Predicate Logic as Working Language

Transcription

1 full.nb 1 Predicate Logic as Working Language Bruno Buchberger Research Institute for Symbolic Computation Johannes Kepler University, Linz, Austria Version: : This manuscript will be published as a book. No part of it may be copied or stored without the written permission of the author. The file may be used by the students of the course "TSW" at the Johannes Kepler University, Linz, Austria, winter semester 2006 for their personal training. However, copying the file, inclusion of parts of the file into other files, and subsequent distribution of the results of such actions is strictly forbidden. If there are any questions on the usage of this file, write directly to buchberger@risc.uni linz.ac.at.

2 full.nb 2 Chapter 4: Eploration Using Quantifier Predicate Logic Quantifier Rules for Proving Eistentially Quantified Formulae in the Knowledge Base Rule ("Skolem Constants") If F Ξ,Η,... is in the knowledge base (where Ξ,Η,... are the only free variables in F) then you can add F Ξ,Η, 0,y0,... to the knowledge base, where 0, y0,... are new object constants (i.e. constants that do neither occur in the knowledge base nor in the goal formula). (One calls these constants "Skolem constants".) Wording Application of this rule is often announced in the following way. "We know F. Ξ,Η,... Therefore we can chose 0, y0,... such that F Ξ,Η, 0,y0,.... Rule ("Skolem Functions") If F Α,Β,... Ξ,Η,...

3 full.nb 3 is in the knowledge base (where Α,Β,Ξ,Η,... are the only free variables in F) then you can add F Ξ,Η, 0 Α,Β,...,,y0 Α,Β,...,... to the knowledge base, where 0, y0,... are new function constants (i.e. function constants that do neither occur in the knowledge base nor in the goal formula). (One calls these constants "Skolem function constants".) Wording Application of this rule is often announced in the following way. "We know F Α,Β,... Ξ,Η,... Therefore we can chose functions 0, y0,... such that F Ξ,Η, 0 Α,Β,...,,y0 Α,Β,...,... Eistentially Quantified Formulae as Proof Goals Rule ("Find Appropriate Terms") If the proof goal is F Ξ,Η,... (where Ξ,Η,... are the only free variables in F) then it suffices to find terms s, t,... such that F Ξ,Η, s,t,... can be proved. Universally Quantified Formulae in the Knowledge Base The appropriate rule for formulae of the form F Ξ,Η,... in the knowledge base was already given in Chapter 3.

4 full.nb 4 Universally Quantified Formulae as Proof Goals The appropriate rule for proof goals of the form F Ξ,Η,... was already given in Chapter 3. The Interplay Between the Quantifier Rules The interplay between the proof rules for universally and eistentially quantified formulae in the knowledge base and as proof goals is the most important and characteristic aspect of full predicate logic. Typically, one has proof goals of the form Α Ξ F or, as special case, proof goals of the form Ξ F and formulae in the knowledge base of the form Β Η G 1 a or the special form Η G 1 b In such situations one normally proceeds by taking "an a0 arbitrary but fied" and tries to find a term s such that F Α a0,ξ s. The term s is then typically constructed from the constants available in the knowledge base and, in particular, from the Skolem constant y0 that can be introduced because of (1a) or (1b), i.e. for which F Η y0 Β or F Η y0 holds.

5 full.nb 5 Eamples The Notion of Limit Definition In the sequel, some of the variables, like, range over the real numbers, some others, like n, over the natural numbers, and some others, like f, over sequences of real numbers (i.e. over functions with natural number input and real number output). Now we define limit f, a f n a Ε, Ε 0 N n N f g n f n g n. definition of limit definition of sequence sum Available Knowledge We assume that we have "all" knowledge available on the arithmetical operations, like +,, <, etc., for eample y y. Proposition We want to prove that limit f, a limit g, b limit f g, a b. limit of sum Proof Let f, g, a, b be arbitrary but fied and assume limit f, a, limit g, b. Af Ag We have to prove limit f g, a b. G By (definition of limit), we have to prove f g n a b Ε. Ε 0 N n N For this, we take Ε arbitrary but fied, assume

