Analogies between Proofs { A Case Study. Erica Melis. Universitat Saarbrucken. Fachbereich Informatik Saarbrucken.

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1 Analogies between Proofs { A Case Study Erica Melis Universitat Saarbrucken Fachbereich Informatik 6600 Saarbrucken melis@cs.uni-sb.de This case study examines in detail the theorems and proofs that are shown by analogy in a mathematical textbook on semigroups and automata, that is widely used as an undergraduate textbook in theoretical computer science at German universities (P. Deussen, Halbgruppen und Automaten, Springer 1971). The study shows the important r^ole of restructuring a proof for nding analogous subproofs, and of reformulating a proof for the analogical transformation. It also emphasizes the importance of the relevant assumptions of a known proof, i.e., of those assumptions actually used in the proof. In this document we show the theorems, the proof structure, the subproblems and the proofs of subproblems and their analogues with the purpose to provide an empirical test set of cases for automated analogy-driven theorem proving. Theorems and their proofs are given in natural language augmented by the usual set of mathematical symbols in the studied textbook. As a rst step we encode the theorems in logic and show the actual restructuring. Secondly, we code the proofs in a Natural Deduction calculus such that a formal analysis This work was supported by a research scholarship of the Deutsche Forschungsgemeinschaft (DFG) 0

2 becomes possible and mention reformulations that are necessary in order to reveal the analogy. 1

3 Introduction Justied analogical reasoning proceeds by transferring an aspect from the base case s to a target case t based on the similarity of these cases with respect to a second aspect. The second and the rst aspect have to be inherently connected. For example, analogical reasoning takes as input the similarity of s and t with respect to their function and the connection between function and structure. Then it yields the commonality of s and t with respect to their structure. Within the context of (automated) theorem proving, problems and proofs are usually inherently connected, in the sense that a basic heuristic assumption stipulates that analogous theorems can be proved analogously also. This assumption is true in many cases. For analogical theorem proving the rst aspect of a connection is the pair (ass; thm) which we call problem, that consists of the set of relevant assumptions ass and of the theorem thm, and the second aspect is the proof of the theorem thm from the assumptions ass. To obtain an empirical test set and in order to gain practical experience with analogical reasoning in mathematical theorem proving we have studied the textbook \Halbgruppen und Automaten" (abbreviated as HUA in the following) [1], since it is particularly rich in proofs that are explicitly stated as analogous to previous proofs by the author. Furthermore the book already served as a test case for automated theorem proving for the Margraf Karl Refutation Procedure [2]. This empirical study is the basis for our own approach to analogy-driven theorem proving that is presented in detail in [5]. This approach is inherently based on the reformulation of the base problem together with the base proof, and on the reformulation of the target problem. The aim of the reformulation is to make the representation of the base and the target problem compatible such that the essential analogy is revealed, carrying over certain reformulated parts of the base proof as parts of a hypothetical target proof. The following study presents all theorems in HUA that are explicitly marked as analogous by the author. Theorems are rst given in English (our 2

4 translation) and coded in predicate logic. The proofs are then coded in a Natural Deduction format, such that a formal analysis becomes possible. The main nding is that a problem P 2 is called analogous to a problem P 1 in the textbook, if P 1 can be reformulated to a problem equal to P 2, or P 1 and P 2 can be reformulated to a common abstraction. Actually, P 2 is often called analogous to P 1 even if only an important subproblem of P 2 is analogous to a subproblem of P 1. Hence the standard approach to automated theorem proving by analogy (e.g., [6]), which is mainly based on symbol mapping of the base case to the target case for a given representation fails in many cases: There is no such symbol map unless the actually given representation is reformulated such that the analogy becomes visible. Why \Halbgruppen und Automaten"? The textbook \Halbgruppen und Automaten" [1] was chosen for this case study since it consists of the three chapters each of which is built upon the previous one, partially by analogies. This is the reason, why this particular textbook is so rich in explicit proofs by analogy and actually very much liked by students because of its uniform structure. The actual chapters are: Semigroups and relations Semigroups and semimoduls Automata. Notation The study is based on Natural Deduction (ND) proofs, since it turned out to be most natural to code the proofs given within the textbook in a proof calculus, whose (primitive) rules are presented in [4] which in turn is based on [3]. The displayed ND-proof lines do not always correspond to primitive rules but can easily splitted into several lines that correspond to primitive ND-rules. The reason is to keep the proofs more readable. In the following we quote the theorems by their original decimal numbering from HUA, for example, Theorem refers to Satz on page 182 in HUA. Sometimes a theorem is not explicitly stated in the textbook but just mentioned as analogous to some previous theorem, for example the existence of certain homomorphisms in semimoduls is directly carried over (i.e. 3

5 is analogous) from the existence of homomophisms in semigroups. Then, for instance, the theorem mentioned as analogous to theorem 5.7 in section 7 of HUA is denoted as As another notational convention, we denote part n of theorem m by m:n. The ND-proofs contain parts that are called relevant assumptions, and these may correspond to applications of the ND-rule called HYP (hypothesis introduction). The relevant assumptions are those hypotheses which cannot be omitted in the proof. As a further renement, the origin of the HYPrule is replaced by the name of the assumption that was introduced by the HYP-rule. For example, ASS means that the formula is an assumption of the problem, DEF points to a denition, and AX means that the hypothesis is an axiom. 4

6 Analogous Theorems and Proofs in HUA The following theorems are marked by the author of HUA to be shown by analogy: Theorem 6.3. (Chapter II) is analogous to theorem 3.3. (Ch.I). Theorem 6.6. (Ch.II) is analogous to theorem 3.6. (Ch.I). Theorem (Ch.II) is analogous to theorem 5.2. (Ch.I). Theorems 4.10, 4.11, and 4.12 (Ch.I) for sets can be taken over for the corresponding (sub-)theorems of 5.6, 5.7, and 5.8 for semigroups. Theorems 5.6, 5.7, and 5.8 (Ch.I) for semigroups are supposed to be carried over analogously to the corresponding theorems 7.5.6, 7.5.7, and for semimoduls (in Ch.II). Theorem 5.3 (Ch.I) is analogous to theorem 4.8 (Ch.I). Theorem is analogous to 9.8 (Ch.II). Theorem 13.7 (Ch.III) is analogous to theorem 6.9 (Ch.II). Theorem (Ch.III) is analogous to theorem 17.6 in the same section (Ch.III). Theorem 17.9 part 2 (Ch.III) is analogous to 17.9 part 1 (Ch.III). Two subproofs of theorem 17.6 are analogous. The more interesting analogies are examined in the following. 5

7 CASE 1: THEOREMS 3.3 and 6.3 The proof of theorem 6.3 is analogous to the proof of theorem 3.3 in HUA, where the respective theorems are given as: Theorem 3.3 Let ft i : i 2 Ig be a family of leftideals in the semigroup F. 1. Then S i T i is a leftideal in F. 2. If T i T i is not empty then T i T i is a leftideal in F. Theorem 6.3 Let ft i : i 2 Ig be a family of F -subsemimoduls in the F -semimodul S. 1. Then S i T i is an F -subsemimodul in S. 2. If T i T i is not empty then T i T i is an F -subsemimodul in S. The analogy of theorem 3.3 and theorem 6.3 is based on the correspondence between the denitions of a leftideal and of a subsemimodul (denition 3.1 and denition 6.2 in HUA) which are given as: Denition 3.1 A nonempty subset T of a semigroup F is called leftideal if F T T, where F T = fft : f 2 F; t 2 T g. Denition 6.2 A nonempty subset T of an F -semimodul S is called F-subsemimodul if F T T, where F T = fft : f 2 F; t 2 T g. Consider the analogy between the subproof of and the subproof of 6.3.1: Corresponding to the denition of a leftideal, it is shown for that S i T i is nonempty, S i T i is a subset of F, and F S i T i S i T i. Thus splitting the theorem into its conjunctive subparts, a straightforward proof structure of is the following: Part 1: Theorem: S i T i is nonempty, i.e. (expanding the denition of 'nonempty'): 9x(x 2 S i T i ). Relevant assumptions: 8i(i 2 I! 9x(x 2 T i )); the denition of S i T i. Part 2: Theorem: S i T i F. Relevant assumptions: the denition of S ; the denition of ; the assumption 6

8 8i(i 2 I! (T i F )). This part is demonstrated in more detail below. Part3: Theorem: F S i T i S i T i, e.g.(after expanding the denition of F T and of ): 8i; x; f(i 2 I ^ x 2 T i ^ f 2 F! f x 2 T i ) Relevant assumptions: the denition of F T ; the denition of ; the denition of S ; the assumption 8i(i 2 I! F T i T i ). Given the denition of an F -subsemimodul, it has to be shown in a proof of that S i T i is nonempty, S i T i is a subset of S, and F S i T i S i T i. Splitting the theorem of into its conjunctive subparts, a straightforward proof structure of becomes: Part 1: Theorem: S i T i is nonempty, i.e.(expanding the denition of `nonemptyness'): 9x(x 2 S i T i ). Relevant assumptions: 8i(i 2 I! 9x(x 2 T i )); the denition of S i T i. Part 2: Theorem: S i T i S. Relevant assumptions of the completed proof: the denition of S ; the denition of ; the assumption 8i(i 2 I! (T i S)). Part3: Theorem: F S i T i S i T i, i.e.(after expanding the denition of F T and of ): 8i; x; f(i 2 I ^ x 2 T i ^ f 2 F! f x 2 T i ) Relevant assumptions: the denition of F T ; the denition of ; the denition of S ; the assumption 8i(i 2 I! F T i T i ). We shall now give the explicit ND proofs of theorem part 2 and of theorem part 2. The rst proof is a translation of the given natural language proof in HUA into an ND-calculus, while the second proof is an analogical reconstruction (it is not given in the textbook but just mentioned as \to be shown analogously"). 7

