Algebra and Grammar: a Peanian Analysis of Everyday English

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Metalogicon (1992) V, 1

Algebra and Grammar: A Peanian Analysis of Everyday English

Michele Malatesta*

It is well known that, by following Leibniz's intuition, Giuseppe Peano constructed the Latino sine flexione, i.e. Latin without inflexion. The rules of LSF (Latino sine flexione) are the following: (1) grammatical cases are superfluous since they can be expressed by prepositions; (2) grammatical genders are superfluous as in Chinese; (3) grammatical numbers are superfluous as in Chinese (N.B. Grammatical numbers not arithmetic numbers are superfluous); (4) conjugations of verbs are superfluous: it is sufficient to express the personal pronouns I, You, He, She, It, etc. as in Chinese; (5) verb tenses are superfluous since the time is expressible by means of adverbs as in Chinese. Note that Peano did not axiomatize LSF.1

* Presented at 3rd International Symposium on Systems Research Informatics and Cybernetics, Baden-Baden (Germany), August 12-18, 1991, in the Plenary Session on Logic and Mathematics of day August 16. 1 An attempt to axiomatize LSF is found in FREGUGLIA [1977] 305-317 and [1978] 72-75. In both works, the author given five general axioms and three specific axioms, proves two theorems. “Assiomi generali A.6.1: Siano , , ecc. simboli per elementi grammaticali, allora se = +

Metalogicon (1992) V, 1

si ha : = - = - A.6.2: Indichi 0 l'elemento grammaticale «de valore nullo» allora se = si ha - = 0 A.6.3: ± 0 = 0 + = A.6.4: - = - + = 0 A.6.5: ± ( ± ) = ( ± ) ± C

Assiomi specifici

A'.6.1: V = es + A = habe + N dove es e habe, imperativi latini dei verbi esse e habere, vengono usati secondo l'interlingua per indicare genericamente i verbi essere e avere. A'.6.2: A = que + V = cum + N = V + -nte A'.6.3: N = V + -ore. Teoremi TEOREMA 6.1: L'elemento grammaticale cum ha la stessa funzione grammaticale dell'elemento que habe (nel senso che possono sostituirsi scambievolmente in una generica frase che contenga o l'uno o l'altro). Dimostrazione: Dall'assioma A3.6.2 si ha: A = cum + N applicando ora lo A.6.1 si ha cum = A - N applicando lo A.6.2 " " cum = A + 0 - N " " A.6.4 " " cum = A - V + V - N " " A.6.5 " " cum = (A - V) + (V - N) (i) Sempre dall'assioma A'.6.2 si ha: A = que + V da cui per A.6.1 si ottiene (ij) que = A - V. A sua volta dall'assioma A'.6.1 si ha : V = habe + N da cui per A.6.1 discende (iij) habe = V - N. Sostituendo (ij) e (iij) in (i) si ha cum = que habe, c.v.d. TEOREMA 6.2: Gli elementi grammaticali que es (colui che è), es -nte, habe -ore hanno valore nullo (nel senso che grammaticalmente non rappresentano niente, cioè sono «eliminabili»). Dimostrazione: Dall'assioma A'.6.1 per A.6.1 si ha es = V - A " " A'.6.2 " " " que = A - V " " A'.6.2 " " " -nte = A - V allora: que es = (A - V) + (V - A) = applicando A.6.5 = A - V + V - A = " A.6.4 = A + 0 - A = " A.6.3 = A - A = " A.6.4

Metalogicon (1992) V, 1

2. English language trend to Chinese.

The English Language is placed at the periphery of the Indo- European Group and, among the Western languages, expresses the maximum trend towards the Chinese. Let we examine the following table:

Classical German Classical Italian English Greek Latin Spanish French C ( ) ( ) ( ) indef ( ) + a Art def + + ( ) ( ) ( ) s Subst + + + ( ) ( ) e Adj + + + ( ) ( ) s — attrib + + ( ) ( ) predic G ( ) ( ) + — indef + e Art def + + ( ) + — n Subst NR NR NR NR NR d Adj + + + — + e attrib + — + + — r predic s G n ( ) ( ) + + indef + r u Art def + + ( ) + — a m Subst + + + + +

m b Adj + + + — + m e attrib + — + + — a r predic t s

= 0 così anche: es -nte = (V - A) + (A - V ) = 0 Dall'assioma A'.6.1 per A.6.1 si ha: habe = V - N " A'.6.3 " " " -ore = N - V per cui analogamente a sopra si ha: habe -ore = (V - N) + (N - V) = 0 c.v.d.”