6 full.nb 6 Ε 0, AΕ and have to find an N0 such that f g n a b Ε. GN0 n N0 Now, by (Af) and (Ag) and (definition of limit), we know that f n a Ε, Ε 0 N n N g n b Ε, Ε 0 N n N and, hence, by (AΕ) and (arithmetic), we know in particular that N N f n a Ε 2, n N g n b Ε 2. n N Hence, we can choose two natural numbers Nf and Ng such that f n a Ε 2, n Nf g n b Ε 2, n Ng ANf ANg We now take N0 : ma Nf, Ng definition of N0 and try to prove (GN0). In fact, for arbitrary but fied n with n N0 An we have f g n a b by definition of sequence sum f n g n a b by arithmetic f n a g n b by arithmetic f n a g n b by An, definition of N0, ANf, ANg, arithmetic Ε 2 Ε 2 by arithmetic

7 full.nb 7 Ε. Annotated Proof We now present the same proof with additional annotations and details (in red color) eplaining the proof techniques applied. Let f0, g0, a0, b0 be arbitrary but fied and assume limit f0, a0, limit g0, b0. Af Ag We have to prove limit f0 g0, a0 b0. G (Note that f0, etc. are new constants and must not be confused with the variables appearing in the (defintion of limit) etc.) By (definition of limit), using the substitutions we have to prove f f0 g0, a a0 b0, f0 g0 n a0 b0 Ε. Ε 0 N n N For this, we take Ε0 arbitrary but fied, assume Ε0 0, AΕ and have to find an N0 such that f0 g0 n a0 b0 Ε0. n N0 GN0 Now, by (Af) and (Ag) and (definition of limit), using the substitutions we know that f f0, a a0 f0 n a0 Ε, Ε 0 N n N and using the substitution we know that f g0, a b0 g0 n b0 Ε, Ε 0 N n N and, hence, by (AΕ) and (arithmetic), we know in particular that

8 full.nb 8 N N f0 n a0 Ε0 2, n N g0 n b0 Ε0 2. n N (The (arithmetic) used here infers Ε from (AΕ). Now f0 n a0 Ε, Ε 0 N n N which is an abbreviation for implies Ε Ε 0 N f0 n a0 Ε, n N Ε0 2 0 N f0 n a0 Ε0 2. n N 2 Now, from (1) and (2), by modus ponens, we obtain ) N f0 n a0 Ε0 2 n N Hence, we can choose two natural numbers Nf and Ng such that f0 n a0 Ε0 2, n Nf g0 n b0 Ε0 2. n Ng ANf ANg We now take N0 : ma Nf0, Ng0 definition of N0 and try to prove (GN0). In fact, for arbitrary but fied n0 with n0 N0 An we have f0 g0 n0 a0 b0 by definition of sequence sum f0 n0 g0 n0 a0 b0 by arithmetic f0 n0 a0 g0 n0 b0

9 full.nb 9 by arithmetic f0 n0 a0 g0 n0 b0 by An, definition of N0, ANf, ANg, arithmetic Ε0 2 Ε0 2 by arithmetic Ε0. (Each of the steps in the above sequence is a symbolic computation proof step. For eample f0 g0 n a0 b0 f n g n a b because, by (definition of sequence sum) using the substitutions f f0, g g0, n n0 we have f0 g0 n0 f0 n0 g0 n0 and, hence, by replacing f0 g0 n0 by f0 n0 g0 n0 in f0 g0 n a0 b0 we obtain f0 g0 n a0 b0 f n g n a b. ) (From the above chain of proof steps, we can conclude f0 g0 n0 a0 b0 Ε in the following way: First, f0 g0 n0 a0 b0 f0 n0 g0 n0 a0 b0 and f0 n0 g0 n0 a0 b0 f0 n0 a0 g0 n0 b0 implies