9 ND proof for theorem part 2. NNo S;D Formula Reason relevant assumptions 1. ; 1 ` 8I; x(x 2 i2i T i $ 9i(i 2 I ^ x 2 T i )) S (DEF ) 2. ; 2 ` 8M; N (M N $ 8x(x 2 M! x 2 N )) (DEF ) 3. ; 3 ` 8i(i 2 I! T i F ) (ASS) S the proof 4. 4; ` t 2 i T i (HYP) 5. 4; 1 ` t 2 i T i $ 9i(i 2 I ^ t 2 T i ) (8D 1) 6. 4; 1 ` t 2 i T i! 9i(i 2 I ^ t 2 T i ) ($ D 5) 7. 4; 1 ` 9i(i 2 I ^ t 2 T i ) (! D 6) 8. 8; ` i 0 2 I ^ t 2 T i0 (HYP) 9. 8; ` t 2 T i0 (^D) 10. 8; ` i 0 2 I (^D) 11. ; 3 ` i 0 2 I! T i0 F (8D) 12. 8; 3 ` T i0 F (! D 10 11) 13. 8; 1, 2, 3 ` t 2 F (8D; $ D;! D ) 14. 4; 1, 2, 3 ` t 2 F (Choice 7 8) 15. ; 1, 2, 3 ` t 2 i TS i! t 2 F (DED 14) 16. ; 1, 2, 3 ` 8x(x 2 i T i! x 2 F ) (8I) 17. ; 2, 3 ` Si T i F ($ D;! D Thm. ; ` Si T i F 16 2) () ND proof for theorem part 2. NNo S;D Formula Reason relevant assumptions 1. ; 1 ` 8I; x(x 2 i2i T i $ 9i(i 2 I ^ x 2 T i )) S (DEF ) 2. ; 2 ` 8M; N (M N $ 8x(x 2 M! x 2 N )) (DEF ) 3. ; 3 ` 8i(i 2 I! T i S) (ASS) S the proof 4. 4; ` t 2 i T i (HYP) 5. 4; 1 ` t 2 i T i $ 9i(i 2 I ^ t 2 T i ) (8D 1) 6. 4; 1 ` t 2 i T i! 9i(i 2 I ^ t 2 T i ) ($ D 5) 7. 4; 1 ` 9i(i 2 I ^ t 2 T i ) (! D 6) 8. 8; ` i 0 2 I ^ t 2 T i0 (HYP) 9. 8; ` t 2 T i0 (^D) 10. 8; ` i 0 2 I (^D) 8

10 11. ; 3 ` i 0 2 I! T i0 S (8D) 12. 8; 3 ` T i0 S (! D 10 11) 13. 8; 2, 3 ` t 2 S (8D; $ D;! D ) 14. 4; 1, 2, 3 ` t 2 (Choice 7 8) 15. ; 1, 2, 3 ` t 2 i TS i! t 2 S (DED 14) 16. ; 1, 2, 3 ` 8x(x 2 i T i! x 2 S) (8I) 17. ; 1, 2, 3 ` Si T i S ($ D;! D Thm. ; ` Si T i S 16 2) () Discussion Conjunctive goal splitting yields the same proof structure for theorem and theorem Expanding the respective denitions then yields the same subtheorems and relevant assumptions for parts 1 and 3, respectively. For these parts the proofs are equal as well, due to the commonality of the subtheorems and of the relevant assumptions. For part 2 the subtheorems in and are not equal right away, however the dierence of the subtheorems and can be removed by replacing the constant F by the constant S. The symbol mapping F ) S applied to theorem 3.3 in order to obtain theorem 6.3 can be extended by matching the assumptions as well. After unfolding the denitions, the proof of theorem 3.3 contains only parts of the denitions 3.1 and 6.2 that correspond directly. These are called assumptions relevant in the base proof. The proof does not use those parts of the denition that actually dier, such as an ideal being contained in a semigroup and a subsemimodul being contained in a semimodul. Since the symbol mapping ff ) Sg is applied to the second subproblem of only, the mapping is consistent (one symbol is mapped to one symbol). The mapped versions of all assumptions relevant in the base proof all occur in the knowledge base or in the assumptions of the target problem 6.3, and this serves as a strong justication for this analogy formation. This example can be dealt with by standard techniques known from the literature on theorem proving by analogy, provided that means for structuring proofs and isolating relevant assumptions are present. 9

11 CASE 2: THEOREM 17.6 and ITS ANALOGUE Theorem 17.6 Let E F, then 1. E is a leftcongruence in the semigroup F, 2. E is compatible with E, 3. For all leftcongruences in F, which are compatible with, we have E, where E is dened in denition 17.5 (see below). Theorem analogue Let E F, then 1. E is a rightcongruence in the semigroup F, 2. E is compatible with E, 3. For all rightcongruences in F, which are compatible with, we have E, where E is dened in a denition analogous to 17.5 (see below). The following denitions are relevant assumptions: Deniton 17.4 Let be an equivalence relation on F and E F. is called compatible with E i for all f 2 F with (f) \ E 6= ; holds (f) E, where (x) = fy : (y; x) 2 g. Denition 5.1 Let be an equivalence relation on a semigroup F. Then is called a leftcongruence i for all g; f 1 ; f 2 2 F holds if (f 1 ; f 2 ) then (gf 1 ; gf 2 ). is called a rightcongruence i for all g; f 1 ; f 2 2 F holds if (f 1 ; f 2 ) then (f 1 g; f 2 g). The particular leftcongruence E in F is dened for E F as. Denition (f; g) 2 E $ ((f 2 E $ g 2 E)! 8h(h 2 F! (hf 2 E $ hg 2 E))). The particular rightcongruence E in F is dened for E F by the following analogous denition. Analogue to denition (f; g) 2 E $ ((f 2 E $ g 2 E)! 8h(h 2 F! (fh 2 E $ gh 2 E))). The proof structure of 17.6 is: 10

12 Part 1: Theorem: E is a leftcongruence in F. Relevant assumptions: the denition of a leftcongruence, the denition of a semigroup, the denition Part 2: Theorem: E is compatible with E. Relevant assumptions: the denition of [; the denition of ; denition 17.4; and denition Part 3: Theorem: leftcongruence() ^ compatible(; E)! E Relevant assumptions: the denition 17.4; the denition of leftcongruence; and the denition of. The proof structure of the analogue of 17.6 is: Part 1: Theorem: E is a rightcongruence in F Relevant assumptions: the denition of a rightcongruence; the denition of a semigroup; the analogue of denition Part 2: Theorem: E is compatible with E. Relevant assumptions: the denition of [; the denition of ; the analogue of denition 17.4; and the analogue of denition Part 3: Theorem: rightcongruence() ^ compatible(; E)! E. Relevant assumptions: the analogue of denition 17.4; the denition of rightcongruence; and the denition of. Discussion The symbol mapping fleftcongruence ) rightcongruence; E ) E g makes the subproblems of 17.6 and those of its analogous theorem equal but in this case the corresponding proofs still dier. This is due to the use of the different denitions of lef tcongruence and rightcongruence within the proofs which belong to the relevant assumptions. 11

13 The denition 17.5 can be transformed to the analogous one by term mapping (i.e. not just symbol mapping, as in the previous example). There are two possibilities for the term mapping that transform the assumptions of 17.6 into the assumptions of its analogue: the concrete term mapping: hf ) fh; hg ) gh; kf ) fk; kg ) gk; hkf ) fhk; hkg ) ghk for constants and variables h; f; g; k or, the term mapping based on the schema term 1 term 2 ) term 2 term 1 which could be used as well. The occurrence of the mapped versions of all relevant assumptions of the base proof in the knowledge base or in the assumptions of the analogue of problem serves as a justication for this analogy formation. This example could also be treated by techniques known from the literature, provided that means for isolating relevant assumptions are used in addition. 12

14 CASE 3: TWO ANALOGOUS PARTS OF This example demonstrates a kind of analogy which is used very often in mathematics. It is shown on the third part of theorem 17.6 of HUA. Theorem Let E F and let be a leftcongruence in F which is compatible with E, then E. The denitions of compatible(; E); leftcongruence(); E, and (x) are relevant assumptions. They have been given in the previous paragraph. The problem is ` E with = flef tcongruence(); compatible(; E); (E F ); semigroup(f )g. Some preparatory steps, usually not expressed explicitly by mathematicians, are necessary for the full ND-proof: 1. Expanding the denition of yields the problem ` 8x; y((x; y) 2! (x; y) 2 E ). 2. Expanding the denition of E yields the problem ` 8x; y((x; y) 2! (x 2 E $ y 2 E) ^ 8f(f 2 F! fx 2 E $ fy 2 E)). 3. Two applications of the Deduction Theorem yield the problem [ f(x 0 ; y 0 ) 2 ; (x 0 2 E $ y 0 2 E)g ` 8f(f 2 F! fx 0 2 E $ fy 0 2 E). 4. Restructuring (splitting) yields the subproblems [ f(x 0 ; y 0 ) 2 ; (x 0 2 E $ y 0 2 E)g ` 8f (f 2 F! f x 0 2 E! f y 0 2 E) [ f(x 0 ; y 0 ) 2 ; (x 0 2 E $ y 0 2 E)g ` 8f (f 2 F! f y 0 2 E! f x 0 2 E). 5. Application of the Deduction Theorem yields the subproblems (a) [ f(x 0 ; y 0 ) 2 ; (x 0 2 E $ y 0 2 E)g [ ff 0 2 F; f 0 x 0 2 Eg ` f 0 y 0 2 E (b) [ f(x 0 ; y 0 ) 2 ; (x 0 2 E $ y 0 2 E)g [ ff 0 2 F; f 0 y 0 2 Eg ` f 0 x 0 2 E. 13