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T Past + + + + — e Present + + + + + n Future + + + + — s e s 14/18 14/18 11.18 12.18 3/18 77.77% 77.77% 61.11% 66.66% 16.66%

This is a synoptic table of some grammatical categories. It considers only the following languages: Classical Greek, German, Classical Latin, Italian, Spanish, French, English. The sign ‘( )’ shows that a given grammatical part does not exist in the corresponding language; the sign ‘+’ indicates that a given grammatical part belongs to the corresponding language and has inflections; the sign ‘ — ’ denotes that a given grammatical part belongs to the corresponding language but is without inflexions; the signe 'NR’ shows that a given grammatical part belongs to the corresponding language but is not remarkable from the inflexion standpoint. Classical Greek has cases, and therefore inflexions, in the definite articles, in substantives, in attributive and predicative adjectives. German has cases, and therefore inflexions, in the indefinite and definite articles, in substantives and in attributive adjectives. Classical Latin has cases, and therefore inflexions, in substantives and in attributive and predicative adjectives. Italian, Spanish, French and English do not have cases. Classical Greek has genders in the definite articles, in substantives, in attributive and predicative adjectives. German has genders in the indefinite and definite articles, in substantives, in attributive adjectives. Classical Latin has genders in substantives, in attributive and predicative adjectives. Italian, Spanish and French have genders in the indefinite and definite articles, in substantives, in attributive and predicative adjectives. English has genders in substantive but they are not relevant

Metalogicon (1992) V, 1 from the inflexion standpoint. Classical Greek has grammatical numbers in the definite articles, in substantives, in attributive and predicative adjectives. German has grammatical numbers in the indefinite and definite articles, in substantives, in attributive adjectives. Classical Latin has grammatical genders in substantives, in attributive and predicative adjectives. Italian, Spanish and French have grammatical numbers in the indefinite and definite articles, in substantives, in attributive and predicative adjectives. English has grammatical numbers only in indefinte articles and in substantives. Classical Greek, German, Classical Latin, Italian, Spanish and French have tense inflexions in the past, in the present and in the future. The English language, apart from the verb to be and the indicative present of to have, has tense inflexions only in the third singular person of the indicative present. With regard to examined grammatical elements and leaving aside the substantive gender which is not remarkable for its own inflexions, note the following: - in Classical Greek the inflexions are 14/18 equal to 77,77 %; if we consider that this language has no indefinite articles, then we obtain the following inflexion number 14/15 equal to 93,33 %; - in German the inflexions are 14/18 equal to 77,77 %; - in Classical Latin the inflexions are 11/18 equal to 61,11 %; if we consider that this language has no indefinite and definite articles, then we obtain the following inflexion number 11/12 equal to 91,66 %; - in Italian, Spanish, French the inflexions are 12/18 equal to 66,66 %; if we consider that these languages have no cases, then we obtain the following inflexion number 12/13 equal to 92,30 %; - in English the inflexions are 3/18 equal to 16,66 %; now, if we consider that this language has no cases, then we obtain the following inflexion number 3/13 equal to 23,07 %. Therefore, whatever statistical criterion we choose when we analyze West languages, English is the one with the lowerst number of

Metalogicon (1992) V, 1 inflexions, and, consequently, the one nearest to Chinese. At this point we see that a Peanian analysis of English is very easy.

3. English without inflexion.

Now we construct an artificial English language by reducing every English verb to its theme, by selecting among relative pronouns the only pronoun that, and by eliminating every inflexion from the verbs to be and to have. We study EWI (English without inflexion) by means of algebraic equations analogous to those in Peano's De derivatione and Algebra de grammatica. We shall give two distinct axiomatic patterns: in the first one we take as axioms four laws of standard algebra without proving their independence, as is usual in abstract algebra handbooks;2 in the second one, the most rigorous, we put as axioms three linguistic equations whose independence we prove.