10 full.nb 10 f0 g0 n0 a0 b0 f0 n0 a0 g0 n0 b0 by the transitivity rule for equality. Second, f0 g0 n0 a0 b0 f0 n0 a0 g0 n0 b0 and f0 n0 a0 g0 n0 b0 f0 n0 a0 g0 n0 b0 implies f0 g0 n0 a0 b0 f0 n0 a0 g0 n0 b0. This can be concluded from the general proposition y y z z by appropriate substitutions and modus ponens. Eercise: Prove the above proposition. Which laws for do you need?) Equivalences and Partitions Overview We will start from the basic notions of set theory like,, etc. and assume that their elementary properties are known (i.e. available in the "knowledge base"). In addition, we will assume that the inference rules for the set quantifiers are known: The set quantifiers allow to construct terms of the following form: and, where is a variable, is a formula, and is a term. We anticipate here the chapter on set theory. We will define two important notions of elementary set theory equivalences and partitions and eplore their interrelation. In the sequel, the variables R, X, P, p, etc. range over arbitrary objects ("sets").

11 full.nb 11 Inference Rules for the Set Quantifiers First note that is an abbreviation for y y T. If one has to prove that S, where S is a term, it suffices to prove that S. Conversely, if S is in the knowledge base, one can also add S to the knowledge base. Hence, in order to prove S one has to find a term U such that Conversely, if S T U U. S is in the knowledge base, one can introduce a new constant Η and add S T Η Η to the knowledge base. Available Knowledge We just give the definitions or aioms for some basic notions of set theory. In addition, we assume that all elementary properties of these notions are known.

12 full.nb 12 A B A B set equality a, b u, v a u b v pair equality A B A B A B A B A B A B U A a A a : : : : U : A B a, b a A b B : a,b Here is an eample of an elementary property that follows from these definitions and aioms: A A. When we use formulae in this knowledge base, we just reference it by the label (set theory). Definition: Equivalences is relation R, X is refleive R, X is equivalence R, X is symmetric R is transitive R is relation R, X R X X is refleive R, X X, R is symmetric R,y, y R y, R is equivalence : is relation : is refleive : is symmetric : is transitive R,y,z, y R y, z R, z R is transitive : Definition: Partitions is subset set P, X are all nonempty P is partition P, X are all disjoint P covers P, X is subset set P, X p X p P are all nonempty P p P p is partition : is subset set : are all nonempty : are all disjoint P p,q P p q p q are all disjoint : covers P, X U P X covers :

13 full.nb 13 Definition: Sets Determined by a Relation sets of relation R, X class, R, X X sets of relation : class, R, X y X y, R y class : Definition: Relation Determined by Sets relation of sets P, X, y,y X p y p relation of sets : p P Theorem We want to prove that is equivalence R, X is partition sets of relation R, X, X partition from equivalence is partition P, X is equivalence relation of sets P, X, X equivalence from partition is equivalence R, X relation of sets sets of relation R, X, X R equivalence of partion of equivalence is partition P, X sets of relation relation of sets P, X, X P partition of equivalence of partition Proof of (partition from equivalence) We take R and X arbitrary but fied, assume and show is equivalence R, X is partition sets of relation R, X, X. For proving this, by (is partition:), we have to prove is subset set P, X, are all nonempty P, are all disjoint P, covers P, X. SS NE DJ CV We only show the proof of (DJ). For the other proofs, see the eercises. Proof of (DJ) For proving (DJ), by (are all disjoint:), we have to prove

14 full.nb 14 p q p q. p,q sets of relation R,X For this, we take p, q arbitrary but fied, assume p sets of relation R, X, q sets of relation R, X, p q, p q NE and show p q. For this, by (set theory), it suffices to prove that p q. For proving this, we assume that Ξ is such that Ξ p, Ξ q, and show a contradiction. In fact, we show that p q. By (set theory), for this it is sufficient to show that p q. For this, we take Η arbitrary but fied, assume and show Η p Η q, and we assume and show Η q Η p. We only show one direction. The other direction is analogous. Now, from (p ) and (q ), by (sets of relation:), we know that p class, R, X X, q class, R, X X.

