WinKE A Pedagogic Tool for Teaching Logic and Reasoning.

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Presentation transcript:

WinKE A Pedagogic Tool for Teaching Logic and Reasoning

WinKE displays a simple KE proof for the equivalence of the two formulas P ® Q and ØQ ® ØP.

The WinKE system features an interactive interface via menus, dialogues, and graphic tools, on- and off-line proof-checking, and is completed by the option to automatically perform proofs (or parts of a proof), and to build up countermodels. User support is provided by bookkeeping facilities, hints, various "undo" utilities, and a detailed on-line help system. The system provides several files with example problems. New ones can be edited directly within WinKE. The interface consists of four windows: the main window providing menus and buttons for quick access to the most basic functions, the graphic window to display and manipulate KE proof trees, a viewer which displays a scaled-down view of the virtual drawing board, allowing to focus on a particular portion of it, and a tool box containing the graphic tools. The design of the interface is close to that of Windows standard software, which makes it very easy to learn how to use the system. WinKE is supportive of an introductory textbook on classical logic (M. D'Agostino, M. Mondadori, Logica, Edizioni Bruno Mondadori, 1997), but may also be used independently. University of Ferrara Imperial College London WinKE is a new interactive theorem proving assistant based on the KE calculus, a refutation system which combines features from Smullyan's analytic tableaux and Gentzen's natural deduction. It has been developed to support teaching logic and reasoning to undergraduate students. The WinKE process of constructing a proof tree is as faithful as possible to the pen-and-paper procedure. Running under Windows 95, it is easy to use and visually satisfying.

Proofs & Countermodels Problems Trees Show that Peirce's law ((P ® Q) ® P) ® P is a tautology. Show that the following set of formulas is inconsistent: P  Q, ØP  Q, P  ØQ, ØP  ØQ. Is Socrates mortal? Rules P  Q ——— P Q (P  Q) ———— P Q (P  Q) ————— P Q P ——— P Alpha & double negation P  Q P ———— Q (P  Q) P ———— Q P  Q P ——— Q P  Q Q ———— P Beta P  Q P ——— Q P  Q P ———— Q (P  Q) P ————— Q (P  Q) P ————— Q Eta  x : P ——— P[x/t]  x : P ———— P[x/t] Gamma for any closed term t  x : P ——— P[x/c]  x : P ———— P[x/c] Delta for a new constant c ———— P  P Principle of Bivalence (PB) Proofs & Countermodels Every country is either red, green, or blue. Countries in the neighborhood of green countries are blue. Countries in the neighborhood of blue countries are red. Show that no country is green. If there are two complementary formulas on a branch, it is said to be closed. A theorem is proven to hold, if the tree generated by its negation is closed. Branches that cannot be closed represent countermodels.

http://dns.unife.it/dgm/WinKE/ Understanding Logic Textbook Lectures WinKE Help x: y: R(x,y)  R(y,x) Marco Mondadori & Marcello D’Agostino, Logica, Edizioni Bruno Mondadori, 1997. For more information visit our web-page: http://dns.unife.it/dgm/WinKE/ Ulle Endriß August 1998

An Example Create a new problem in WinKE Imagine you're in Transylvania and you're meeting a suspicious-looking passenger. You ask him what he knows about Dracula. He answers: "Everybody is afraid of Dracula and I am the only person Dracula is afraid of." What can you conclude from this statement? Transform the problem into First Order Logic " x : IsAfraidOf(x,Dracula) " x : IsAfraidOf(Dracula,x) ® SamePerson(x,Passenger) Building up countermodels is very easy in KE. All you have to do is to collect the literals on a saturated branch. automated deduction Apply KE rules to expand the proof tree Countermodel IsAfraidOf(Dracula,Dracula) SamePerson(Dracula,Passenger) g applied to no. 1 manual deduction g applied to no. 2 This means: Dracula and the passenger are the same person! b applied to no. 4 & 3 The branch is saturated, i.e. there are no further possible proof steps (that could lead to a closure). Therefore, a countermodel exists.