4. First axiomatic version.

3.1 Syntax

Let ‘A ’, ‘B ’ be symbols of grammatical elements. Let ‘A’ (Adjective), ‘V’ (Verb), ‘S’ (Abstract Substantive), ‘Es’ (Ending of Abstract Substantive) be the names of grammatical elements. The signs ‘0’, ‘+’, ‘-’, ‘=’ belong to the language.

RULES OF FORMATION

- 0 is a grammatical element;

2 Amongst others see P. J. HIGGINS [1975] 7 sgg. Freguglia also does not give any proof of independence of the algebraic axioms of Peano's Algebra de grammatica. See FREGUGLIA [1977] and [1978] cit.

Metalogicon (1992) V, 1

- all the adjectives, verbs and abstract substantives belonging to the English language are grammatical elements and the words ‘that ’, ‘with ’, ‘to’ are grammatical elements; the sign ‘Es’ i.e. “Ending of Abstract Substantive” is a grammatical element; - the names of grammatical elements are grammatical elements; - if A is a grammatical element, then +A and -A are grammatical elements; - if A and B are grammatical elements, then A +B and A -B are grammatical elements; - if A and B are grammatical elements, then A =B is a linguistic equation; - if A =B is a linguistical equation, and A = 0 or B= 0, then A =B is a satisfied linguistic equation; - there are no other grammatical elements nor linguistic equations nor satisfied linguistic equations.

RULES OF DEDUCTION

I. If in an axiom we substitute in every occurrence a grammatical element for a symbol of grammatic element, then we obtain a correct linguistic equation. II. If A +B = A +B is asserted, then A +B - A = B or A +B -B = A can be asserted. III. If A =B is asserted, and C =D is asserted, and C is a grammatical element of A =B, then A =B where C has been replaced by D can be asserted. IV. If A =B and C =D are asserted, then A +C = B +D can be asserted. V. If A =B is asserted, then B =A can be asserted.

AXIOMS

Metalogicon (1992) V, 1

1. A =A 3 2. A ± 0 = A 3. A -A = 04 4. A ± B = B ± A

DEFINITIONS i A = that + V ii A = with + S iii V = be + A iv V = have + S v S = V + Es vi S = (to + be) + A

3 Do not forget that Axiom 1. is the first axiom of Alessandro Padoa's Logica. “apparentemente sterile, esso è invece la fonte più ricca e più varia di verità [...] diremo che 8 + 7 = 3 ∞ 5 è un'identità aritmetica, perché la sua verità non può essere consociuta se non da chi conosca i simboli aritmetici in essa adoperati. Ma, per chi li conosca, donde scaturisce la certezza della sua verità?Dal fatto che, eseguendo le operazioni in essa indicate, essa si converte nell'identità logica [...] gli scettici che negano la validità dell'Assioma 1 ignorano l'uso incessante che ne viene fatto”, PADOA [1957] 30-33.Also in Mathematical Logic which precedes Arithmetic in PEANO [1902-03] the law of identity holds the first place. See p. 3. 4 If instead of 2. we write 2. bis A + 0 = 0 + A = A and instead of 3. we write 3.bis A + (-A ) = (-A ) + A = 0 2.bis and 3.bis are laws of neutral element and of inverse element respectively. The definition of a group is wasteful in the sense that not all the conditions are independent. Thus, for exemple, one can omit the conditions 2.bis and 3.bis “since these relations can be deduced from the remaining conditions”, HOWSON [1972] 25.

Metalogicon (1992) V, 1

DEDUCTION

Lemma 0.01

V - A = be

Proof

(1) be + A = be + A (Ax 1, R I, A /be + A) (2) be + A - A = be (by (1), R II) (3) V - A = be (by (2), Df iii, R III)

Lemma 0.02

A - V = that

Proof

(1) that + V = that + V (Ax 1, R I, A /that + V) (2) that + V - V = that (by (1), R II) (3) A - V = that (by (2), Df i, R III)

Theorem I

be + that = 0

Proof

(1) V - A + A - V = be + that (by L 0.01 and L 0.02, R IV) (2) V + A - A + V = be + that (by (1), Ax 4 A /-A, B//+A, RIII) (3) V + 0 - V = be + that (by (2), Ax 3 A /A, R III) (4) V - V + 0 = be + that (by (3), Ax 4 A /+0, B//-V, R III) (5) 0 + 0 = be + that (by (4), Ax 3 A /V, R III) (6) 0 = be + that (by (5), Ax 2 A /0, R III)