15 Thus we have obtained the subproblems (a) and (b) which are supposed to be proved analogously in HUA. We present the two ND-proofs in the following and discuss the respective transformation afterwards. 14

16 ND Proof of theorem part (a) NNo S;D Formula Reason assumptions 1. ; 1 ` 8R; x; y8f(leftcongr(r) $ ((x; y) 2 R ^ f 2 F! (DEF) (fx; fy) 2 R)) 2. ; 2 ` leftcongr() (ASS) 3. ; 3 ` 8x; y(x 2 (y) $ (x; y) 2 ) (DEF) 4. ; 4 ` 8M; N : set; 8x(x 2 (M \ N ) $ x 2 M ^ x 2 N ) (DEF) 5. ; 5 ` 8M : set(nonempty(m ) $ 9x(x 2 M )) (DEF) 6. ; 6 ` 8E(compatible(; E) $ 8x(nonempty((x) \ E)! (DEF) (x) E)) 7. ; 7 ` compatible(; E) (ASS) 8. ; 8 ` 8M; N : set8x(m N! (x 2 M! x 2 N )) (DEF) 9. ; 9 ` 8R8x; y; z(equivrelr $ (x; x) 2 R ^ ((x; y) 2 R! (y; x) 2 R)...) (DEF equivrel) 10. ; 10 ` equivrel() (ASS) 11. ; 11 ` (x 0 ; y 0 ) 2 (ASS) 12. ; 12 ` f 0 2 F (ASS) 13. ; 13 ` f 0 x 0 2 E (ASS) proof 14. ; 9, 10 ` 8x; y; z((x; x) 2 ^ ((x; y) 2! (y; x) 2 )...) (8D,$D 9) 15. ; 9, 10 ` 8x((x; x) 2 ) (^D 14) 16. ; 1, 2, 12 ` (x 0 ; y 0 ) 2! (f 0 x 0 ; f 0 x 0 ) 2 (8D,$D,!D ) 17. ; 1, 2, 11, ` (f 0 x 0 ; f 0 y 0 ) 2 (! D 11 16) ; 3 ` (f 0 x 0 ; f 0 y 0 ) 2! f 0 x 0 2 (f 0 y 0 ) (8D,$D 3) 19. ; 1, 2, 3, 11, ` f 0 x 0 2 (f 0 y 0 ) (! D 17 18) ; 1, 2, 3, 11, ` f 0 x 0 2 (f 0 y 0 ) ^ f 0 x 0 2 E (^I 19 13) 12, ; 13, 1, 2, 3, ` f 0 x 0 2 ((f 0 y 0 ) \ E) (8D,$D,!D 4, 11, ) 22. ; 13, 1, 2, 3, ` 9x(x 2 ((f 0 y 0 ) \ E)) (9I 21) 4, 11, ; 5, 13, 1, 2, ` nonempty((f 0 y 0 ) \ E) (8D, 3, 4, 11, 12 $D,!D 5 22) 24. ; 6, 7 ` 8x(nonempty((x) \ E)! (x) E) ($D,!D ; 5, 6, 7, 13, 1, 2, 3, 4, 11, 12 7) ` (f 0 y 0 ) E (8D!D 23 24) 15

17 26. ; 8, 5, 6, 7, 13, 1, 2, 3, 4, 11, 12 ` f 0 y 0 2 (f 0 y 0 )! f 0 y 0 2 E (8D,!D 8 25) 27. ; 9, 10 ` (f 0 y 0 ; f 0 y 0 ) 2 (8D 15) 28. ; 3, 9, 10 ` f 0 y 0 2 (f 0 y 0 ) (8D,$ 29. ; 3, 9, 10, 8, 5, 6, 7, 13, 1, 2, 3, 4, 11, 12 D,!D 3 27) ` f 0 y 0 2 E (! D 28 26) Thm. ; ` f 0 y 0 2 E () 16

18 ND Proof of theorem part (b) NNo S;D Formula Reason assumptions 1. ; 1 ` 8R; x; y8f(leftcongr(r) $ ((x; y) 2 R ^ f 2 F! (DEF) (fx; fy) 2 R)) 2. ; 2 ` leftcongr() (ASS) 3. ; 3 ` 8x; y(x 2 (y) $ (x; y) 2 ) (DEF) 4. ; 4 ` 8M; N : set; 8x(x 2 (M \ N ) $ x 2 M ^ x 2 N ) (DEF) 5. ; 5 ` 8M : set(nonempty(m ) $ 9x(x 2 M )) (DEF) 6. ; 6 ` 8E(compatible(; E) $ 8x(nonempty((x) \ E)! (DEF) (x) E)) 7. ; 7 ` compatible(; E) (ASS) 8. ; 8 ` 8M; N : set8x(m N! (x 2 M! x 2 N )) (DEF) 9. ; 9 ` 8R8x; y; z(congruencer $ (x; x) 2 R ^ ((x; y) 2 R! (y; x) 2 R)...) (DEF equivrel) 10. ; 10 ` congruence() (ASS) 11. ; 11 ` (x 0 ; y 0 ) 2 (ASS) 12. ; 12 ` f 0 2 F (ASS) 13. ; 13 ` f 0 y 0 2 E (ASS) proof 14. ; 9, 10 ` 8x; y; z((x; x) 2 ^ ((x; y) 2! (y; x) 2 )...) (8D,$D 9) 15. ; 9, 10 ` 8x((x; x) 2 ) (^D 14) 16. ; 9, 10 ` (y 0 ; x 0 ) 2! (x 0 ; y 0 ) 2 (8D,$D,^D 9 10) 17. ; 9, 11 ` (y 0 ; x 0 ) 2 (! D 11 16) 18. ; 1, 2, 12 ` (y 0 ; x 0 ) 2! (f 0 y 0 ; f 0 x 0 ) 2 (8D,$D,!D ) 19. ; 1, 2, 9, 11, ` (f 0 y 0 ; f 0 x 0 ) 2 (! D 17 18) ; 3 ` (f 0 y 0 ; f 0 x 0 ) 2! f 0 y 0 2 (f 0 x 0 ) (8D,$D 3) 21. ; 1, 2, 3, 9, ` f 0 y 0 2 (f 0 x 0 ) (! D 19 20) 11, ; 1, 2, 3, 9, ` f 0 y 0 2 (f 0 x 0 ) ^ f 0 y 0 2 E (^I 21 13) 11, 12, ; 13, 1, 2, 3, ` f 0 y 0 2 ((f 0 x 0 ) \ E) (8D,$D,!D 4, 9, 11, ) 24. ; 13, 1, 2, 3, ` 9x(x 2 ((f 0 x 0 ) \ E)) (9I 23) 4, 9, 11, ; 5, 13, 1, 2, ` nonempty((f 0 x 0 ) \ E) (8D,$D,!D 3, 4, 9, 11, 5 24) ; 6, 7 ` 8x(nonempty((x) \ E)! (x) E) ($D,!D 6 7) 17

19 27. ; 5, 6, 7, 13, 1, 2, 3, 4, 9, 11, ; 8, 5, 6, 7, 13, 1, 2, 3, ` (f 0 x 0 ) E (8D!D 25 26) ` f 0 x 0 2 (f 0 x 0 )! f 0 x 0 2 E (8D!D 8 27) 4, 9, 11, ; 9, 10 ` (f 0 x 0 ; f 0 x 0 ) 2 (8D 15) 30. ; 3, 9, 10 ` f 0 x 0 2 (f 0 x 0 ) (8D,$ D,!D 3 29) 31. ; 9, 10, 8, 5, ` f 0 x 0 2 E (! D 30 28) Thm. 6, 7, 13, 1, 2, 3, 4, 11, 12 ; ` f 0 x 0 2 E () 18

20 Discussion An attempt to translate the rst subproof to the second subproof by the symbol mapping fx 0 ) y 0 ; y 0 ) x 0 g fails, since the relevant assumptions dier in ((x 0 ; y 0 ) 2 ) and ((y 0 ; x 0 ) 2 ) after this mapping, respectively. In order to obtain equal assumptions, ((x 0 ; y 0 ) 2 ) is to be replaced by ((y 0 ; x 0 ) 2 ) within the assumptions. ((y 0 ; x 0 ) 2 ) becomes a new subtheorem, which is proven by the subproof of (b) that consists of the lines 15 and

