A Brief Introduction to the Intuitionistic Propositional Calculus
... Problem 1 Prove that α ⇒ (β ⇒ γ) `I (α ∧ β) ⇒ γ. Problem 2 Show that α ⇒ β 6`I ¬α ∨ β by demonstrating that there exists a Kripke model K = (W, ≤, |=) and a world w ∈ W such that w |= α ⇒ β, but w 6|= ¬α ∨ β. Problem 3 Show that world w1 in the simple Kripke model in Section 4 does not satisfy Peirc ...
... Problem 1 Prove that α ⇒ (β ⇒ γ) `I (α ∧ β) ⇒ γ. Problem 2 Show that α ⇒ β 6`I ¬α ∨ β by demonstrating that there exists a Kripke model K = (W, ≤, |=) and a world w ∈ W such that w |= α ⇒ β, but w 6|= ¬α ∨ β. Problem 3 Show that world w1 in the simple Kripke model in Section 4 does not satisfy Peirc ...
Lecture Notes 12: Cognition and Computation
... to reply appropriately to everything that may be said in its presence, as even the lowest type of man can do…” ...
... to reply appropriately to everything that may be said in its presence, as even the lowest type of man can do…” ...
Proof Theory in Type Theory
... Another question is if we really need an inductive definition with “uncountable branching” (from the reference [1, 4], one should expect that it is not needed). This should come from an analysis of the given proof. For instance, it seems that we are really working in the fragment of (S0 ) with only ...
... Another question is if we really need an inductive definition with “uncountable branching” (from the reference [1, 4], one should expect that it is not needed). This should come from an analysis of the given proof. For instance, it seems that we are really working in the fragment of (S0 ) with only ...
PDF
... ∀X:FORM. ∀T :TableauxX . ∀U6=∅. ∀I:PredX →Rel(U). U,I|=origin(T ) 7→ ∃θ:path(T ). U,I|=θ where U,I|=θ ≡ ∀Y:S-FORM. Y on θ 7→ (U,I)|=Y. This is similar to what we had in the propositional case. However, I is now a first-order valuation over U instead of a boolean valuation and the definition of |=, ...
... ∀X:FORM. ∀T :TableauxX . ∀U6=∅. ∀I:PredX →Rel(U). U,I|=origin(T ) 7→ ∃θ:path(T ). U,I|=θ where U,I|=θ ≡ ∀Y:S-FORM. Y on θ 7→ (U,I)|=Y. This is similar to what we had in the propositional case. However, I is now a first-order valuation over U instead of a boolean valuation and the definition of |=, ...
Lecture01 - Mathematics
... Set Theory: Informally we define a set as a collection of objects. The resulting theory of how one can operate on sets is known as naïve set theory. It is naïve because the informal definition leads to subtle paradoxes. A more careful definition of set removes these paradoxes and leaves the conclusi ...
... Set Theory: Informally we define a set as a collection of objects. The resulting theory of how one can operate on sets is known as naïve set theory. It is naïve because the informal definition leads to subtle paradoxes. A more careful definition of set removes these paradoxes and leaves the conclusi ...
Some Principles of Logic
... (a) If I pass my exams, I will be a graduate (b) If I win the pools, I will become rich ...
... (a) If I pass my exams, I will be a graduate (b) If I win the pools, I will become rich ...
Document
... The significance of Russell's paradox can be seen once it is realized that, using classical logic, all sentences follow from a contradiction. For example, assuming both P and ~P, any arbitrary proposition, Q, can be proved as follows: from P we obtain P Q by the rule of Addition; then from P Q and ~ ...
... The significance of Russell's paradox can be seen once it is realized that, using classical logic, all sentences follow from a contradiction. For example, assuming both P and ~P, any arbitrary proposition, Q, can be proved as follows: from P we obtain P Q by the rule of Addition; then from P Q and ~ ...
Decidable fragments of first-order logic Decidable fragments of first
... In a nutshell, the idea for obtaining a decision procedure asserted by the proposition is the following: for a given sentence ϕ search in parallel for a finite model of ϕ and for a proof of ¬ϕ. ...
... In a nutshell, the idea for obtaining a decision procedure asserted by the proposition is the following: for a given sentence ϕ search in parallel for a finite model of ϕ and for a proof of ¬ϕ. ...
Exam 2 study guide
... Proving a formula of the form φ→ ψ, where the conditional φ→ψ is provable: first prove the conditional φ→ψ, then Necessitate, then distribute the over the → using K . Proving a formula of the form ◊φ→◊ψ, where the conditional φ→ψ is provable. As above, but use K◊. Proving a formula of the form φ ...
... Proving a formula of the form φ→ ψ, where the conditional φ→ψ is provable: first prove the conditional φ→ψ, then Necessitate, then distribute the over the → using K . Proving a formula of the form ◊φ→◊ψ, where the conditional φ→ψ is provable. As above, but use K◊. Proving a formula of the form φ ...
Multi-Agent Only
... But technically things were surprisingly cumbersome! The problem lies in the complexity in what agents consider possible: ...
... But technically things were surprisingly cumbersome! The problem lies in the complexity in what agents consider possible: ...
Robot Morality and Review of classical logic.
... have either rice pilaf or baked potato with your dinner. In such circumstances, he plainly does not mean either rice pilaf or baked potato or both. You have to choose. So this use of “or” doesn’t fit the definition of disjunction given above. ...
... have either rice pilaf or baked potato with your dinner. In such circumstances, he plainly does not mean either rice pilaf or baked potato or both. You have to choose. So this use of “or” doesn’t fit the definition of disjunction given above. ...
Solutions to Workbook Exercises Unit 16: Categorical Propositions
... Hx: x howls U.D.: animals Bx: x barks Cx: x is a cat Lx: x likes to walk Dx: x is a dog Mx: x meows Fx: x likes canned food Wx: x wags its tail (a) Some dogs howl. ∃x (Dx • Hx) x ...
... Hx: x howls U.D.: animals Bx: x barks Cx: x is a cat Lx: x likes to walk Dx: x is a dog Mx: x meows Fx: x likes canned food Wx: x wags its tail (a) Some dogs howl. ∃x (Dx • Hx) x ...
Propositional and Predicate Logic - IX
... Theorem For every theory T and sentence ϕ, if ϕ is valid in T , then ϕ is tableau provable from T , i.e. T |= ϕ ⇒ T ` ϕ. Proof Let ϕ be valid in T . We will show that an arbitrary finished tableau (e.g. systematic) τ from a theory T with the root entry F ϕ is contradictory. If not, then there is som ...
... Theorem For every theory T and sentence ϕ, if ϕ is valid in T , then ϕ is tableau provable from T , i.e. T |= ϕ ⇒ T ` ϕ. Proof Let ϕ be valid in T . We will show that an arbitrary finished tableau (e.g. systematic) τ from a theory T with the root entry F ϕ is contradictory. If not, then there is som ...