Counting false entries in truth tables of bracketed formulae connected by m-implication

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1 arxiv: v1 [mathco] 1 Mar 01 Counting false entries in truth tables of bracketed formulae connected by m-implication Volkan Yildiz ah05146@qmulacuk or vo1kan@hotmailcouk Abstract Inthispaperwecount thenumberofrows y n withthevalue false in the truth tables of all bracketed formulae with n distinct variables connected by the binary connective of modified-implication We find a recurrence and an asymptotic formulae for y n We also determine the parity of y n Keywords: Propositional logic, m-implication, Catalan numbers, parity, asymptotics, Catalan tree AMS classification: 05A15, 05A16, 03B05, 11B75 1 Introduction In this paper we study enumerative and asymptotic questions on formulae of propositional logic which are correctly bracketed chains of m-implications, where the letter m stands for modified For brevity, we represent truth values of propositional variables and formulae by 1 for true and 0 for false For background information on propositional logic the reader can refer to the following books, [6], and [3], or to the introduction page of, [4] In-fact 1

2 this paper is an extension of [4] In [4], we have shown that the following results are true: Theorem 11 Let f n be the number of rows with the value false in the truth tables of all bracketed formulae with n distinct propositions p 1,,p n connected by the binary connective of implication Then and for large n, f n n 1 f n = ( i C i f i )f n i, with f 1 = 1 (1) i=1 ( ) 3n πn 3 Where C i is the ith Catalan number A number of new enumerative problems arise if we modify the binary connective of implication as in below cases Case(i) Use instead of, where defined as follows φ ψ φ ψ For any valuation ν, { ν(φ ψ) = 0 if ν(φ) = 1 and ν(ψ) = 1, 1 otherwise Case(ii) Use instead of, where defined as follows φ ψ φ ψ For any valuation ν, { ν(φ ψ) = 0 if ν(φ) = 0 and ν(ψ) = 0, 1 otherwise Case(iii) Use instead of, where defined as follows φ ψ φ ψ For any valuation ν, { ν(φ ψ) = 0 if ν(φ) = 0 and ν(ψ) = 1, 1 otherwise Let s n, h n be the number of rows with the value false in the truth tables of all bracketed formulae with n distinct propositions p 1,,p n connected by the binary connective of m-implication, in the case (iii) and (ii), respectively

3 11 Case(iii) A row with the value false comes from an expression ψ χ where ν(ψ) = 0 andν(χ) = 1 If ψ contains ivariables, then χ contains n i, andthenumber of choices is given by the summand: n 1 s n = s i ( n i C n i s n i ), where s 0 = 0,s 1 = 1 () i=1 The recurrence relation () is equivalent to the recurrence relation (1), so all the results we have in [4], and [8] hold for the case(iii) too 1 Case(ii) A row with the value false comes from an expression ψ χ where ν(ψ) = 0 andν(χ) = 0 If ψ contains ivariables, then χ contains n i, andthenumber of choices is given by the summand: n 1 h n = h i h n i, where h 0 = 0,h 1 = 1 (3) i=1 The recurrence relation (3) is very well known; it is the recurrence relation for Catalan numbers Corollary 1 Suppose we have all possible well-formed formulae obtained from p 1 p p n by inserting brackets, where p 1,,p n are distinct propositions Then each formula defines the same truth table Example 13 Here are the truth tables, (merged into one), for the bracketed m-implications, in n = 3 variables p 1 p p 3 p 1 (p p 3 ) (p 1 p ) p

4 13 Case(i) We are interested in bracketed m-implications, case(i), which are formulae obtained from p 1 p p n by inserting brackets so that the result is well-formed, where p 1,,p n are distinct propositions Proposition 14 Let y n be the number of rows with the value false in the truth tables of all brackted m-implications, case(i), with n distinct variables Then n 1 ( ) y n = ( i C i y i )( n i C n i y n i ), with y 0 = 0, y 1 = 1 (4) i=1 Proof A row with the value false comes from an expression φ ψ, where ν(φ) = 1 and ν(ψ) = 1 If φ contains i variables, then ψ contains n i variables, and the number of choices is given by the summand in the proposition Example 15 and y 1 = 1,y = ( 1 C 1 y 1 )( 1 C 1 y 1 ) = 1 y 3 = ( 1 C 1 y 1 )( C y )+( C y )( 1 C 1 y 1 ) = 3+3 = 6 Example 16 Here are the truth tables, (merged into one), for the two bracketed m-implications, case(i), in n = 3 variables Where the corresponding rows with the value false are in blue: p 1 p p 3 p 1 (p p 3 ) (p 1 p ) p which coincides with the result we had from Example 15 4