21 CASE 4: THEOREMS 5.7 and Let us look at the analogy that provides a proof of theorem 5.7 that is based on the proof of theorem by examining the proofs of theorem and theorem 5.7: A stronger reformulation technique works for these examples, namely, abstraction based on the denition of a homomorphism. Theorem Let S; T 1 ; T 2 be F -semimoduls. Let 1 : S 7! T 1 ; 2 : S 7! T 2 be two homomorphisms into the F -semimoduls T 1 and T 2, respectively, and let 1, 2 be the respectively induced leftcongruences. 1. If there exists a homomorphism : T 1 7! T 2 with 1 = 2, then Let 1 2 and if 1 is surjective, then there is a unique homomorphism : H 1 7! H 2 with 1 = 2. If in addition, 2 is surjective, then is surjective as well. Theorem 5.7 Let S 0 ; H 1 ; H 2 be semigroups. Let 1 : S 0 7! H 1 ; 2 : S 0 7! H 2 be two homomorphisms into the semigroups H 1 and H 2, respectively, and let 1, 2 be the respectively induced congruences. 1. If there exists a homomorphism : H 1 7! H 2 with 1 = 2, then Let 1 2 and if 1 is surjective, then there is a unique homomorphism : H 1 7! H 2 with 1 = 2. If in addition, 2 is surjective, then is surjective as well. The proofs are based on the denitions of a homomorphism in semigroups and a homomorphism in F-semimoduls, respectively: Denition 2.1 Let F and H be semigroups. A mapping : F 7! H is called a homomorphism (from F to H) i 8f; g(f; g 2 F! (f g) = (f) (g). Denition 7.1 Let S and T be F -semimoduls. A mapping : F 7! H 1 This analogy is harder to nd than the transformation of the proof of 5.7 to a proof of

22 is called a homomorphism (from S to T ) i 8f; s(f 2 F ^ s 2 S! (f s) = (f) (s). The proofs of theorem and of theorem 5.7 can now be structured as follows. The proof structure of becomes: Part 1: Theorem: 1 2 Relevant assumptions: the denition of 1 mapping with 1 = 2. and of 2 ; existence of a Part 2a: Theorem: There exists a function with (8z(z 2 S! 1 (z) = 2 (z)) ^ 8x9y((x 2 T 1! y 2 T 2 ) ^ (x) = y). Relevant assumptions: 2 is a mapping S 7! T 2 ; 1 is a mapping S 7! T 1 ; 1 is surjective; the comprehension axiom; = is an equivalence relation; the denition of 1 and 2 ; 1 2 ; the representation of functions as relations. Part 2b: Theorem: is the only mapping for which the theorem of 2a holds, i.e., 8 0 8x; y((x 2 S! 0 ( 1 (x)) = 2 (x))! (y 2 T 1! (y) = 0 (y))) Relevant assumptions: the denition of : ( 1 ) = 2 ; surjectivity of 1 ; transitivity of =. Part 2c: Theorem: is a F -semimodul-homomorphism, i.e., 8f; x(x 2 T 1 ^ f 2 F! (f x) = f (x)). Relevant assumptions: surjective 1 ; 1 is a homomorphism in an F - semimodul; 2 is a homomorphism in an F -semimodul; the denition of ; theorem of 2a. Part 2d: Theorem: If 2 is surjective then is surjective. The proof structure of 5.7 becomes: 21

23 Part 1: Theorem: 1 2 Relevant assumptions: the denition of 1 and of 2 ; the existence of a mapping with 1 = 2. Part 2a: Theorem: There exists a function with 8z(z 2 S 0! 1 (z) = 2 (z)) ^ 8x9y((x 2 H 1! y 2 H 2 ) ^ (x) = y). Relevant assumptions: 1 : F 7! H 1 is a mapping from a semigroup into a semigroup; 2 : F 7! H 2 is a mapping from a semigroup into a semigroup F ) H 2 ; 1 is surjective; the comprehension axiom; = is an equivalence relation; the denitions of 1 and 2 ; 1 2 ; the representation of functions as relations. Part 2b: Theorem: is the only mapping for which the theorem of 2a holds, i.e., 8 0 8x; y((x 2 S 0! 0 ( 1 (x)) = 2 (x))! (y 2 H 1! (y) = 0 (y))) Relevant assumptions: the denition of : ( 1 ) = 2 ; 1 is surjective; the transitivity of =. Part 2c: Theorem: is a semigroup-homomorphism, i.e., 8x; y(x 2 H 1 ^ y 2 H 1! (x y) = (x) (y)). Relevant assumptions: 1 is surjective; 1 is a semigroup-homomorphism; 2 is a semigroup-homomorphism; the denition of ; the theorem of 2a. Part 2d: If 2 is surjective then is surjective. Note that, the parts 1, 2a, 2b, 2d of theorem are equal to the corresponding parts of theorem 4.11 of HUA as well, and in general play an important r^ole. The crucial point for the transformation of the proof of theorem to the proof of theorem are the relevant assumptions of the respective part 1 of theorem 5.7 and of theorem 7.5.7, which dier in symbols only. Hence, they become equal by the symbol mapping ff ) S, H 1 ) T 1, and H 2 ) T 2 g. This symbol mapping is to be applied to the whole proof of

24 The proofs of parts 2c of theorem and theorem 5.7 are given next. For simplicity, let be a polymorphic function. 23

25 ND Proof of theorem part 2c NNo S;D Formula Reason relevant assumptions 1. ; 1 ` 8x; y; f(x 2 T 2^y 2 T 2 ^f 2 F ^x = y! f x = f y) (T 2 is semimodul) 2. ; 2 ` 8x; f(f 2 F ^ x 2 S! (f x) 2 S) (S is semimodul) 3. ; 3 ` 8x; y; f(x 2 T 1^y 2 T 1 ^f 2 F ^x = y! f x = f y) (Ax =T 1 -issemimodul) 4. ; 4 ` hom from S() $ 8f8x(f 2 F ^ x 2 S! (f x) = f (x)) (DEF hom from S ) (hom from T 1 ) 5. ; 5 ` hom from T 1 () $ 8f8x(f 2 F ^ x 2 T 1! (f x) = f (x)) 6. ; 6 ` 8x; f(f 2 F ^ x 2 T 1! f x 2 T 1 ) (T 1 -issemimodul) 7. ; 7 ` 8x; y; z(x = y ^ y = z! x = z) (Ax =transitive) 8. ; 8 ` 8x(x 2 S! 2 (x) 2 T 2 ) (Def 2 ) 9. ; 9 ` 8x(x 2 T 1! (x) 2 T 2 ) (lemma 2a) 10. ; 10 ` 8x(x 2 S! 1 (x) 2 T 1 ) (Def 1 ) 11. ; 11 ` 8x; y(x 2 T 1 ^ y 2 T 1 ^ x = y! (x) = (y)) (lemma 2a) 12. ; 12 ` 8x(x 2 S! ( 1 (x)) = 2 (x)) (lemma 2a) 13. ; 13 ` 8y(y 2 T 1! 9x(x 2 S ^ 1 (x) = y)) (ASS surjective 1 ) 14. ; 14 ` hom from S( 1 ) (ASS) 15. ; 15 ` hom from S( 2 ) (ASS) The Proof 16. ; 4, 14 ` 8f8x(f 2 F ^ x 2 S! 1 (f x) = f 1 (x)) ($ D;! D 4 14) 17. ; 4, 15 ` 8f8x(f 2 F ^ x 2 S! 2 (f x) = f 2 (x)) ($ D;! D 4 15) ; ` f 2 F (HYP) ; ` x 0 2 T 1 (HYP) (*) ; 13 ` 9y(y 2 S ^ 1 (y) = x 0 ) (8D;! D 19 13) ; 13 ` a 2 S ^ 1 (a) = x 0 (9D 20) ; 10, 13 ` a 2 S ^ 1 (a) = x 0 ^ 1 (a) 2 T 1 (8D; ^D; $ D; ^I 10 21) ; 10, 13 ` 1 (a) 2 T 1 ^ 1 (a) = x 0 (^ D 22) , 18; 10, 13 ` x 0 2 T 1 ^ 1 (a) 2 T 1 ^ f 2 F ^ 1 (a) = x 0 (^ I ) 24

26 25. 18, 19; 10, 13, , 18; 10, 13, , 18; 10, 3, 4, 11, , 19; 10, 3, 4, 14, , 18; 10, 3, 4, 2, , 19; 10, 3, 4, , 18; 10, 3, 4, 12, 1, 8, , 18; 10, 3, 4, 11, , 18; 10, 3, 4, 14, 15, 11, 7, 2, 8, 9, 6, ; 3, 4, 14, 4, 15, 11, 7, 2, 10, 8, 9, 6, 13, 12, ; 3, 4, 14, 15, 11, 7, 2, 10, 8, 9, 6, 13, 12, ; 3, 4, 14, 15, 11, 7, 2, 10, 8, 9, 6, 13, 12, 1, 5 ` f 1 (a) = f x 0 (8D;! D 3 24) ` a 2 S ^ 1 (a) = x 0 ^ 1 (a) 2 T 1 ^ f 1 (a) = f x 0 (^ I 25 22) (**) ` (f x 0 ) = (f 1 (a)) (^D;! D ) ` (f 1 (a)) = ( 1 (f a)) (8D;! D ) ` ( 1 (f a)) = 2 (f a) (8D,! D ) ` 2 (f a) = f 2 (a) (8D;! D ) ` f 2 (a) = f ( 1 (a)) (8D ) ` f ( 1 (a)) = f (x 0 ) (! D ) ` (f x 0 ) = f (x 0 ) (! D ) ` f 2 F ^ x 0 2 T 1! (f x 0 ) = f (x 0 )) (DED 33) ` 8f8x(f 2 F ^ x 2 T 1! (f x) = f (x)) (8I 34) ` hom from T 1 () ($ D;! D; 8D 5 35) ND Proof of theorem 5.7 part 2c NNo S;D Formula Reason m relevant assumptions 1. ; 1 ` 8x 1 ; x 2 ; y 1 ; y 2 (x 1 ; x 2 ; y 1 ; y 2 2 H 2 ^ x 1 = x 2 ^ y 1 = (Ax semigroup y 2! x 1 y 1 = x 2 y 2 ) H 2 ) 2. ; 2 ` 8x 1 ; x 2 (x 1 2 S 0 ^ x 2 2 S 0! x 1 x 2 2 S 0 ) (Ax S' semigroup) 25