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(7) be + that = 0 (by(6), R V)

Theorem II

that + be = 0

Proof

(1) that + be = 0 (by Th I, Ax 4 A /be, B /that, R III)

Lemma 0.03

V - S = have

Proof

(1) have + S = have + S (Ax 1, R I, A/ have + S) (2) have + S - S = have (by (1), R II) (3) V - S = have (by (2), Df iv, R III)

Lemma 0.04

S - V = Es

Proof (1) V + Es = V + Es (Ax 1, R I, A/ V + Es) (2) V + Es - V = Es (by (1) R II) (3) S - V = Es (by (2), Df v, R III)

Theorem III

have + Es = 0

Proof

(1) V - S + S - V = have + Es (by L 0.03 and L 0.04, R IV)

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(2) V + S - S - V = have + Es (by (1), Ax 4 A/ -S, B/ + S, R III) (3) V + 0 - V = have + Es (by (2), Ax 3 A/ S, R III) (4) V - V + 0 = have + Es (by (3), Ax 4 A/ + 0, B/ - V, R III) (5) 0 + 0 = have + Es (by (4), Ax 3 A/ V, R III) (6) 0 = have + Es (by (5), Ax 2 A/ 0, R III) (7) have + Es = 0 (by (6), R V)

Theorem IV

Es + have = 0

Proof

(1) Es + have = 0 (by Th III, Ax 4, A/ have, B/ Es, R III)

Lemma 0.05

A - S = with

Proof

(1) with + S = with + S (Ax 1, R I, A/ with + S) (2) with + S - S = with (by (1), R II) (3) A - S = with (by (2), Df ii, R III)

Lemma 0.06

S - A = (to + be )

Proof

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(1) (to + be ) + A = (to + be ) + A(Ax 1, R I, A/with + S) (2) (to + be ) + A - A = (to + be ) (Ax (1), R I, A/ (to = be ) + A) (3) S - A = (to + be ) (by (2), Df vi, R III)

Theorem V

with + (to + be ) = 0

Proof

(1) A - S + S - A = with + (to + be ) (by L 0.05 and L 0.06, R IV) (2) A + S - S - A = with + (to + be ) (by (1), Ax 4 A/ - S, B/ + S, R III) (3) A + 0 - A = with + (to + be ) (by (2), Ax 3 A/ S, R III) (4) A - A + 0 = with + (to + be ) (by (3), Ax 4 A/ + 0, B/ - A, R III) (5) 0 + 0 = with + (to + be ) (by (4), Ax 3 A/ A, R III) (6) 0 = with + (to + be ) (by (5), Ax 2 A/ 0, R III) (7) with + (to + be ) = 0 (by (6), R V)

Theorem VI

(to + be ) + with = 0

Proof

(1) (to + be ) + with = 0 (by Th V, Ax 4 A/ with, B/ (to + be ), R III)

3.2 Semantics

3.2.1 In semantics we go back from English without inflexion to Everyday English language, analogously to a computer that

Metalogicon (1992) V, 1 works with binary arithmetic, but visualizes the results in terms of decimal arithmetic.

Put:

T (Theme, stem, root) Ea (Ending of Adjective) Ev (Ending of Verb) Es (Ending of Substantive) has been already considered, by the Peanian viewpoint, in rules of formations.

NEW DEFINITIONS

(a) A = T + Ea e.g. stud + ent act + ing arm + full fleet + ( ) (b) V = T + Ev stud + y act + ( ) arm + ( ) fleet + ( ) (c) S = T + Es stud + y act + ion arm + ( )

3.2.2 Let us remember the definitions i-vi : i A = that + V ii A = with + S iii V = be + A iv V = have + S v S = to + V vi S = (to + be) + A

3.2.3 Let ‘infl’ be symbol for inflexion; Now if we read ‘that’, ‘with’, ‘be ’, ‘have’, etc. metalinguistically as “the relative pronoun that”, “the preposition with”, “the verb be”, “the verb have” etc., and if we substitute in definiens of i - vi the symbols A, V, S by definitions (a), (b), (c), we obtain: i a A = that + (T + Ev) ii a A = with + (T + Es)

Metalogicon (1992) V, 1 iii a V = (be + infl) + (T + Ea) iv a V = (have + infl) + (T + Es) v a S = to + (T + Ev) vi a S = (to + be) + (T + Ea)

SEMANTICAL RULES

If Ea = 0, then A = T. If Ev = 0, then V = T. If Es = 0, then S = T.

