1. We will now study some special kinds of non-standard quantifiers. Definition 4. Let  (x),  (x) be two fixed formulae of a language L n such that x is the only free variable in both of them and they don’t have common predicates. Let M and N be two models. Then we have the following two four-fold tables: We define: N is associational better than M if a 2  a 1, d 2  d 1, c 1  c 2, b 1  b 2. Moreover, a binary quantifier  is associational if, for all formulae  (x) and  (x), all models M, N: if v M (  (x)  (x)) = TRUE, N associational better than M, then v N (  (x)  (x)) = TRUE. Obviously, the quantifier of simple association is associational: this follows by the fact that, under the given circumstances, a 2 d 2  a 1 d 1 >b 1 c 1  b 2 c 2. Church quantifier of implication is associational, too. Indeed, given a model M such that v M (  (x) => C  (x)) = TRUE, the corresponding four-fold table has a form Thus, any model N that is associational better than M has a form Thus, v N (  (x) => C  (x)) = TRUE. Quantifiers of founded p-implication are associational: if a 2  a 1  n, b 1  b 2, then a 2 b 1  a 1 b 2, therefore a 2 a 1 + a 2 b 1  a 2 a 1 + a 1 b 2 and finally, I (Today called : Basic implication)

2. Definition 5. Let  (x),  (x) be two fixed formulae of a language L n such that x is the only free variable in both of them and they don’t have common predicates. Let M and N be two models. Then we have the following two four-fold tables: We define: N is implicational better than M if a 2  a 1, b 1  b 2. Moreover, a binary quantifier  is implicational if, for all formulae  (x) and  (x), all models M, N: if v M (  (x)  (x)) = TRUE, N implicational better than M, then v N (  (x)  (x)) = TRUE. Church quantifier of implication is implicational, quantifiers of founded p-implication are implicational [proof: by a similar argument that they are associational]. However, quantifier of simple association is NOT implicational: consider the following counter example: Clearly, N is implicational better than M, and a 1 d 1  c 1 b 1 thus, v M (  (x)~  (x)) = TRUE. However, a 2 d 2 <c 2 b 2, thus v N (  (x)~  (x)) = FALSE. Therefore ~ is not implicational. Lemma. Let  be an implicational quantifier. Then  is associational. Proof. Let  be implicational and v M (  (x)  (x)) = TRUE If N is associational better than M, then N is clearly also implicational better than M, so v N (  (x)  (x)) = TRUE. Therefore  is associational, too. 

3. Theorem 2. Let  (x),  (x),  (x) be formulae, and let  be an implicational quantifier. Then  (x)   (x)  (x)  [  (x)   (x)] is a sound rule of inference. Proof. Let M be a model such that v M (  (x)   (x)) = TRUE and We realise that a 1 = #{x | v M (  (x)) = v M (  (x)) = TRUE}  #{x | v M (  (x)   (x)) = v M (  (x))=TRUE} = a 2 and b 1 = #{x | v M (  (x)) = v M (  (x)) = TRUE }  #{x | v M (  (x)   (x)) = v M (  (x)) = TRUE} = #{x | v M (  (  (x)   (x))) = v M (  (x)) = TRUE }. Thus, we have Since is implicational we conclude that v M (  (x)  [  (x)   (x)]) = TRUE, too.  In the proof we used an obvious fact: for all implicational quantifiers , if v M (  (x)   (x)) = TRUE and Then, for any other formulae  *(x),  *(x) such that we have v M (  *(x)   *(x)) = TRUE, too. We will use this fact in the next Theorem, too.

4. Theorem 3. Let  (x),  (x),  (x) be formulae, and let  be an implicational quantifier. Then [  (x)   (x)]   (x)  (x)  [  (x)   (x)] is a sound rule of inference. Proof. Let M be a model such that v M( ([  (x)   (x)]   (x)) = TRUE and We realise that a 1 = #{x | v M (  (x)   (x))) = v M (  (x)) = TRUE}  #{x | v M (  (x)   (x))) = v M (  (x)) = TRUE} = a 2 and b 1 = #{x | v M (  (x)   (x))) = v M (  (x)) = TRUE} = #{x | v M (  (x)) = v M (  (x)   (x)))= TRUE } = #{x | v M (  (x)) = v M (  (  (x)   (x)))= TRUE } = b 1. Thus, we have in the model M Since is implicational we conclude that v M (  (x)  [  (x)   (x)]) = TRUE, too. Theorem 4. Let  (x) and  (x) be formulae, and let ~ be the simple association quantifier. Then  (x) ~  (x)  (x) ~  (x) and  (x) ~  (x)  (x) ~  (x) SYM NEG are sound rules of inference. Exercises 13. Prove Theorem 4. 14. Prove that Theorem 4 does not hold for founded p-implication quantifiers.