27 3. ; 3 ` 8x 1 ; x 2 ; y 1 ; y 2 (x 1 ; x 2 ; y 1 ; y 2 2 H 1 ^ x 1 = x 2 ^ y 1 = y 2! x 1 y 1 = x 2 y 2 ) (Ax semigrouph 1 ) 4. ; 4 ` hom from S 0 () $ 8x8y(x 2 S 0 ^ y 2 S 0! (x y) = (x) (y)) (DEF hom from S' ) 5. ; 5 ` hom from H 1 () $ 8x8y(x 2 H 1 ^ y 2 H 1! (x y) = (x) (y)) (DEF hom from H 1 ) 6. ; 6 ` 8x; y(x 2 H 1 ^ y 2 H 1! x y 2 H 1 ) (Ax semigroup H 1 ) 7. ; 7 ` 8x; y; z(x = y ^ y = z! x = z) (Ax= transitive) 8. ; 8 ` 8x(x 2 S 0! 2 (x) 2 H 2 ) (ASS Def 2 ) 9. ; 9 ` 8x(x 2 H 1! (x) 2 H 2 ) (lemma 2a) 10. ; 10 ` 8x(x 2 S 0! 1 (x) 2 H 1 ) (Def 1 ) 11. ; 11 ` 8x; y(x 2 H 1 ^ y 2 H 1 ^ x = y! (x) = (y)) (lemma 2a) 12. ; 12 ` 8x(x 2 S 0! ( 1 (x)) = 2 (x)) (lemma 2a) 13. ; 13 ` 8y(y 2 H 1! 9x(x 2 S 0 ^ 1 (x) = y)) (ASS surj 1 ) 14. ; 14 ` hom in S 0 ( 1 ) (ASS) 15. ; 15 ` hom in S 0 ( 2 ) (ASS) The Proof 16. ; 4, 14 ` 8y8x(y 2 S 0 ^ x 2 S 0! 1 (x y) = 1 (x) 1 (y) ($ D;! D 4 14) 17. ; 4, 15 ` 8y8x(y 2 S 0 ^ x 2 S 0! 2 (x y) = 2 (x) 2 (y) ($ D;! D 4 15) ; ` x 10 2 H 1 (HYP) ; ` x 20 2 H 1 (HYP) (*) ; 13 ` 9y(y 2 S 0 ^ 1 (y) = x 10 ) (! D; 8D 18 13) ; 13 ` 9z(z 2 S 0 ^ 1 (z) = x 20 ) (! D; 8D 19 13) ; 13 ` a 1 2 S 0 ^ 1 (a 1 ) = x 10 (9D 20) ; 13 ` a 2 2 S 0 ^ 1 (a 2 ) = x 20 (9D 21) ; 10, 13 ` a 1 2 S 0 ^ 1 (a 1 ) = x 10 ^ 1 (a 1 ) 2 H 1 (^D; 8D; ^I;! D 22 10) ; 10, 13 ` a 2 2 S 0 ^ 1 (a 2 ) = x 20 ^ 1 (a 2 ) 2 H 1 (^D; 8D; ^I;! D 23 10) ; 10, 13 ` 1 (a 1 ) 2 H 1 ^ 1 (a 1 ) = x 10 (^ D 24) ; 10, 13 ` 1 (a 2 ) 2 H 1 ^ 1 (a 2 ) = x 20 (^ D 25) ; 10, 13 ` x 10 2 H 1 ^ 1 (a 1 ) 2 H 1 ^ 1 (a 1 ) = x 10 (^ I 18 26) ; 10, 13 ` x 20 2 H 1 ^ 1 (a 2 ) 2 H 1 ^ 1 (a 2 ) = x 20 (^ I 19 26) , 19; 10, 13 ` x 10 2 H 1 ^ 1 (a 1 ) 2 H 1 ^ 1 (a 1 ) = x 10 ^ x 20 2 H 1 ^ 1 (a 2 ) 2 H 1 ^ 1 (a 2 ) = x 20 (^ I 28 29) 26

28 31. 18, 19; 10, 13, , 19; 10, 13, , 19; 10, 3, 13, 11, , 19; 10, 3, 13, 4, 14, 11, 6, , 19; 10, 3, 13, 2, , 19; 10, 3, 13, 4, , 19; 10, 3, 13, 12,??, 8, , 19; 10, 3, 13, 11, 1, , 19; 10, 3, 13, 1, 4, 14, 15, 11, 7, 2, 8, 9, 12, ; 3, 13, 1, 4, 14, 15, 11, 7, 2, 8, 9, 12, 6, ; 3, 13, 1, 4, 14, 15, 11, 7, 2, 8, 9, 12, 6, ; 3, 13, 1, 4, 14, 15, 11, 7, 2, 8, 9, 12, 6, 5, 10 ` 1 (a 1 ) 1 (a 2 ) = x 10 x 20 (8D;! D 3 30) ` a 1 2 S 0 ^ 1 (a 1 ) = x 10 ^ 1 (a 1 ) 2 H 1 ^ a 2 2 S 0 ^ (^ I (a 2 ) = x 20^ 1 (a 2 ) 2 H 1^ 1 (a 1 ) 1 (a 2 ) = x 10 x 20 25) (**) ` (x 10 x 20 ) = ( 1 (a 1 ) 1 (a 2 )) (^D;! D ) ` ( 1 (a 1 ) 1 (a 2 )) = ( 1 (a 1 a 2 )) (^D; 8D;! D ) ` ( 1 (a 1 a 2 )) = 2 (a 1 a 2 ) (^D; 8D,! D ) ` 2 (a 1 a 2 ) = 2 (a 1 ) 2 (a 2 ) (^D; 8D;! D 17 32) ` 2 (a 1 ) 2 (a 2 ) = ( 1 (a 1 )) ( 1 (a 2 )) (^D; 8D ) ` ( 1 (a 1 )) ( 1 (a 2 )) = (x 10 ) (x 20 ) (^D;! D ) ` (x 10 x 20 ) = (x 10 ) (x 20 ) (^D;! D ) ` x 10 2 H 1 ^ x 20 2 H 1! (x 10 x 20 ) = (x 10 ) (x 20 ) (DED 39) ` 8x 1 8x 2 (x 1 2 H 1 ^ x 2 2 H 1! (x 1 x 2 ) = (x 1 ) (x 2 )) (8I 40) ` hom from H 1 () ($ D;! D; 8D 5 41) Discussion The subtheorems 2c of and 5.7 dier in more than one corresponding symbol. Thus symbol mapping is not sucient to obtain equal theorems and 27

29 assumptions of 5.7.2c and c. Term mapping, e.g., ff term(x) ) term(x) term(y)g is not sucient either, since part (*) would dier after the term mapping and a proof checker would not accept the transformation as a proof of 5.7.2c. Probably more importantly however, the theorem and the assumptions of 5.7.2c and c contain dierent subformulae of the form x 2 M and quantiers which have to be modied by the mapping as well. This problem is due to fact that the mapping of terms is essentially an abstraction by which some irrelevant symbols disappear. The actual justication for this abstraction is the occurrence of the denition of a homomorphism within the relevant assumptions. An analogy based on pure term mapping is not justied at all and hence the abstracting reformulation has to be preferred. Another reason for this preference is that less modication is to be done after this analogy formation. An reformulation of theorem and its proof to theorem 5.7 and its proof consists of three steps. 1. Abstraction of both problems (i.e., theorem and assumptions) c and c based on the meaning of the two respective denitions of homomorphism. The key is a reformulation of terms of the form f term(x) to terms Op(term(x)) for c and term1term2 to Op(term1,term2) with a function variable Op. This reformulation aects the denitions of homomorphism within the relevant assumptions: 8f8x(f 2 F ^ x 2 S! (f x) = f (x)) becomes 8x(x 2 S! (Op(x)) = Op((x))) by the mapping f term )Op(term) 8x; y(x 2 S 0 ^ y 2 S 0! (x y) = (x) (y)) becomes 8x; y(x 2 S 0 ^ y 2 S 0! (Op 0 (x; y)) = Op 0 ((x); (y))) by the mapping term1term2 )Op(term1,term2). The reformulation aects also the corresponding terms within the whole proof. Certain subformulae and quantiers become superuous and, hence, can be omitted. As a result we obtain the theorems and reformulated proofs c 0 and c The problems c 0 and c 0 are not equal yet. Their comparison suggests another reformulation of c 0 to c 00 in order to obtain equal abstracted assumptions and theorems, which increases 28

