Logical Algebra 1. By Vern Crisler Copyright 2000; 2013

Transcription

1 Logil Alger 1 By Vern Crisler Copyright 2000; Introdution 2. Addition in Logil Alger 3. Bsi Rules for Logil Alger 4. Exerises: Solve Using the Rules 5. Sommers & Term Logi 6. Logil Arithmeti & Logil Alger 7. The Logi of Complex Terms 8. Vlid Argument Forms 9. Multi-termed rguments 10. Duplited Terms 11. Frtionl Logi An Option 1. Introdution Logil lger works with either simple terms suh s x, or y, or z, or with omplex terms suh s xy, or yz, or ze. In textook logi, the usul wy of proving n rgument is y stting logil opertions nd justifitions elow the onlusion line to show wht logi rule is involved. Logil lger uses n ddition method of finding solutions rther thn geometry method. Speil Note: logiins often use different symols to express logil opertions. For instne negtion is represented y minus sign or tilde ~ or sign. For implition, it n e either the rrow or the sign, or sometimes <. We ll disuss these lter, ut for right now, plese note tht in logil lger we re using the sign for therefore rther thn the usul sign. Nowdys logi of terms is expressed using vriles suh s x, y, z,,,, e, m, n, or A, B, C, et. A logi of propositions is lmost lwys expressed with suh letters s p, q, r, nd so on. In terms of deep struture oth term logi nd propositionl logi re identil. You my rememer tht in logil rithmeti, we found tht polrizing the middle terms symolizing them s plus or minus helped us nel them so tht we ould derive vlid onlusion. In this pper, we will use the ddition method insted. 2. Addition in Logil Alger: We introdue here the ddition wy of symolizing logil prolems. Insted of the trditionl sign, the + sign for disjuntion will e used, nd we set up the prolem s though it were n ddition prolem in fmilir grdeshool mth: 1 ~ y + z 2. + y + ~x 0 + z + ~x 1

2 The logil opertions involved in ddition will e explined elow, ut for now just notie tht the ddition method involves dding the symols, nd whtever is left goes to the solution line. As you n see, the y s dded together nel nd the other letters re just y themselves, so they reh the onlusion. The result of the ddition is: z + ~x. Also, putting zero on the onlusion line is optionl. It ould just s esily e left lnk, nd we will leve it out from now on. The onlusion z + ~x is equivlent to the sme expression using the vel sign, z ~x. The remining expression n e re-symolized in textook nottion s either ~x z, or x z or ~z ~x. The lgeri symolism ove is symolism for terms or onepts. As noted, we would use p, q, or r if we were symolizing propositions. Also, s noted, term n e omplex term mde up of more thn one vrile. For instne, while x is simple term nd y is simple term, (ze) is lso term, nd (f g) is lso term. Both re omplex terms. In the ove ses, the omplements of terms suh s (ze) or (f g) re other omplex terms tht re negted, (ze) or (f g). This would show tht the omplex terms re tully linking terms tht n e nelled from the rgument. (We ll tlk more out this lter under the setion on the logi of omplex terms.) Exmple: Invlidity In logil ddition, the following is n exmple of n invlid rgument, with x, y, nd z stnding in this se for sujet, middle, nd predite terms: 1. x y 2. z y z x Here is the expression in ddition form, showing the lk of polriztion: 1. ~x + y 2. + y + ~z no onlusion As you n see, the y s do not nel eh other out euse they re oth positive, thus undistriuted. Without distriution there n e no onlusion. You n hek the rgument in nother wy, too. Use the sme set-up ut dd n extr line ontining the denil of the onlusion. In this se, we dd line ~x + y 2. + y + ~z 3. x + ~z 4. ~x z & The denil of z x is z~x. If the onlusion line results in zero s for ll letters, it s vlid. If ll terms hven t een nelled, it s not vlid. In this rgument. when we dd z~x to the rgument on line 4, only two of the pths to the onlusion re loked. Beuse one pth still remins open, the originl onlusion is invlid. 1 Exmple: y s nd z s nd x s: Now, let us see how universl 1 This method of heking for vlidity or invlidity is somewht similr to the tree methods devised y Lewis Crroll in his Symoli Logi, Brtley ed., pp. 279ff; or y Rihrd C. Jeffrey, Forml Logi: Its Sope nd Limits, 1967, pp. 63ff. 2

3 sttement n e omined with simple onjuntive expression: 1. y x 2. z & y z & x In logil ddition: 1. ~y + x 2. +y & z x & z The presene of the onjuntion sign & in one of the premises requires tht the onlusion will e prtiulr. The onjuntion sign & lwys trumps the disjuntion sign + in the rgument nd in the onlusion. If & is in the premises, it must go to the onlusion, whih is then rendered onjuntive onlusion. If we wnt to hek vlidity y denying the onlusion line, we d set up the rgument in this wy: 1. ~y + x 2. +y & z 3. x & z 4. ~x + ~z The denil of x & z is ~(x & z), whih is ~x + ~z. The result is tht ll terms re nelled s illustrted on line 5. With zeroes on ll rows, this mens the originl rgument is vlid. Exmple: Per Aidens Is the following rgument vlid? 1. y x 2. y z z & x If we try to solve it with ddition we find tht the rgument ppers to e invlid. Why is this so? Here is wht it looks like: 1. ~y + x 2. ~y + z? As you n see, we hve n unpolrized middle term. In this form the rgument is not vlid. However, we n onvert one of the premises y wy of per idens resulting in z & y. As you n see in premiss 2 elow, we hnged y z to its prtiulr form, whih is z & y, or y & z: 1. ~y + x 2. +y & z x & z The onlusion x & z is onvertile simply s z & x. Exmple: Negtivity Is the following rgument vlid? 1. y ~z 2. y ~x ~x & ~z In ddition form: 1. ~y + ~z 2. ~y + ~x? Here, the middle terms re not polrized. Cheking for vlidity or invlidity y denying the onlusion in the ddition method we hve: 3