5. We have introduced deduction rules (i.e. sound rules of inference) mainly to minimise the amount tautologies, called hypothesis i.e. outputs in practical GUHA data mining tasks. For example, Theorem 1 says that if  is an implicational quantifier and  (x)   (x) is true in a given model M, so is  (x)  [  (x)   (x)] true. Thus, we do not have to print  (x)  [  (x)   (x)] as a data mining result. Next we will study some other useful deduction rules. Consider elementary conjunctions EC and elementary disjunctions ED, i.e. open formulae of a form   P 1 (x) ...   k P k (x) and   P 1 (x) ...   k P k (x), where  i :s are either ‘  ’ or empty sign. For example, P 1 (x)   P 5 (x) and P 1 (x)  P 3 (x)   P 5 (x) are EC’s P 2 (x)   P 3 (x)  P 4 (x) and P 2 (x)  P 4 (x) are ED’s. Denote EC’s or T by symbols        … (maybe empty) and denote ED’s or  by symbols     2   3  … (maybe empty). Definition 6. An elementary association is a sentence of the form , where  is a quantifier and ,  are disjoint, i.e. have no common predicates. Let     and    2 be elementary associations. We say that     results from    2 by specification if either     and    2 are identical or there is an ED  0 disjoint from  1 such that  2 and  0   1 are logically equivalent (i.e. have always the same truth value) and   is logically equivalent to     0. [We say also:     despecifies to    2 ] Example. P 1 (x)  P 3 (x)   P 5 (x)  P 2 (x)  P 4 (x) results from P 1 (x)   P 5 (x)  P 2 (x)   P 3 (x)  P 4 (x) by specification [indeed,  0 =  P 3 (x)]

6. Moreover, we say that     results from    2 by reduction [or     dereduces to    2 ] if   is   and  1 is a subdisjunction of  2 Example.[P 1 (x)   P 5 (x)]  [P 2 (x)   P 3 (x)  P 4 (x)] results from [P 1 (x)   P 5 (x)]  [P 2 (x)   P 3 (x)  P 4 (x)   P 6 (x)  P 7 (x)] by reduction [indeed,  2 =  P 6 (x)  P 7 (x)]. We introduce the despecifying-dereduction rules (SpRd-rules); they are of the form  22 where     results from    2 by successive reduction and specification, i.e. there is a ED  3 (a sub-ED of  2 ) such that    1 despecifies to    3 and    3 dereduces to    2. Example. [P 1  P 3   P 5 ]  [P 2  P 4 ] despecifies to [P 1   P 5 ]  [P 2   P 3  P 4 ] and [P 1   P 5 ]  [P 2   P 3  P 4 ] dereduces to [P 1   P 5 ]  [P 2   P 3  P 4   P 6  P 7 ] Thus, we have an SpRd-rule [P 1  P 3   P 5 ]  [P 2  P 4 ] [P 1   P 5 ]  [P 2   P 3  P 4   P 6  P 7 ] Theorem 5. For any implicational quantifier , SpRd-rules are sound rules of inference. Proof. In a same manner than Theorem 4 and Theorem 5.ž Remark. Theorem 5 can be reformulated in the following way: whenever        2 is a SpRd-rule, then  (     )  (    2 ) [i.e. is a tautology].

7. Theorem 5. SpRd-rules are transitive, that is, if     and    2 then    1    2    3    3 Proof. The result is obvious as soon as we realise that the order of despecification and dereduction can be reverted, i.e. (       )    dereduces to (       )   (     2 ) despecifies to    (     2    ) despecifies to    (      ) dereduces to We introduce two more types of quantifiers:  p - equivalence quantifiers, where 0 < p  1. (today: Basic equivalence) For any model M, v(x (  (x)  p  (x))) = TRUE iff (a+d)  p(a+b+c+d), in particular, in a model M such that b+c > 0, v(x (  (x)  p  (x))) = FALSE  p - equivalence quantifiers, also called  -double quantifiers, where 0 < p  1. (Basic double implication) For any model M, v(x (  (x)  p  (x))) = TRUE iff a  p(a+b+c), in particular, in a model M such that d > 0, v(x (  (x)  p  (x))) = FALSE Exercises. Prove that 15.  p - equivalence quantifiers and 16.  p - equivalence quantifiers are associational  II