30 the number of arguments of Op in c. 0. This reformulation causes several additional changes within the reformulated proof. 3. Finally, to return to the original theorem and assumptions of c, a reversion of the abstraction of c has to be applied to c 00. All these reformulations have to be applied to the whole proofs and not only to the assumptions and the theorem. 29

31 CASE 5: THEOREMS 4.8 and 5.3 Theorem 4.8 Let and be two equivalence relations, then 1. ( \ ) is an equivalence relation and 2. ( [ ) t is the smallest equivalence relation, containing and. Theorem 5.3 Let and be two leftcongruences, then 1. ( \ ) is a leftcongruence and 2. ( [ ) t is the smallest leftcongruence containing and. The proofs of theorem 4.8 and theorem 5.3 can be structured as follows. The proof structure for theorem 4.8 becomes: Part 1: 1. Theorem: ( \ ) is an equivalence relation Subtheorem: reexivity of ( \ ), 1.2. Subtheorem: symmetry of ( \ ), 1.3. Subtheorem: transitivity of ( \ ). Part 2: 2. Theorem: ( [ ) t is an equivalence relation Subtheorem: reexivity of ( [ ) t, 2.2. Subtheorem: symmetry of ( [ ) t, 2.3. Subtheorem: transitivity of ( [ ) t. Part 3: Theorem: ( [ ) t is the smallest equivalence relation. The proof structure of 5.3 becomes: Part 1: 1.a. Subtheorem: ( \ ) is a leftcongruence. 1.a.1. Subsubtheorem: reexivity of ( \ ), 1.a.2. Subsubtheorem: symmetry of ( \ ), 1.a.3. Subsubtheorem: transitivity of ( \ ), 1.b. Subtheorem: (f 1 ; f 2 ) 2 ( \ )! (gf 1 ; gf 2 ) 2 ( \ ). Part 2: 2. Theorem: ( [ ) t is a leftcongruence 30

32 2.a. Subtheorem: ( [ ) t is an equivalence relation. 2.a.1. Subsubtheorem: reexivity of ( [ ) t, 2.a.2. Subsubtheorem: symmetry of ( [ ) t, 2.a.3. Subsubtheorem: transitivity of ( [ ) t, 2.b. Subtheorem: (f 1 ; f 2 ) 2 ( [ ) t! (gf 1 ; gf 2 ) 2 ( [ ) t. Part 3: Theorem: ( [ ) t is the smallest equivalence relation. Discussion This example illustrates particularly well the importance of structuring theorems and proofs for analogy-driven theorem proving: Some parts of the proofs become identical. For example, the proofs of the parts 1a, 2a, and 3 of theorem 5.3 are identical to the corresponding subproofs of theorem 4.8 since the problems have identical theorems and assumptions. Looking at the remaining parts it turns out that b can be proved analogously to and b can be proved analogously to These proofs are given in the following: ND Proof of theorem 4.8 part 1.2 NNo S;D Formula Reason relevant assumptions 1. ; 1 ` 8R(symm(R) $ 8x 1 ; x 2 ((x 1 ; x 2 ) 2 R! (x 2 ; x 1 ) 2 R)) (DEFsymm) 2. ; 2 ` 8R 1 ; R 2 ; x((x 2 (R 1 \ R 2 ) $ (x 2 R 1 ^ x 2 R 2 )) (DEF-\) 3. ; 3 ` symm() (ASS) 4. ; 4 ` symm() (ASS) The proof 5. 5; ` (f 1 ; f 2 ) 2 ( \ ) (HYP) 6. ; 2 ` (f 1 ; f 2 ) 2 ( \ )! (f 1 ; f 2 ) 2 ^ (f 1 ; f 2 ) 2 (8D; $ D 2) 7. 5; 2 ` (f 1 ; f 2 ) 2 (! D; ^D 6) 8. 5; 2 ` (f 1 ; f 2 ) 2 (! D; ^D 6) 9. ; 1 ` symm()! 8x 1 ; x 2 ((x 1 ; x 2 ) 2! (x 2 ; x 1 ) 2 ) (8D; $ D 1) 10. ; 1, 3 ` 8x 1 ; x 2 ((x 1 ; x 2 ) 2! (x 2 ; x 1 ) 2 ) (! D 9 3) 11. ; 1, 3 ` (f 1 ; f 2 ) 2! (f 2 ; f 1 ) 2 (8D 10) 12. ; 1 ` symm()! 8x 1 ; x 2 ((x 1 ; x 2 ) 2! (x 2 ; x 1 ) 2 ) (8D; $ D 1) 13. ; 1, 4 ` 8x 1 ; x 2 ((x 1 ; x 2 ) 2! (x 2 ; x 1 ) 2 ) (! D 12 4) 31

33 14. ; 1, 4 ` (f 1 ; f 2 ) 2! (f 2 ; f 1 ) 2 (8D 13) 15. 5; 2, 1, 3 ` (f 2 ; f 1 ) 2 (! D 11 7) 16. 5; 2, 1, 4 ` (f 2 ; f 1 ) 2 (! D 14 8) 17. 5; 2, 1, 4, 3 ` (f 2 ; f 1 ) 2 ( \ ) (8D; $ D 2 16) 18. ; 2, 1, 4, 3 ` (f 1 ; f 2 ) 2 ( \ )! (f 2 ; f 1 ) 2 ( \ ) (DED 17) 19. ; 2, 1, 4, 3 ` 8f 1 ; f 2 ((f 1 ; f 2 ) 2 ( \ )! (f 2 ; f 1 ) 2 ( \ )) (8I 18) 20. ; 2, 1, 4, 3 ` symm( \ ) (8D; $ D;! D 1 19) Thm. ; ` symm( \ ) () 32

34 ND Proof of theorem 5.3 part 1.b NNo S;D Formula Reason relevant assumptions 1. ; 1 ` 8R(leftcongruence(R) $ 8g; x 1 ; x 2 (g 2 H ^ (x 1 ; x 2 ) 2 R! (gx 1 ; gx 2 ) 2 R)) (DEFleftcongruence) 2. ; 2 ` 8R 1 ; R 2 ; x((x 2 (R 1 \ R 2 ) $ (x 2 R 1 ^ x 2 R 2 )) (DEF-\) 3. ; 3 ` leftcongruence() (ASS) 4. ; 4 ` leftcongruence() (ASS) The proof 5. 5; ` (f 1 ; f 2 ) 2 ( \ ) (HYP) 6. ; 2 ` (f 1 ; f 2 ) 2 ( \ )! (f 1 ; f 2 ) 2 ^ (f 1 ; f 2 ) 2 (8D; $ D 2) 7. 7; ` g 0 2 H (HYP) 8. 5; 2 ` (f 1 ; f 2 ) 2 (! D; ^D, 6) 9. 5; 2 ` (f 1 ; f 2 ) 2 (! D; ^D, 6) 10. ; 1 ` leftcongruence() $ 8g; x 1 ; x 2 (g 2 H ^ (x 1 ; x 2 ) 2 (8D; $ D 1)! (gx 1 ; gx 2 ) 2 ) 11. ; 1 ` 8x 1 ; x 2 ((x 1 ; x 2 ) 2! (g 0 x 1 ; g 0 x 2 ) 2 ) (! D 10 3) 12. ; 1 ` leftcongruence() $ 8g; x 1 ; x 2 (g 2 H ^ (x 1 ; x 2 ) 2 (8D; $ D 1)! (gx 1 ; gx 2 ) 2 ) 13. 7; 1 ` 8x 1 ; x 2 ((x 1 ; x 2 ) 2! (g 0 x 1 ; g 0 x 2 ) 2 ) (! D 12 4) 14. 7; 1 ` (f 1 ; f 2 ) 2! (g 0 f 1 ; g 0 f 2 ) 2 (8D 11) 15. 5, 7; 2, 1, 3 ` (g 0 f 1 ; g 0 f 2 ) 2 (! D, 14 8) 16. 7; 1 ` (f 1 ; f 2 ) 2! (g 0 f 1 ; g 0 f 2 ) 2 (8D 13) 17. 5, 7; 2, 1, 4 ` (g 0 f 1 ; g 0 f 2 ) 2 (! D, 16 9) 18. 5, 7; 2, 1, 4, ` (g 0 f 1 ; g 0 f 2 ) 2 ( \ ) (8D; $ D ) 19. 7; 2, 1, 4, 3 ` g 0 2 H ^ (f 1 ; f 2 ) 2 ( \ )! (g 0 f 1 ; g 0 f 2 ) 2 ( \ ) (DED 18) 20. ; 2, 1, 4, 3 ` 8g; f 1 ; f 2 (g 2 H ^ (f 1 ; f 2 ) 2 ( \ )! (gf 1 ; gf 2 ) 2 (8I 19) \ ) 21. ; 2, 1, 4, 3 ` leftcongruence( \ ) (8D; $ D;! D, 1 20) Thm. ; ` leftcongruence( \ ) () Discussion Symbol- or term mappings are not sucient for a transformation of to b. For example, the symbol mapping fsymm ) lef tcongrunceg is not sucient, since the denitions of symm and lef tcongruence (which are part 33