GENERAL RULE

If the ending of a grammatical element is equal to 0, then this grammatical element is equal to theme. This rule is sufficient to construct computer programmes to translate everyday languages without inflexions into everyday languages without inflexions by means of very elementary semantic dictionaries.5

5. Second axiomatic version

5.1 Syntax

RULES OF FORMATION

5 One could improve Peano's algebra of grammar by means of the substitution of the Peanian definition v S = V + Es by definition v bis S = to + V and, consequently, by a new version of lemma 0.04 and theorems III and IV.

Metalogicon (1992) V, 1

The same as the first version.

RULES OF DEDUCTION

All the rules of standard algebra are rules of deduction on EWI.

AXIOMS

1 ° A = that + V = with + S 2 ° V = be + A = have + S 3 ° S = V + Es = (to + be) + A

INDEPENDENCE OF THE AXIOMS

In order to prove the independence of one axiom it is sufficient to find a system of objects which satisfies all the axioms except the one under consideration, and which does not satisfy the latter.

Independence of the first axiom. Let us assign to eight elements of axioms 1 ° - 3 ° the following values

V = 1 A = 2 (to + be) = 3 Es = 4 S = 5 be = -1 with = -3 have = -4

It is evident that equations 2 ° and 3 ° are satisfied but it is impossible to satisfy equation 1 °. To satisfy equation 1 ° one ought to assign to the grammatical element ‘that ’ the number 1 which has already been assigned to the grammatical element ‘V’ and in the same context a determinate numerical value cannot be assigned to two different grammatical elements. In fact we have for axioms 2 ° and 3 ° the following solutions

2 ° 1 = -1 + 2 = -4 + 5 3 ° 5 = 1 + 4 = 3 + 2

Metalogicon (1992) V, 1 but we have no solution for axiom 1 °

1 ° 2 = * + 1 = -3 + 5 where the sign ‘*’ shows that it is impossible to assign to the corresponding grammatical element a numerical value different from those already assigned.

Independence of the second axiom. Let us assign to eight elements of axioms 1 ° - 3 ° the following values

A = 1 V = 2 Es = 3 (to + be) = 4 S = 5 that = -1 have = -3 with = -4

It is evident that equations 1 ° and 3 ° are satisfied but it is impossible to satisfy equation 2 °. To satisfy equation 2 ° one ought to assign to the grammatical element ‘be’ the number 1 which has already been assigned to the grammatical element ‘A’ and in the same context a determinate numerical value cannot be assigned to two different grammatic elements. In fact we have for axioms 1 ° and 3 ° the following solutions

1 ° 1 = -1 + 2 = -4 + 5 3 ° 5 = 3 + 2 = 4 + 1 but we have no solution for axiom 2 °

2 ° 2 = * + 1 = -3 + 5 where the sign ‘*’ shows that it is impossible to assign to the corresponding grammatical element a numerical value different from those already assigned.

Independence of the third axiom. Let us assign to eight elements of axioms 1 ° - 3 ° the following values

A = 1 S = 2 have = 3 be = 4 V = 5

Metalogicon (1992) V, 1

with = -1 Es = -3 that = -4

It is evident that equations 1 ° and 2 ° are satisfied but it is impossible to satisfy equation 3 °. To satisfy equation 3 ° one ought to assign to the grammatical element ‘(to + be)’ the number 1 which has already been assigned to the grammatical element ‘A’ and in the same context a determinate numerical value cannot be assigned to two different grammatic elements. In fact we have for axioms 1 ° and 2 ° the following solutions

1 ° 1 = -4 +5 = -1 + 2 2 ° 5 = 4 + 1 = 3 + 2 but we have no solution for axiom 1 °

1 ° 2 = -3 + 5 = * + 1 where the sign ‘*’ shows that it is impossible to assign to the corresponding grammatical element a numerical value different from those already assigned.