15 full.nb 15 Because of this we can take Α and Β such that p class Α, R, X, q class Β, R, X. Now, from this and from Ξ p, by (class:), we know that Ξ,Α R, Ξ,Β R, Η,Α R. Hence, from is equivalence[r,x], Η,Β R, i.e., by (class:), i.e. Η class Β, R, X, Η q. Proof of (partition from equivalence) with Comments on the Proof Techniques We take R and X arbitrary but fied, assume and show is equivalence R, X is partition sets of relation R, X, X. (Note that R and X are now considered to be "new" constants.) For proving this, by (is partition:), we have to prove is subset set P, X, are all nonempty P, are all disjoint P, covers P, X. SS NE DJ CV We only show the proof of (DJ). For the other proofs, see the eercises. Proof of (DJ) For proving (DJ), by (are all disjoint:), we have to prove p q p q. p,q sets of relation R,X

16 full.nb 16 For this, we take p, q arbitrary but fied, assume p sets of relation R, X, q sets of relation R, X, p q, p q NE and show p q. For this, by (set theory), it suffices to prove that p q. (For proving... we assume... and show a contradiction. By the assumption it Ξ and assume that... holds for Ξ.)..., we are allowed to take an object, call For proving this, we assume that Ξ is such that Ξ p, Ξ q, and show a contradiction. In fact, we show that p q. By (set theory), for this it is sufficient to show that p q. For this, we take Η arbitrary but fied, assume and show Η p Η q, and we assume and show Η q Η p. We only show one direction. The other direction is analogous. Now, from (p ) and (q ), by (sets of relation:), we know that p class, R, X X,

17 full.nb 17 q class, R, X X. (p class, R, X X means that p t t class, R, X X and this means that p has the property characterizing this set, i.e. p class, R, X X. Hence, we are allowed to take an object, call it Α and assume that p class Α, R, X, Α X. ) Because of this we can take Α and Β such that p class Α, R, X, q class Β, R, X. Now, from this and from Ξ p, by (class:), we know that Ξ,Α R, Ξ,Β R, Η,Α R. Hence, from is equivalence[r,x], Η,Β R, i.e., by (class:), Η class Β, R, X, i.e. Η q. (Here, we may further detail the steps that lead from Ξ,Α R, Ξ,Β R, Η,Α R, to

18 full.nb 18 In fact, from Η,Β R. is equivalence R, X, by (is equivalence:), we know that is symmetric R, X, is transitive R, X, and, hence, by (is symmetric:) and (is transitive:),, y R y, R 1,y, y R y, z R, z R. 2,y,z Now, from Ξ,Α R, by (1), Α,Ξ R. Now, from Η,Α R and Α,Ξ R, by (2), Η,Ξ R. And now, from Η,Ξ R and Ξ,Β R, again by (2), ) Η,Β R. Structuring Proofs Unfolding Definitions As shown in the above eamples, proofs proceed by "unfolding" the definitions of notions occuring in the goal formula and in the formulae in the knowledge base. In many proofs, hardly any difficult idea is needed ecept unfolding the definitions and playing with the constants introduced in the various stages of the proofs for combining suitable terms that fulfill the conditions specified in eistentially quantified formulae. However, proofs of theorems may become quite long. Also, one uses to reference long formulae occurring in proofs by labels. Thus, long proofs may become hard to read because one tends to lose the overview and it is sometimes quite annoying to jump back and forth a couple of pages for finding the formulae referenced in a given proof step. Also, understanding proofs has two very different aspects both of which are equally important: understanding the main idea of a proof independent of formal details, being able to check the correctness of each step of the proof and the completeness of the proof.

19 full.nb 19 Therefore it is very important to develop techniques for structuring proofs with the goal to invent and understand proofs more easily. We discuss three techniques for structuring proofs: develop the key ideas of proofs in eamples, in particular graphic illustrations, theory eploration instead of isolated theorem proving, introducing only one quantifier at a time, "backward proof presentation" versus "forward proof presentation".