4 1. ~y + ~z 2. ~y + ~x 3. ~z ~x& 4. z + x & As you n see from line 5, not ll pths hve een loked, mening there is no vlid onlusion. However, vlid rgument n e derived if one of the originl premises is onverted per idens, hene: 1. ~y + ~z 2. y & ~x 3. ~z & ~x 3. Bsi Rules for Logil Alger In the ddition method of solving logil prolems the si rules for determining vlidity or invlidity re summrized s follows: (1) As long s linking terms re polrized, universl sentene omined with nother universl sentene leds to universl onlusion. (2) As long s linking terms re polrized, universl sentene omined with prtiulr sentene leds to prtiulr onlusion. (3) A prtiulr sentene omined with nother prtiulr sentene results in n invlid onlusion (terms nnot e polrized). These rules will help you use the ddition method without introduing invlidity into the rgument. hek results y denil of the onlusion. In tht se, ll you need to do is dd the denil to the premises, then hek to see if ll pths to the new onlusion re loked. If so, the onlusion is vlid. If ny pth remins open, the onlusion is not vlid. Historil Note: Gottfried Leiniz ws the first to mke the ttempt to develop logil lulus, nd his work ntiipted muh tht ws to follow, inluding representing logi y lines, y irles, nd y numers (f., Frege, Venn, nd Godel respetively). 4. Exerises: Using the Rules: Determine if the following rguments re vlid or invlid: (1) y z x y x z (2) y ~z x y x ~z (3) y z xy xz (4) y ~z xy x~z (4) This is n extr rule if you wish to 4

5 (5) z ~y x y x ~z (6) z y x ~y x ~z (7) z ~y xy x~z (8) z y x~y x~z (9) y z y x xz (10) yz y x xz (11) y z yx xz (12) y ~z y x x~z (13) y~z y x x~z (14) y ~z yx x~z (15) z y y x xz (16) z y y ~x x ~z (17) zy y x xz (18) z ~y y x x~z (19) z ~y yx x~z Solutions: The 19 vlid syllogisms I hve used the forms of the 19 vlid syllogisms of trditionl logi here in order to help the reder () to memorize the rules of logil lger y working through the prolems using the ddition method; nd () to see n isomorphism etween syllogisti logi on the one hnd, nd term or propositionl logi on the other. We will see lter, however, tht the methods of solution differ. Here re the solutions: 5

6 1. Brr: y z x y x z 1. ~y + z 2. +y + ~x z + ~x For the onlusion, z + x is the sme s x + z, whih is the sme s x z. 2. Celrent y ~z x y x ~z 1. ~y + ~z 2. +y + ~x ~z + ~x Conlusion z + x, or ~x + ~z, whih is x z 3. Drii y z xy xz 1. ~y + z 2. +y & x z & x Note tht rule 2 is stisfied. Also, z & x is the sme s x & z. Additionlly, the & sign in the seond premiss hs dominne in the onlusion over the + sign of the first premiss. The stndrd geometri method would solve this rgument y simplifying the seond premiss, from x & y to y, whih then sets up modus ponens to derive z. One you hve z, then x nd z n e derived y onjuntion, hene x & z. 4. Ferio y ~z xy x~z 1. ~y + ~z 2. +y & x ~z & x Rule 2 stisfied. 5. Cesre z ~y x y x ~z 1. ~z + ~y 2. +y + ~x ~z + ~x In terms of rule 1, we re le to polrize the linking terms, ~y nd +y. This enles us to drw the onlusion is x + z, or x z. Using the stndrd geometri method, the first premiss of the rgument would e hnged to y ~z through trnsposition. From x, we would go to y, then from y we would go to ~z. This would e modus ponens nd from there we ould derive the onlusion x z. 6

7 6. Cmestres z y x ~y x ~z 1. ~z + y 2. ~y + ~x ~z + ~x The onlusion is x + z, whih is x z. Using the stndrd geometri method, the seond premiss would e trnsposed, whih sets up modus ponens to solve the rgument. 7. Festino z ~y xy x~z 1. ~z + ~y 2. x & y ~z & x The onlusion z & x is onvertile to x & z. Rell tht x in the seond premiss of the rgument is ssoited with the onjuntion sign &. As we stted erlier, in deriving the onlusion the & sign trumps lterntion (either the + sign or the v or vel sign). Using the stndrd geometri method, the first premiss is trnsposed to y ~z. The seond premiss is simplified from x & y to just x, llowing the use of modus ponens to derive ~z. Conjuntion rings ~z nd x together in the onlusion of the rgument. 8. Broko z y x~y x~z 1. ~z + y 2. x & ~y ~z & x Conlusion, x & z. In the stndrd geometri method, the first premiss would e trnsposed to ~y ~z, the seond would e simplified to ~y, whih sets up modus ponens. The onlusion would e derived using the onjuntion of x nd ~z. 9. Drpti y z y x xz 1. ~y + z 2. +y & x z & x The onlusion z & x n e onverted to x & z.in its originl form, the rgument is invlid under logil lger simply euse the linking term y nnot e polrized. If we try to use lterntion on oth premises, we end up with two ~y s in our premises. However, if the seond premiss is onverted to its per idens form, going from y x to xy or to x & y, then modus ponens n e used y omining the first premiss with the new onjuntive expression, nd the resulting rgument would e vlid. 7