DEDUCTION

Lemma 1a

V - A = be

Proof

(1) V = be + A (Ax 2 °) (2) V - A = be (by transfer of element A to the left-side of the equation)

Lemma 2a

Metalogicon (1992) V, 1

A - V = that

Proof

(1) A = that + V (Ax 1 °) (2) A - V = that (by transfer of element V to the left-side of the equation)

Theorem Ia

be + that = 0

Proof

(1) V - A + A - V = be + that (by adding left- and right-sides of the equations 1a and 2a) (2) V + 0 - V = be + that (by rule -a +a = 0) (3) V - V = be + that (by rule a + 0 = a) (4) 0 = be + that (by rule a - a = 0) (5) be + that = 0 (by commutation of Left- and right-sides of an equation)

Theorem IIa

that + be = 0

Proof

(1) that + be = 0 (from Th. Ia by rule a + b = be + a)

Lemma 3a

V - S = have

Proof

(1) V = have + S (Ax 2 °)

Metalogicon (1992) V, 1

(2) V - S = have (by transfer of element S to the left- side of the equation)

Lemma 4a

(1) S = V + Es (Ax 3 °) (2) S - V = Es (by transfer of element V to the left- side of the equation)

Theorem IIIa

have + Es = 0

Proof

(1) V - S + S - V = have + Es (by adding left- and right-sides of the equations 3a and 4a) (2) V + 0 - V = have + Es (by rule -a +a = 0) (3) V - V = have + Es (by rule a + 0 = a) (4) 0 = have + Es (by rule a - a = 0) (5) have + Es = 0 (by commutation of left- and right-sides of an equation)

Theorem IVa

Es + have = 0

Proof

(1) Es +have = 0 (from Th. IIIa by rule a + b = be + a)

Lemma 5a

A - S = with

Metalogicon (1992) V, 1

Proof

(1) A = with + S (Ax 1 °) (2) A - S = with (by transfer of element S to the left- side of the equation)

Lemma 6a

S - A = (to + be)

Proof

(1) S= (to + be)+ A (Ax 3 °) (2) S - A = (to + be) (by transfer of element A to the left- side of the equation)

Theorem Ia

with + (to + be) = 0

Proof

(1) A - S + S - A = with + (to + be) (by adding left- and right-sides of the equations 5a and 6a) (2) A + 0 - A = with + (to + be) (by rule -a +a = 0) (3) A - A = with + (to + be) (by rule a + 0 = a) (4) 0 = with + (to + be) (by rule a - a = 0) (5) with + (to + be) = 0 (by commutation of left- and right-sides of an equation)

Theorem VIa

(to + be) + with = 0

Proof

Metalogicon (1992) V, 1

(1) (to + be) + with = 0 (from Th.Va by rule a + b = b + a)

5.2 Semantics

5.2.1 as 3.2.1.

5.2.2 Let us draw the definitions i - vi from axioms 1 ° - 3 °: i A = that + V ii A = with + S iii V = be + A iv V = have + S v S = to + V vi S = (to + be) + A

5.2.3 as 3.2.3.

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FREGUGLIA P., Giuseppe Peano ed i prodromi della linguistica matematica, “Physis” (1977), XIX, 305-317.

FREGUGLIA P., L'algebra della logica. Un profilo storico, Roma, Editori Riuniti, 1978.

HIGGINS P. J., A First Course in Abstract Algebra, New York - Cincinnati - Toronto, Van Nostrand Reinhold Company, 1975.

HOWSON A. G., A Handbook of Terms Used in Algebra and Analysis, Cambridge, University Press, 1972.

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LEIBNIZ G. W., Sämtliche Schriften und Briefe, ed. Preussische Akademie der Wissenschaften, Darmastadt, Otto Reichl, 1923 sgg.

PADOA A., Logica, in BERZOLARI - VIVANTI - GIGLI [1957], vol. I, 5-79.

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PEANO G., Il latino quale lingua ausiliare internazionale, in Atti della Reale Accademia delle Scienze di Torino, vol. XXXIX (1903-04) 273-283.

PEANO G., De derivatione, Academia pro Interlingua, Discussione, Tomo III (1912) 20-43.

PEANO G., Algebra de Grammatica, “Schola et Vita”, vol. V (1930) 323-336.

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