35 of the relevant assumptions) are not equal after this mapping. An additional term mapping would have to be restricted to certain occurrences of terms, because the overall mapping f(f 2 ; f 1 ) ) (g f 1 ; g f 2 )g or f(term1; term2) ) (g term1; g term2)g also yields f(f 1 ; f 2 ) ) (gf 2 ; gf 1 )g which is not desired at all. Furthermore, the reformulated proof cannot be veried for b because of missing sort declarations and quantiers. Furthermore, the theorem and the assumptions of and b contain subformulae of the form x 2 M and quantiers which have to be modi- ed by the mapping. This problem is due to fact that the necessary mapping of terms is essentially an abstraction by which some symbols irrelevant for the proof disappear, just as in the previous case. The actual justication for this abstraction is the occurrence within the relevant assumptions of the denitions of symm and lef tcongruence, which have the same characteristic structure. An analogy formation based on pure term mapping is not justied at all and hence the abstracting reformulation has to be preferred. A successful transformation is composed of an abstraction followed by a symbol mapping, and a subsequent reverse abstraction. 1. The abstraction of problem that yields problem ' changes the pairs (term 2 ; term 1 ), which are determined by the denition of symm, to terms f rev (term 1 ; term 2 ). The abstraction of problem b that yields problem b' transforms the pairs (g term 1 ; g term 2 ) to f g (term 1 ; term 2 ). These reformulations aect the pairs contained in the denition of symm(r) and lef tcongruence(r), respectively. It aects the derived terms within the whole proof and in addition, certain formulae and quantiers have to be removed. 2. The symbol mapping fsymm ) leftcongruence, f rev ) f g g is applied to problem and yields problem which is equal to problem b Finally, to return to the original problem b, a reversion of the abstraction of problem b has to be applied to problem In the following the ND-proofs of theorem 4.8 part 2.2 and of theorem 5.3 part 2.b are given. 34

36 ND Proof for theorem 4.8 part 2.2 NNo S;D Formula Reason relevant assumptions 1. ; 1 ` 8R(symm(R) $ 8x 1 ; x 2 ((x 1 ; x 2 ) 2 R! (x 2 ; x 1 ) 2 R)) (DEF symm) 2. ; 2 ` 8x(x 2 ( [ ) $ x 2 _ x 2 ) (DEF-[) 3. ; 3 ` 8x; y; k(k 2 N! ((x; y) 2 R 1! (y; x) 2 R 1 ) ^ (((x; y) 2 R k! (y; x) 2 R k )! ((x; y) 2 R k+1! (y; x) 2 R k+1 ))! 8n(n 2 N! (x; y) 2 R n! (y; x) 2 R n )) (Induction- AX) 4. ; 4 ` 8x; y((x; y) 2 ( [ ) 1 $ (x; y) 2 ( [ )) (DEF- ( [ ) 1 ) 5. ; 5 ` 8n; x; y(n 2 N! ((x; y) 2 ( [ ) n+1 $ 9z((x; z) 2 (DEF ( [ ( [ ) n ^ (z; y) 2 ( [ ) 1 ) _ ((z; y) 2 ) n+1 ) ( [ ) n ^ (x; z) 2 ( [ ) 1 )) 6. ; 6 ` 8x; y(x; y) 2 ( [ ) t $ 9n(n 2 N ^ (x; y) 2 ( [ ) n ) (DEF- ( [ ) t ) 7. ; 7 ` symm() (ASS) 8. ; 8 ` symm() (ASS) induction base 9. 9; ` (f 1 ; f 2 ) 2 ( [ ) 1 (HYP) 10. 9; ` (f 1 ; f 2 ) 2 _ (f 1 ; f 2 ) 2 (8D;! D 2 9) ; ` (f 1 ; f 2 ) 2 (HYP) ; 1, 7 ` (f 2 ; f 1 ) 2 (8D;! D ) ; 1, 7, 2 ` (f 2 ; f 1 ) 2 ( [ ) (8D; $ D;! D 2 12) ; 1, 7, 2, ` (f 2 ; f 1 ) 2 ( [ ) 1 (8D; $ 4 D;! D 13 4) 15. ; 1, 7, 2, 4 ` (f 1 ; f 2 ) 2! (f 2 ; f 1 ) 2 ( [ ) 1 (DED 14) ; ` (f 1 ; f 2 ) 2 (HYP) ; 1, 8 ` (f 2 ; f 1 ) 2 (8D;! D ) ; 1, 2, 8 ` (f 2 ; f 1 ) 2 ( [ ) (8D; $ D;! D 2 17) ; 1, 2, 8, ` (f 2 ; f 1 ) 2 ( [ ) 1 (8D; $ 4 D;! D 18 4) 20. ; 1, 2, 8, 4 ` (f 1 ; f 2 ) 2! (f 2 ; f 1 ) 2 ( [ ) 1 (DED 19) 35

37 21. 9; 1, 2, 8, 7, ; 1, 2, 8, 7, 4 ` (f 2 ; f 1 ) 2 ( [ ) 1 (_D ) ` 8f 1 ; f 2 ((f 1 ; f 2 ) 2 ( [ ) 1! (f 2 ; f 1 ) 2 ( [ ) 1 (DED,8I 21) The proof induction step ; ` k 2 N (HYP) ; ` 8x 1 ; x 2 ((x 1 ; x 2 ) 2 ( [ ) k! (x 2 ; x 1 ) 2 ( [ ) k ) (Induction- HYP) ; ` (f 1 ; f 2 ) 2 ( [ ) k+1 (HYP) , 23; 5 ` 9z(((f 1 ; z) 2 ( [ ) k ^ (z; f 2 ) 2 ( [ ) 1 ) _ ((z; f 2 ) 2 ($ D;! D ( [ ) k ^ (f 1 ; z) 2 ( [ ) 1 )) 5 25) ; ` ((f 1 ; x 0 ) 2 ( [ ) k ^ (x 0 ; f 2 ) 2 ( [ ) 1 ) _ ((x 0 ; f 2 ) 2 (HYP) ( [ ) k ^ (f 1 ; x 0 ) 2 ( [ ) 1 ) case ; ` (f 1 ; x 0 ) 2 ( [ ) k ^ (x 0 ; f 2 ) 2 ( [ ) 1 (HYP) ; ` (f 1 ; x 0 ) 2 ( [ ) k (^D 28) ; ` (x 0 ; f 2 ) 2 ( [ ) 1 (^D 28) , 23; ` (x 0 ; f 1 ) 2 ( [ ) k (8D;! D 24 29) , 24; ` (f 2 ; x 0 ) 2 ( [ ) 1 (8D;! D 24 30) , 28, 23; ` (f 2 ; x 0 ) 2 ( [ ) 1 ^ (x 0 ; f 1 ) 2 ( [ ) k ) (^I 31 32) 34. ; 24, 28, 23 ` (f 2 ; x 0 ) 2 ( [ ) 1 ^ (x 0 ; f 1 ) 2 ( [ ) k _ (f 2 ; x 0 ) 2 (_I 33) ( [ ) k ^ (x 0 ; f 1 ) 2 ( [ ) , 24, 23; ` 9z((f 2 ; z) 2 ( [ ) 1 ^ (z; f 1 ) 2 ( [ ) k _ (f 2 ; z) 2 (^I; 9I 34) ( [ ) k ^ (z; f 1 ) 2 ( [ ) 1 ) , 24, 23; 5 ` (f 2 ; f 1 ) 2 ( [ ) k+1 (8D; $ D;! D 35 5) case ; ` (x 0 ; f 2 ) 2 ( [ ) k ^ (f 1 ; x 0 ) 2 ( [ ) 1 (HYP) ; ` (x 0 ; f 2 ) 2 ( [ ) k (^D 37) ; ` (f 1 ; x 0 ; ) 2 ( [ ) 1 (^D 37) , 23; ` (f 2 ; x 0 ) 2 ( [ ) k (8D;! D 24 38) , 24; ` (x 0 ; f 1 ) 2 ( [ ) 1 (8D;! D 24 39) , 37, 23; ` (f 2 ; x 0 ) 2 ( [ ) 1 ^ (x 0 ; f 1 ) 2 ( [ ) k ) (^I 40 41) 43. ; 24, 37, 23 ` (f 2 ; x 0 ) 2 ( [ ) 1 ^ (x 0 ; f 1 ) 2 ( [ ) k ) _ (f 2 ; x 0 ) 2 (_I 42) ( [ ) k ^ (x 0 ; f 1 ) 2 ( [ ) 1 ) , 24, 23; ` 9z((z; f 1 ) 2 ( [ ) 1 ^ (f 2 ; z) 2 ( [ ) k _ (f 2 ; z) 2 (9I 43) ( [ ) k ^ (z; f 1 ) 2 ( [ ) 1 ) , 24, 23; 5 ` (f 2 ; f 1 ) 2 ( [ ) k+1 (8D; $ D;! D 44 5) 36