8 10. Dismis yz y x xz 1. y & z 2. ~y + x z & x The onlusion z & x n e onverted to x & z. 11. Dtisi y z yx xz 1. ~y + z 2. +y & x z & x The onlusion z & x n e simply onverted to x & z. 12. Felpton y ~z y x x~z 1. ~y + ~z 2. +y & x ~z & x The onlusion z & x n e onverted to x & z. One of the premises would hve to e onverted for this rgument to e vlid. Hene, y x ws onverted to x & y, or y & x, nd this n e simplified to y, whih n then e used in modus ponens. 13. Bokrdo y~z y x x~z 1. y & ~z 2. ~y + x x & ~z Linking term nels, nd mpersnd dominne requires tht the onlusion e prtiulr. 14. Ferison y ~z yx x~z 1. ~y + ~z 2. +y & x ~z & x Conlusion, z & x or x & z. 15. Brmntip z y y x xz 1. ~z + y 2. ~y + x ~z + x 8

9 The full onlusion is ~z + x or z x. However, the onlusion would e z & x or x & z. This involves moving from the universl to the prtiulr whih is vlid move under logil lger. 16. Cmenes z y y ~x x ~z 1. ~z + y 2. ~y + ~x ~z + ~x The onlusion is x + z, or x z. In the stndrd geometri method, oth premises would hve to e trnsposed, nd this sets up modus ponens to derive the onlusion. 17. Dimris zy y x xz 1. z & y 2. y + x z & x The onlusion z & x is onvertile into x & z. 18. Fespo z ~y y x x~z 1. ~z + ~y 2. x & y x & z One premiss hs to e onverted per idens in order for the rgument to e vlid. 19. Fresison z ~y yx x~z 1. ~z + ~y 2. x & y x & ~z The onlusion is onjuntive euse one of the premises is onjuntive. 5. Sommers & Term Logi The lgeri formt used in this pper ws inspired y the term logi of Fred Sommers, ut is relevntly different. The lgeri logi given in this essy is more of middle ground pproh etween Boole nd Sommers. It still lrgely retins Boolen-type struture, wheres Sommers hs moved frther wy from Boole. In my opinion, Sommers plus nd minuses seem to e ering gret del of symoli weight. Clifton MIntosh sys with respet to Sommers version of term logi, TFL: TFL [trditionl forml logi] ontins the full logi of propositions, ll of syllogisti resoning, portion of the logi of identity, 9

10 nd some of the logi of reltions. All of this is represented with terms nd the two lgeri signs + nd. This requires + nd to do mny jos. 2 From my perspetive, they re doing ltogether too mny jos. In nother ontext, Sommers sys, The logi [of terms] will mke use of stndrd rithmeti opertions to test for vlidity. As with Boole, rithmeti is epted s given. There re no rules of inferene. Vlidity is rithmetil. 3 We would point out, however, tht logi is not relly rithmetil, not relly mthemtis. Even though we n orrow symols or words from mthemtis to use in logil opertions, these re only ment metphorilly. It is fundmentl mistke to think tht logi n e turned into mthemtis or tht mthemtis n e turned into logi. There is only n nlogy etween the two, not n identity. Contrry to Sommers, the rules of logil inferene re still opertive in oth term logi nd Boolen logi, even if not lwys up front nd enter. There is simply no wy to get long without them, no mtter how mthemtil one tries to e. In the following, Sommers logil symolism is presented using the trditionl nmes of the vlid syllogisms. As n e seen, Sommers is le to polrize the M s in mny of the equtions nd rrive t the orret onlusion, ut nnot do so for the ones in oxes: 2 Clifton MIntosh, Appendix F, in Fred Sommers, The Logi of Nturl Lnguges, p Fred Sommers, The Clulus of Terms, in ed. G. Engelretsen, The New Syllogisti, p. 14. Figure 1 1. Brr, M + P S + M = S + P 2. Celrent, M P S + M = S P 3. Drii, M + P + S + M = + S + P 4. Ferio, M P + S + M = + S P Figure 2 5. Cesre, P M S + M = S P 6. Cmestres, P + M S M = S P 7. Festino, P M + S + M = + S P 8. Broko, P + M + S M = + S P Figure 3 9. Drpti, M + P M + S =? 10. Dismis, + M + P M + S = + S + P 11. Dtisi, M + P + M + S = + S + P 12. Felpton, M P M + S =? 13. Bokrdo, + M P M + S = + S P 14. Ferison, M P + M + S = + S P Figure Brmntip, P + M M + S =? 16. Cmenes, P + M M S = S P 17. Dimris, + P + M M + S = + S + P 18. Fespo, P M M + S =? 19. Fresison, P M + M + S = + S P Sommers rule for determining vlidity is tht: An inferene is vlid if nd only if (i) the sum of the premises equls the onlusion, nd (ii) the numer of prtiulr onlusions equls the numer of prtiulr premises. 4 If purely rithmetil proedure is used with the ove stted rules, four of the rguments of trditionl logi will not work. Also, s we shll see, Sommers needs more thn the ove rules in order for his proedure to work. First of ll, (9) Drpti, (12) Felpton, nd (18) Fespo hve unpolrized middles, nd the vlid onlusions for these 4 Fred Sommers, quoted in G. Engleretsen, Something to Rekon With: The Logi of Terms, p

