∪ {α} ⊢ β then Γ ⊢ (α → β).
