Second-Order Propositional Order - Lecture 14 | CS 4860, Study notes of Reasoning

Download Second-Order Propositional Order - Lecture 14 | CS 4860 and more Study notes Reasoning in PDF only on Docsity! Applied Logic Lecture 14: Second-Order Propositional Logic (Semantics and proof rules) CS 4860 Spring 2009 Thursday, March 5, 2009 These are preliminary notes, containing only the necessary formalities. If I ever get around to it I will add more explanations 14.1 Assignments Let Var be the type of propositional variables, and let B = {f, t} be the booleans (with f meaning false and t meaning true). An assignment is a function v : Var → B. Given an assignment v, a boolean b, and a propositional variable p, the “updated” assignment v|pb is the function (in Var → B) defined by v|pb (q) = { b if q = p v(q) otherwise 14.2 Semantics of P2 Let A be a P2-formula and let v be an assignment; let v[A] (an abbreviation of value(A, v)) be the notation for the (boolean) value of A under v, and let v[A] : B be defined recursively as follows: v[⊥] = f v[p] = v(p) v[A⊃B] = (¬B v[A]) B v[B] v[(∀p)A] = (v|pf)[A] ∧B (v| p t )[A] where ¬B : B→B, B : B×B→B, and ∧B : B×B→B are the standard boolean operators. For a finite set of formulas Γ, we define v ∧ [∆] = ∧ B { v[A] | A ∈∆ } and define v ∨ [Γ] = ∨ B { v[A] | A ∈Γ }, where ∧ B S is the conjunction of the boolean values in the set S and ∨ B S is their disjunction. (By convention, ∧ B ∅ = t and ∨ B ∅ = f .) The value v[∆⊢Γ] of a sequent can now be defined as (¬B v ∧ [∆]) B v ∨ [Γ]. Examples: let v(p0) = t, v(p1) = f, v(p2) = f v[(p0⊃p1)]= (¬B v[p0]) B v[p1] = (¬B t) B f = f v[(p0⊃(p0⊃p1))]= (¬B v[p0]) B v[p0⊃p1] = (¬B t) B f = f v[(p0⊃p0)]= (¬B v[p0]) B v[p0] = (¬B t) B t = t v[(p0⊃(∀p0(p0⊃p0)))]= (¬B v[p0]) B (v| p0 f )[p0⊃p0] ∧B (v| p0 t )[p0⊃p0] = f B (v[f⊃f ]) ∧B (v[t⊃t]) = f B (t ∧B t) = t The semantics of P2 can also be defined by reducing a P2-formula into an ordinary propositional formula. Since a variable can only assume two possible values, we can replace every universally quantified formula by (∀p)A by the formula A[⊤/p] ∧ [⊥/p], where ⊤ ≡ ⊥ ⊃ ⊥.1 1This reduction technique only works with P2. It cannot be used to reduce first-order logic to propositional logic, 1 14.3 Rules of P2 The multiple-conclusioned sequent proof rules for P2 are as follows ⊥L : ∆,⊥⊢Γ ⊃L : ∆, A⊃B⊢Γ ∆⊢A, Γ ∆⊢A⊃B, Γ ∆, A⊢B, Γ ⊃R ∆, B⊢Γ ∀L(B) : ∆, ∀pA⊢Γ ∆, ∀pA, A[B/p]⊢Γ ∆⊢∀pA, Γ ∆⊢A[q/p], Γ ∗∗ ∀R(q) axiom : ∆, A⊢A, Γ thinL : ∆, A⊢Γ ∆⊢Γ ∆⊢A, Γ ∆⊢Γ thinR ∗∗ this is only legal if q 6∈FV (∆, Γ, ∀pA). The rules for ∃ can be derived from the rules given above: ∃L : ∆, ∃pA⊢Γ ∆, A|pq⊢Γ ∗∗ ∆⊢∃pA, Γ ∆⊢A|pB, Γ ∃R The familiar rules for ∧ , ∨ , and ∼ can also be derived. An example proof: ⊢(∀p.p)⊃⊥ ∀p.p⊢⊥ ⊃R ⊥⊢⊥ ∀L(⊥) Here is a proof that the two definitions of conjunction given above are actually equivalent. ⊢A ∧B ⊃ (∀p)((A⊃B⊃p)⊃p) ⊃R A ∧B ⊢ (∀p)((A⊃B⊃p)⊃p) ∀R(P ) A ∧B ⊢ (A⊃B⊃P )⊃P ⊃R A ∧B, (A⊃B⊃P ) ⊢ P ⊃L 1.A ∧B ⊢ A, P ∧L A, B ⊢ A, P axiom 2.A ∧B, B⊃P ⊢ P ⊃L 2.1.A ∧B ⊢ B, P ∧L A, B ⊢ B, P axiom 2.2.A ∧B, P ⊢ P axiom since variables may assume infinitely many values. 2