11 rguments nnot e derived in ny strightforwrd mnner with Sommers term logi. Respetively, the onlusions should hve een, (9) + S + P (12) + S P (18) + S P But the unpolrized middles prevent summtion (or nelltion of the middles), thus violting rule (i), even though these re vlid rguments in trditionl logi. The suggestion hs een mde tht for 9, 12, nd 18, we dd tutologil premiss, for instne, m + m for the following: 1. M + P 2. M + S 3. + M + M 4. + S + P This onlusion + S + P is the vlid onlusion of Drpti, ut onsider the following: 1. M + P 2. M + S 3. + M + M 4. S + P Here we hve two universl premises, now with the ddition of the tutologil premiss. We sueed in nelling ll the M s, ut the onlusion is invlid euse there were two negtive premises in the originl rgument. Apprently, the ddition of the extr premiss n only work for positive rther thn negtive premises, so rule out two negtive premises hs to e dopted in ddition to Sommers first rule. following to derive onlusion? 1. + M + P 2. + M + S 3. M M 4. +S + P Oviously, while the dditionl premiss polrizes the M s, the new premiss is tully ontrdition, i.e., No M is M. Moreover, in ordne with Sommers rule 2, the onlusion is invlid euse we hve two prtiulr premises in the originl rgument. Thus, in ddition to purely rithmetil proedure, more rules hve to e dopted, nmely, () new tutologil premiss n only e positive nd prtiulr in form, nd () two prtiulr premises do not result in vlid onlusion. So muh for the lk of ny rules for inferene. There is nother prolem with Sommers method. The rgument (15) Brmntip hs polrized middles under Sommers rule 1, ut this does not relly help, nor do the other rules of preserving vlidity help. Arithmetilly, we derive the onlusion s: (15) + S P But the orret onlusion is (15)* + S + P. The lim tht we n dd tutologil premiss does not work in the se of Brmntip, sine the prolem is not with the middle terms ut with negtive term opertor on one of the extreme terms. Consider: Why n t we dd new premiss to the 11

12 1. P + M 2. M + S 3. + M + M 4. + S P The M s do not nel out, nd the vlid onlusion should hve een + S + P. Wht rule should e dopted for this diffiulty? Here is the solution of Brmntip using our own symolism: 1. P + M 2. M + S 3. P S Or P + S, whih is All P is S. We n then use onversion per idens to derive S & P, or Some S is P. Cn this e done with Sommers nottion? Consider gin Brmntip using Sommer s rithmetil proedure: 1. P + M 2. M + S 3. P + S Here we should note tht the position of the terms in the onlusion is just s importnt s ny rule out the sum of the premises. If S nd P were reversed, i.e., (+ S P), we would hve Some S is not P rther thn the orret All P is S. In Boolen-type symolism, the order of the signs does not mtter. For instne, the form P + S is equivlent to S + P. However, in Sommers nottion, the position of the terms is ruil for the onlusions in Ferio, Festino, Broko, Felpton, Bokrdo, Ferison, Brmntip, Fespo, nd Fresison. If we were to reverse the + S P onlusions of these rgument to P + S, we would not hve the sme onlusions. 5 So, in ddition to the rules lredy mentioned, we would need nother rule stting tht the S, P form should e mintined. However, the only wy this ould e done for Brmntip is y onversion per idens, reduing P + S to S + P, nd y this preserving the S, P form of the onlusion. Thus, we hve four perfetly good syllogisms tht do not work in Sommers lger of terms, nd for two different resons, one the lk of polriztion in 9, 12, nd 18, nd the other simply euse the efore the predite nnot e hnged to + nd rought over to the onlusion in 15. Therefore, in order for them to work, Sommers needs to dd more rules to his rithmetil proedure, nd his lim tht the rules of inferene re not opertive in his lger of terms ws ertinly misdireted. The differenes etween Sommers symolism nd our symolism n e seen in the following omprtive hrt: Tle 3: Logil Alger ompred English Sommers Logil Alger 1. All S is P (S) + (P) (S) + (P) 2. No S is P (S) (P) (S) + (P) 3. Some S is P + (S) + (P) (S) & (P) 4. Some S is not P + (S) (P) (S) & (P) In Logil Alger, the + sign represents disjuntion (lterntion). It is merely repling the vel or v sign in mny representtions of logi found in 5 This type of prolem ws notied y John Venn in nother onnetion, f. Symoli Logi, p