38 46. 37, 28, 24, ` (f 2 ; f 1 ) 2 ( [ ) k+1 (_D , 27; 5 27) , 24, 37, ` (f 2 ; f 1 ) 2 ( [ ) k+1 (CHOICE 25, 28; ) , 24, 28, ` 8f 1 ; f 2 ((f 1 ; f 2 ) 2 ( [ ) k+1! (f 2 ; f 1 ) 2 ( [ ) k+1 ) (DED,8I 47) 37; ; 5 ` k 2 N ^ 8x 1 ; x 2 (((x 1 ; x 2 ) 2 ( [ ) k! (x 2 ; x 1 ) 2 (DED 48) ( [ ) k )! 8f 1 ; f 2 ((f 1 ; f 2 ) 2 ( [ ) k+1! (f 2 ; f 1 ) 2 ( [ ) k+1 )) 50. ; 1, 2, 3, 4, ` 8f 1 ; f 2 8n(n 2 N ^ (f 1 ; f 2 ) 2 ( [ ) n! (f 2 ; f 1 ) 2 (8I; ^I;! 5, 7, 8 ( [ ) n ) D ) for ( [ ) t ; ` (f 1 ; f 2 ) 2 ( [ ) t (HYP) ; 6 ` 9n(n 2 N ^ (f 1 ; f 2 ) 2 ( [ ) n ) (8D; $ D;! D 6 51) ; ` m 0 2 N ^ (f 1 ; f 2 ) 2 ( [ ) m0 (HYP) ; 1, 2, 3, 4, 5, 7, 8 ` m 0 2 N ^ (f 2 ; f 1 ) 2 ( [ ) m0 (8D;! D 53 50) ; 1, 2, 3, ` 9n(n 2 N ^ (f 2 ; f 1 ) 2 ( [ ) n ) (9I 54) 4, 5, 7, ; 1, 2, 3, ` (f 2 ; f 1 ) 2 ( [ ) t (! D 6 55) 4, 5, 7, ; 6, 1, 2, 3, 4, 5, 7, 8 ` f 2 ; f 1 ) 2 ( [ ) t (CHOICE ) 58. ; 6, 1, 2, 3, ` (f 1 ; f 2 ) 2 ( [ ) t! (f 2 ; f 1 ) 2 ( [ ) t (DED 57) 4, 5, 7, ; 6, 1, 2, 3, ` 8f 1 ; f 2 ((f 1 ; f 2 ) 2 ( [ ) t! (f 2 ; f 1 ) 2 ( [ ) t ) (8I 58) 4, 5, 7, ; 6, 1, 2, 3, ` symm( [ ) t ($ D 59 1) 4, 5, 7, 8 Thm. ; ` symm( [ ) t () 37

39 ND Proof for theorem 5.3 part 2.b NNo S;D Formula Reason relevant assumptions 1. ; 1 ` 8R(leftcongruence(R) $ 8x 1 ; x 2 ; g(g 2 F (x 1 ; x 2 ) 2 R! (gx 1 ; gx 2 ) 2 R)) (DEFleftcongruence) 2. ; 2 ` 8x(x 2 ( [ ) $ x 2 _ x 2 ) (DEF-[) 3. ; 3 ` 8k; x; y; x 1 ; y 1 ; x 2 ; y 2 ; x 3 ; y 3 ; g(k 2 N ^ g 2 F! ((x; y) 2 R 1! (gx; gy) 2 R 1 ) ^ (((x 1 ; y 1 ) 2 R k! (Induction- AX) (gx 1 ; gy 1 ) 2 R k )! ((x 2 ; y 2 ) 2 R k+1! (gx 2 ; gy 2 ) 2 R k+1 ))! 8n(n 2 N! (x 3 ; y 3 ) 2 R n! (gx; gy) 2 R n )) 4. ; 4 ` 8x; y((x; y) 2 ( [ ) 1 $ (x; y) 2 ( [ )) (DEF- 5. ; 5 ` 8n; x; y(n 2 N! ((x; y) 2 ( [ ) n+1 $ 9z((x; z) 2 ( [ ) n ^ (z; y) 2 ( [ ) 1 ) _ ((z; y) 2 ( [ ) n ^ (x; z) 2 ( [ ) 1 ))) ( [ ) 1 ) (DEF ( [ ) n+1 ) 6. ; 6 ` 8x; y(x; y) 2 ( [ ) t $ 9n(n 2 N ^ (x; y) 2 ( [ ) n ) (DEF- ( [ ) t ) 7. ; 7 ` leftcongruence() (ASS) 8. ; 8 ` leftcongruence() (ASS) induction base 9. 9; ` g 0 2 F (HYP) ; ` (f 1 ; f 2 ) 2 ( [ ) 1 (HYP) ; 2 ` (f 1 ; f 2 ) 2 _ (f 1 ; f 2 ) 2 (8D;! D 2 10) ; ` (f 1 ; f 2 ) 2 (HYP) , 9; 1, 7 ` (g 0 f 1 ; g 0 f 2 ) 2 (8D; $ D; ^I;! D ) , 9; 1, 2, , 9; 1, 2, 4, 7 ` (g 0 f 1 ; g 0 f 2 ) 2 ( [ ) (8D; $ D;! D 2 13) ` (g 0 f 1 ; g 0 f 2 ) 2 ( [ ) 1 (8D; $ D;! D 14 4) 16. 9; 1, 2, 4, 7 ` (f 1 ; f 2 ) 2! (g 0 f 1 ; g 0 f 2 ) 2 ( [ ) 1 (DED 15) ; ` (f 1 ; f 2 ) 2 (HYP) , 9; 1, 8 ` (g 0 f 1 ; g 0 f 2 ) 2 (8D; $ D; ^I! D ) , 9; 1, 2, 8 ` (g 0 f 1 ; g 0 f 2 ) 2 ( [ ) (8D; $ D;! D 2 18) 38

40 (DED 35) (DED 43) , 9; 1, 2, ` (g 0 f 1 ; g 0 f 2 ) 2 ( [ ) 1 (8D; $ 4, 8 D;! D 19 4) 21. 9; 1, 2, 4, 8 ` (f 1 ; f 2 ) 2! (g 0 f 1 ; g 0 f 2 ) 2 ( [ ) 1 (DED 20) , 9; 1, 2, ` (g 0 f 1 ; g 0 f 2 ) 2 ( [ ) 1 (_D , 7, 8 21) 23. ; 1, 2, 4, 7, ` 8f 1 ; f 2 ; g(g 2 F ^ (f 1 ; f 2 ) 2 ( [ ) 1! (gf 1 ; gf 2 ) 2 (DED,8I 22) 8 ( [ ) 1 ) induction step ; ` k 2 N (HYP) ; ` 8x 1 ; x 2 ((x 1 ; x 2 ) 2 ( [ ) k! (gx 1 ; gx 2 ) 2 ( [ ) k ) (InductionHYP) , 24; ` (f 1 ; f 2 ) 2 ( [ ) k+1 (HYP) , 24; 5 ` 9z(((f 1 ; z) 2 ( [ ) k ^ (z; f 2 ) 2 ( [ ) 1 ) _ ((z; f 2 ) 2 (8D; $ ( [ ) k ^ (f 1 ; z) 2 ( [ ) 1 )) D;! D 5 26) , 26; 5 ` ((f 1 ; x 0 ) 2 ( [ ) k ^ (x 0 ; f 2 ) 2 ( [ ) 1 ) _ ((x 0 ; f 2 ) 2 (9D 27) ( [ ) k ^ (f 1 ; x 0 ) 2 ( [ ) 1 ) case ; ` (f 1 ; x 0 ) 2 ( [ ) k ^ (x 0 ; f 2 ) 2 ( [ ) 1 (HYP) , 24; ` (f 1 ; x 0 ) 2 ( [ ) k (^D 29) , 24; ` (x 0 ; f 2 ) 2 ( [ ) 1 (^D 29) , 24, 25; ` (gf 1 ; gx 0 ) 2 ( [ ) k (! D 25 30) , 24, 25; ` (gx 0 ; gf 2 ) 2 ( [ ) 1 (! D 25 31) , 24, 25; ` 9z((gf 1 ; z) 2 ( [ ) k ^ (z; gf 2 ) 2 ( [ ) 1 ) (^I; 9I 32 33) , 24, 25; 5 ` (gf 1 ; gf 2 ) 2 ( [ ) k+1 (^I; 8D; $ D;! D ) , 25; 5 ` (f 1 ; x 0 ) 2 ( [ ) k ^ (x 0 ; f 2 ) 2 ( [ ) 1! (gf 1 ; gf 2 ) 2 ( [ ) k+1 case ; ` (x 0 ; f 2 ) 2 ( [ ) k ^ (f 1 ; x 0 ) 2 ( [ ) 1 (HYP) , 24; ` (f 1 ; x 0 ) 2 ( [ ) 1 (^D 37) , 24; ` (x 0 ; f 2 ) 2 ( [ ) k (^D 37) , 24, 25; ` (gf 1 ; gx 0 ) 2 ( [ ) 1 (! D 25 38) , 24, 25; ` (gx 0 ; gf 2 ) 2 ( [ ) k (! D 25 39) , 24, 25; ` 9z((gf 1 ; z) 2 ( [ ) 1 ^ (z; gf 2 ) 2 ( [ ) k ) (^I; 9I 40 41) , 24, 25; 5 ` (gf 1 ; gf 2 ) 2 ( [ ) k+1 (^I; 8D; $ D;! D; _I ) , 25; 5 ` (f 1 ; x 0 ) 2 ( [ ) 1 ^ (x 0 ; f 2 ) 2 ( [ ) k! (gf 1 ; gf 2 ) 2 ( [ ) k , 26, 25; 5 ` (gf 1 ; gf 2 ) 2 ( [ ) k+1 (_D ) 39