13 textooks. It is the sme + sign found in ooks on digitl or Boolen logi nd mens the very sme thing, tht is or. In the Logil Alger olumn, the + or & stnding in etween the S nd P represent the logil onnetives of disjuntion nd onjuntion, not term opertors. I hve pled prentheses round the terms in the Logil Alger olumn so there won t e ny onfusion with Sommers use of plus nd minus signs in his symolism. Notie tht in Sommers 2 nd 4, Logil Alger hs the + nd & signs oming in the middle ut not stnding for term negtion. Tht is wht the is next to the prentheses, term opertor. Sommers hs his + nd doing doule duty, stnding for is in 1 nd 3, nd for un in 2 nd 4. Moreover, they re not reversile, s we hve seen. For ll these resons, our onlusion is tht Sommers term logi, while novel, suggestive, nd refreshing, does not offer relly perspiuous nd esy to use method of solving logi prolems. 6. Logil Arithmeti & Logil Alger Compre this syllogism in logil lger with its solution y logil rithmeti: z~y x y Solving y ddition we hve: 1. ~x + y 2. ~y & z ~x & z If we trnslte this into logil rithmeti here is wht we hve: ZoY + XY =? As you n see, neither the first or seond premises distriute the middle term, nd if we hnge ZoY to its first figure form (ZoY; Zi~Y; ~YiZ; ~Yo~Z), we would hve unpolrized middle terms: ~Yo~Z + Xe~Y =? In this se, oth Y s hve negtive term opertors. Is the rgument vlid under logil lger, ut invlid under logil rithmeti? Let s look t nother exmple: y z x ~y Solving y ddition we hve: 1. ~y + z 2. ~y + ~x z & ~x In logil rithmeti the rgument would e YZ + XeY =? As you n see, this rgument would e invlid in the first figure sine the middle terms nnot e polrized. Is this nother onflit etween logil rithmeti nd logil lger? The short nswer is no, there is no onflit. This is euse the equtions in logil lger nnot relly e heked y logil rithmeti (or the rules of the syllogism). The reson is euse term nd propositionl logi re indifferent with respet to figure nd mode of syllogism. In the syllogism, rules of vlidity hve referene to the figure (1 st through 4 th ), nd mode (positive or negtive). These re ll importnt in deriving vlid onlusion in syllogism, or in heking the vlidity of syllogism. On the other hnd, propositionl logi nd term logi do not relly hve figures nd modes, nd thus 13

14 hve different set of rules for vlidity to follow. 7. The Logi of Complex Terms: For the logi of omplex terms we use some stndrd logi symols nd lso introdue ouple of new ones. I divide one premiss from nother with the semiolon sign. Hene (xy); (z + e); g; ~m; y z; et., re seprte premises. For implition we use the tried nd true horseshoe symol. It derives ultimtely from the letter C for ontined in ut logiins love to use kwrd letters, so they turned the C round so tht it looks like kwrd C, nd thus we got our horseshoe symol. For disjuntion, we use the + sign rther thn the norml sign. For onjuntion, we usully leve out ny sign, nd only provide one when the premises re prt of prolem in logil ddition. In tht se we use the & sign. We ould use the dot or the sign, ut everyone knows wht & mens, so we ll use it insted. For negtion nd equivlene, we ll ontinue to use the ~ sign nd the sign respetively. Conlusions of rguments egin with therefore nd sine logiins don t like to use the implition sign to signl the onlusion, they ve ome up with the sign. In ple of this, I prefer the old nd dventurous symol, whih reples the shy nd retiring sign. Here is omprison of the symolism of logil lger with ommon textook symols: Tle 1: Logi Symols Nme Logil Alger Textook Textook 1. Implition 2. Disjuntion (inlusive) + 3. Disjuntion (exlusive) 4. Conjuntion & 5. Negtion ~ 6. Equivlene 7. Therefore 14

15 8. Vlid Arguments Forms We should tke the time to stte some of the more generl lws of logil lger. These lws n e found in ny logi textook, ut the following disussion shows wht they look like when used long with the ddition method (see next pge): Tle 2: Rules of Inferene 1) Modus Ponens x y; x; = y 1. ~x + y 2. +x y 2) Modus Tollens x y; ~y; = ~x 1. ~x + y 2. +~y ~x 3) Hypothetil Syllogism x y; y z; = x z 1. ~x + y 2. ~y + z ~x + z 4) Disjuntive Syllogism x + y; ~x; = y 1. x + y 2. ~x + y 5) Construtive Dilemm (x y)(r z); x + r; = y + z 1. ~x + y 2. ~r + z 3. x + r y + z 6) Destrutive Dilemm (x y)(r z); ~y + ~z; = ~x + ~r 1. ~x + y 2. ~r + z 3. ~y + ~ z ~x + ~r 7) Simplifition xy; = x (xy)(ze); = (xy) (x + y)(z + e); = (x + y) 8) Conjuntion x; y; = xy (xy); (ze); = (xy)(ze) (x + y); (z + e); = (x + y)(z + e) 9) Addition x; = x + y (xy); = (xy) + (ze) (x + y); = (x + y) + (z + e) 15

16 The following re logilly equivlent sttements: 10) De Morgn s Theorems ) ~(xy) (~x + ~y) ) ~(x + y) (~x~y) 11. Commuttion ) x + y y + x ) xy yx 12) Assoition ) x + (y + z) (x + y) + z 13) Distriution ) x(y + z) (xy) + (xz) ) x + (yz) (x + y)(x + z) 14) Doule negtion x ~ ~x 15) Trnsposition x y ~y ~x ~x + y y + ~x 16) Implition & Disjuntion x y ~x + y 17) Equivlene ) (x y) (x y)(y x) ) (x y) (xy) + (~x~y) 18) Exporttion (xy) z x (y z) ~(xy) + z = ~x + (~y + z) 19) Tutology ) x (x + x) ) x (xx) In ddition to these re the following rules: 20. Rules of 1 nd 0: x(0) 0 x + 0 x x x(1) x x(1) + y(0) x 21. Contrdition x~x 0; x(y~y) 0 y~y 0; y(x~x) 0 16

17 22. Exluded Middle (x + ~x) Development: 6 x x + x (x + x)(1) (x + x)(y + ~y) xy + x~y + xy + x~y 23. Development (ontinued) x x + x (x + x)(1)(1) (x + x)(y+~y)(z + ~z) xyz + xy~z + x~yz + x~y~z 24. Asorption ) x + (xy) x ) x(x + y) x ) x + (~xy) x + y d) x(~x + y) y e) (xy) + (x~y) x f) (x + y)(x + ~y) x 6 This will e disussed t greter length in Logil Alger 2. 17

18 9. Multi-termed Arguments Consider the following rgument: 1. ~(d + n) m 2. (k~) 3. ~(l ~e) ~m 4. ~k (~d + ~h) 5. ~ ~(~hl) 6. ~(h m) 7. n ( + ~) 8. ~( m) ~e Is there n esy wy to solve this? The nswer is yes, ut efore solving prolems like the ove, plese keep priniple #12 in mind, the onept of ssoition. This priniple tells us tht we n tke ny omintion of disjuntions nd group them freely. Hene, n expression suh s x + y + z n e omined either s: (x + y) + z, or (z + x) + y, or x + (y + z) This goes lso for onjuntions suh s xyz: (xy)z, or (xz)y, or x(yz) The seond priniple to keep in mind is #16, where the expression x y is equivlent to the expression ~x + y. The third priniple to follow is lled De Morgn s Theorems where you n negte onjuntion to rrive t disjuntive terms nd vie vers. This is priniple #10 nd you might wnt to prtie it few times if you don t understnd how it works. Lerning to swith k nd forth etween onjuntion nd disjuntion is so si to mny logil opertions tht it s good ide to get it under your elt efore going forwrd. Now if you n keep these three priniples in mind, we n put the ove rgument with eight premises into the ddition method to rrive t onlusion. We first onvert the expressions into disjuntions using the three priniples lredy noted: 1. d + n + m 2. ~k ~l + ~e + ~m 4. ~d + ~h + k 5. h + ~l + 6. ~h + m ~ + ~n 8. ~ + m + ~e ~l + ~e + Now tht we ve got ll of them into disjuntive form, we n set them up for logil ddition. I use ddition for this rgument ut if nyone wnts to set up these longer rguments into logi mtrix or logi grid, they re welome to do so. In logi mtrix, eh term gets its own olumn. The vertil lines represent disjuntion nd tke the ple of the + sign in logil ddition. Eh horizontl line represents onjuntion, nd orresponds to the line numers in logil ddition. The linking terms re nelled s usul, nd the remining terms go down to the onlusion. Here is the stked rgument under ddition, nd lso n exmple of logil mtrix: 18

19 1. d + n + m 2. + ~k ~m + ~l + ~e 4. ~d + k + ~h ~l + h 6. + m + ~h ~n + + ~ 8. + m + ~ + ~e + + ~l + ~e And here is the logil mtrix: 1. d n m 2. ~k 3. ~m ~l ~e 4. ~d k ~h 5. ~l h 6. m ~h 7. ~n ~ 8. m ~ ~e ~l ~e hek ~ & l & e & OK Beuse we were le to group freely, we n dd up ll the polrized terms (d, ~d, h ~h, et.). The onlusion is + ~l + ~e. We n rerrnge the onlusion nywy we wnt, for instne ~l + ~e +. This n then e onverted to ~(l & e) +, nd even further into (l & e). We n lso hek the vlidity of the rgument y denying the onlusion. Sine ll rods led to zero, the rgument is vlid. 7 The ove eight premises re lso trnsltle into logil lger, s follows: (~d~n) m + (k~) + (le) e m + (dh) k + (~hl) (h~m) + (~) e n + (~m) e e 7 It should e noted tht one need not use ll 8 lines for the mtrix. Only two lines re relly needed: one for the positive letter nd the other for the negtive letter. 19

20 And from here, it is firly simple to get it k into Lewis Crroll s originl sorites prolem, given in his prtiulr logil nottion (or s est s I n render it) 8 : d n 1m 0 k 1 0 le 1 m 0 dh 1 k 0 h l 0 hm 1 0 n 0 m 1e 0 Now tht we ve seen how to solve multitermed rguments, the next thing to do is to solve rguments with omplex terms. Here s textook exmple, showing how omplex term n e onstruted : (xy) + (ze); x ~x; = z In solving this rgument, rell tht x ~x is the sme s (~x + ~x) whih is the sme s ~x y rule 19. Also, y mens of disjuntive ddition we n uild linking term. In this se ~y n e dded to ~x to give us (~x + ~y), or onjuntively ~(xy). This is then polrized with respet to (xy) nd n e nelled. Also, the onlusion ze n e simplified y rule 7 to z. 1. xy + ze 2. ~(xy) + ze As noted, the term ze n e simplified to z, whih is the sme onlusion s the originl rgument. This is why those rules ove re so importnt. They help us to uild omplex terms from simple terms. Continuing with some more textook exmples: x y; y z; z e; = x + e Trnsforming the premises into their equivlent disjuntive forms we hve: 1. x + y 2. y + z 3. z + e ~x + e This ws simple modus ponens rgument esily solved y logil ddition. Here is n rgument using mteril equivleny: 1. (x y) + ~z 2. e m 3. m n 4. n r 5. [(xy) + (xy)] + ~r e z In logil ddition: 1. (x y) + z 2. + e + m 3. m + n 4. n + r 5. (x y) + ~r z + ~e The onlusion is ~z + ~e, whih n e onverted to ~e + ~z, nd further onverted to e ~z. The rgument is vlid. Note tht we were le to onvert premiss 5 from ~(xy + ~x~y) to ~(x y) y using Rule 17. Here s n exmple using onstrutive dilemm: 1. x + (y + z) 2. (y e) & (z m) 3. (e + m) (x + z) 4. x z 8 Crroll, p

21 In the seond premiss we hve (y e) & (z m). Sine we lso hve x in the fourth premiss, the remining disjunt of the first premiss is (y + z) nd this goes to the seond premiss s onstrutive dilemm. The result would e written (y + z) (e + m). In the third premiss, the (e + m) (x + z) is trnsformed into its equivlent expression, (e + m) + (x + z). All of this gives us linking terms, inluding omplex linking terms, whih n e nelled to rrive t the vlid onlusion of z. Note tht ~x stnding y itself in the fourth premiss n e used nywhere nother x ours, e.g., on line 5 elow: 1. x + (y + z) 2. ~(y + z) + (e + m) 3. ~(e + m) + (x + z) 4. ~x 5. x + z 6. ~x z Here s n exmple using onjuntion: x, y; ~(xy) + ~z; = ~z Aording to rule 8, if we re given x nd y in the premises, we n form the onjuntion xy. One this is done, it n e solved y logil ddition: 1. x & y 2. ~(x & y) + ~z ~z Here s n exmple using simplifition, onjuntion, nd destrutive dilemm: 1. f x 2. g y 3. (x + y)(f + z) 4. ~(~f + ~g) + h h We first hve to set up the prolem so tht it n e resolved y logil ddition. We follow rule 7 in simplifying the third premiss to (x + y). By rule 8 we onjoin the first nd seond premises, thus (g y)(f x). This sets up wht s lled destrutive dilemm, s per rule 6. Applying (~y + ~x) we hve the result (y + x) (g + f), whih n e trnsformed to (x + y) + (f + g). We re now redy to solve: 1. (~x + ~y) 2. ~(~x + ~y) + (~f + ~g) 3. ~(~f + ~g) + h h It s importnt to rememer tht while this ddition method of logi helps you solve rguments with severl premises, it does not set up the prolem for you in the first ple. Tht s where lerning the rules of simplifition, onjuntion, nd so on will help you to onstrut omplex terms. One simple nd omplex terms re set up, linking terms n e polrized, nd n e neled s usul. Here s n exmple from Copi showing just how omplex linking term n eome: 1. (m n) (f g) 2. (x ~y) (n ~y) 3. {[(x ~y) + (z e)](f + g)} [(z e) (m n)] 4. (x ~y) + (z e) 5. f + g (n ~y) + (f g) 21

22 Conjoin the fourth nd fifth premises, hene, {[(x ~y) + (z e)](f + g)}. As n e seen, this is equivlent to the eginning of the third premiss. We n onvert the third premiss to disjuntive form, hene ~{[(x ~y) + (z e)]f + g)} + [(z e) (m n)]. This llows us to nel most of the monstrous third premiss, s shown elow. The rest of the linking terms re omplex terms, with only two of them remining in the onlusion. 1. {[(x ~y) + (z e)](f + g)} 2. ~{[(x ~y) + (z e)](f + g)} + ~(z e) + (m n) 3. ~(m n) + (f g) 4. (z e) + (x ~y) 5. ~(x ~y) + (n ~y) (f g) + (n ~y) 10. Duplited Terms: In using logil ddition, the gol is to nel out linking terms nd send the results to the onlusion line. On the front end, you sometimes need to use simplifition, onjuntion, et, to set up the prolem. On the k end, there s one more step efore you n e sure the rgument is vlid. The se of omplex terms tht re duplited in more thn one premiss n rete prolems. The esiest wy to hek the vlidity of n rgument with duplited terms is to tke the duplited terms nd redue them to onjuntions. If the resulting onjuntions polrize, then the omplex terms n e nelled s usul. If not, the rgument is invlid. Suppose you hve the following terms in more thn one premiss: 1 st premiss: ~x + ~y 2 nd premiss: xy Given wht we ve sid out duplited terms in more thn one premiss, how would you go out determining the vlidity of this rgument? First, you would onvert the disjuntive terms of the premises to onjuntive form, resulting in ~(xy) on the one hnd nd (xy) on the other. Are they polrized? Yes, in this se they re, for (xy) is omplex term whih is the omplement of ~(xy). Beuse these omplex terms omplement one nother, they n e nelled s esily s ny simple term. Now onsider the following: x (y + z); x (y e); ~e; = ~x Here is wht it would look like using logil ddition: 1. ~x + y + z 2. ~y + ~z + e 3. ~e ~x As you n see, premises 1 nd 2 ontin duplited terms, y nd z. Tht mens you d need to hek them to mke sure they n e polrized. Let s onvert 22

23 them into onjuntive form s follows: First: y + z = ~(~y~z) Seond: ~y + ~z = ~(yz). Upon nlysis, we find tht ~(~y~z) nd ~(yz) re not linking terms. The orret polriztion of ~(yz) is simply (yz), not ~(~y~z), nd the orret polriztion of ~(~y~z) is simply (~y~z). The result is tht the onlusion nnot e vlidly derived. Another exmple: 1. ~x + y 2. ~x + ~y This would e (~x + y) (~x + ~y), whih under rule 13 n e expnded to (~x~x + ~x~y + y~x + y~y), whih is ~x + ~x~y + ~xy + 0. This simplifies to ~x + ~x (~y + y) + 0, whih is simplified further to (~x + ~x) (1), then (~x + ~x), whih redues to ~x. This lst form of duplited terms n e found in Crroll s rgument ove under Setion 9, loted in premises 3 nd 8, with terms m nd e. The si thing to do is to look out for duplite terms when you hve two or more premises ontining the sme letters. The following is the form you wnt to wth out for: 1. x + y 2. ~x + ~y Premises 1 & 2 present the illusion tht you n just nel the terms, ut we hve seen tht simple trnsltion to their onjuntive forms shows tht the linking terms re not polrized. Now onsider the following. If these terms were prt of n rgument, would you e le to nel them nd derive vlid onlusion? ~x + y; x + ~y; =? By logil ddition: 1. ~x + y 2. x + ~y? It looks little like the previous exmple, duplited terms nd ll. Also, when we redue the premises to onjuntive form the expression ~(x~y)~(y~x) doesn t look like it is polrized. Be reful, though. In tul ft, the expression (~x + y)(~y + x) n e expnded to ~x~y + ~xx + y~y + xy, with the terms ~xx nd y~y eing 0 s or ontrditions, leving xy + ~x~y. This turns out to e the definition of equivlene, x y. So, the terms re definitely linked in the rgument nd n e nelled to reh vlid onlusion. When you see duplited terms in different premises, just rememer to redue them to onjuntions, nd if they polrize, fine, or if they represent equivleny, tht s oky, too. If neither one of the ove, onsider the rgument invlid. 11. Frtionl Logi -- An Option In order to represent logil lger with frtionl symols, we d need to mke hnge to the trditionl disjuntion symol (the v or + sign). As the 23

24 nme implies, the frtionl symol for disjuntion would e division sign: x y This mens x or y. If you wnt to put the frtionl expression on one line, the following would e orret: x/y. Similrly, x y, ( If x then y ) would e trnsformed to If not x, then y, symolized s: ~ x y This is wht would e lled mteril implition, nd it trnsltes from the x y form. Historil Note: It ppers tht ertin Mr. Johnson (pprently W.E. Johnson) ws one of the first to suggest using the frtion sign to represent lterntion. 1 He lso suggested tht the vertil line should represent onjuntion. Keynes provides exmples of the use of the signs. The expression AB or CD ws represented y: AB CD The expression (A or C) nd (B or D) ws represented y: A B C D Tehnilly, in the seond expression there should e division signs etween A nd C nd etween B nd D, ut Keynes only used the vertil line in order to emphsize the sign of onjuntion. As will eome ler, these signs were developed in different wy y Johnson nd Keynes from the wy we re using them here. To use frtionl symolism for ny rgument, here is how it would work: Exmple: Simple frtionl rgument: All y is z: y z All x is y: x y Therefore, All x is z x z In frtionl nottion, this would e: ~ y ~ x & z y ~ x z As n e seen, the y s re polrized, one positive, one negtive, nd fter nelling, it s esy to derive the onlusion involving whtever terms re leftover, suh s x nd z. Moreover, the frtionl representtion of the onlusion n lwys e onverted k to the textook representtion y simple trnsltion: ~x + z to x z, ( If x then z ). It might e helpful to put rkets round the polrized linking terms. ~ y z & ~ x [y] ~ x z By nelling the rketed terms, the resulting frtion ~x/z forms the onlusion, nd it n e onverted to x z. 24

25 25 Exmple: Keynes & simplifition: More omplited prolems n lso e solved using frtionl representtion. Here is logil expression whih needs to e simplified: + ~~ + ~ + ~~. 9 In frtionl nottion the expression would look like this Simplify s follows: ) ( ) ( ) ( 9 This exmple is from John N. Keynes, Studies nd Exerises in Forml Logi, 1906, p The fourth simplifition ould e solved in different wy if the is retined nd the is removed from premiss 3, s follows: 3 4* 5* 6* ) (

26 Exmple: Keynes & Equivleny: Keynes sks us to show the equivleny etween two expressions: or d or d is equl to or d. 10 This is nother simplifition prolem, nd the solution is: d d d d (1) d d d d d d ( ) d ( d) ( ) d d ( d) 1 d 1 d (1) d(1) d Thus, the solution is equivlent to Keynes solution or d. Frtionl d representtion would e helpful to those interested in digitl iruits or swithing lger. The struture of frtionl equtions looks somewht similr to stndrd representtions of logi gtes Keynes, p See espeilly, J. E. Whitesitt, Boolen Alger nd Its Applitions, New York: Dover, 1995, pp. 75ff; lso J. D. Lenk, Hndook of Logi Ciruits, VA: Reston Pu., 1972; nd J. R. Gregg, Ones nd Zeros: Understnding Boolen Alger, Digitl Ciruits, nd the Logi of Sets, NY: IEEE Press,