5 Using Proposition 14, it is straightforward to calculate the values of y n for small n The first values are {y n } n 1 = 1,1,6,9,16,978,6156,40061,67338,181938, , , , , , , , , , , , , Generating Function Recall from [4], that the number of bracketings of a product of n terms is the Catalan number with the generating function C n = 1 ( ) n, with C 0 = 0, C n x n = (1 1 4x)/ n n 1 respectively (see also [, page 61]) Let g n be the total number of rows in all truth tables for bracketed m- implications, case(i), with n distinct variables It is clear that g n = n C n, with g 0 = 0 Let Y(x) and G(x) be the generating functions for y n, and g n, respectively That is, Y(x) = n 1 y nx n, and G(x) = n 1 g nx n Since, Then, n 1 y n = i=1 n 1 ( ) ( i C i y i )( n i C n i y n i ), where y 0 = 0, y 1 = 1 n x n 1y n = x+ n 1 i C i n i C n i x n n 1 i C i y n i x n n 1 i=1 n 1 i=1 n 1 y i i C n i x n i + n 1 y i y n i x n i=1 n 1 i=1 n 1 Now it is straightforward to get the following result: Y(x) = x+(g(x) Y(x)) (5) 5

6 where G(x) can be obtained from the generating function of C n by replacing x by x: that is, G(x) = (1 1 8x)/ (6) Substituting (6) into (5) gives the following quadratic equation: Y(x) +Y(x)( 1 8x )+(1 1 8x x) = 0 (7) Solving equation (7) gives the following proposition: Proposition 1 The generating function for the sequence {y n } n 1 is given by Y(x) = 1 8x 3 4x 1 8x (As with the Catalan numbers, the choice of sign in the square root is made to ensure that Y(0) = 0) With the help of Maple we can obtain the first terms of the above series, and hence give the first values of y n ; these agree with the values found from the recurrence relation 3 Asymptotic Analysis In this section we want to get an asymptotic formula for the coefficients of the generating function Y(x) from Proposition 1 We use the following result [1, page 389]: Proposition 31 Let a n be a sequence whose terms are positive for sufficiently large n Suppose that A(x) = n 0 a nx n converges for some value of x > 0 Let f(x) = ( ln(1 x/r)) b (1 x/r) c, where c is not a positive integer, and we do not have b = 0 and c = 0 Suppose that A(x) and f(x) each have a singularity at x = r and that A(x) has no singularities in the interval [ r,r) Suppose further that lim x r exists and has nonzero A(x) value γ Then a n f(x) γ ( ) n c 1 n (lnn) b r n, if c 0, γb(lnn) b 1 n, if c = 0 6

7 Note 3 We also have ( n c 1 where the standard gamma-function Γ(x) = 0 It follows that Γ( 1/) = π/ n ) n c 1 Γ( c), t x 1 e t dt, with Γ(x+1) = xγ(x), Γ(1/) = π Recall that G(x) = (1 1 8x)/, therefore Y(x) = (1+G(x)) (1+4G(x)) 4x Asin[4], beforestudyingy(x), wefirststudyg(x) ThisG(x)couldeasilybe studied by using the explicit formula for its coefficients, which is n( n n 1) /n But our aim is to understand how to handle the square root singularity A square root singularity occurs while attempting to raise zero to a power which is not a positive integer Clearly the square root, 1 8x, has a singularity at 1/8 Therefore by Proposition 31, r = 1/8 We have G(1/8) = 1/, so we would not be able to divide G(x) by a suitable f(x) as required in Proposition 31 To create a function which vanishes at 1, we simply look at 8 A(x) = G(x) 1/ instead That is, let Then f(x) = (1 x/r) 1/ = (1 8x) 1/ A(x) γ = lim = 1 x 1/8 1 8x Now by using Proposition 31 and Note 3, g n 1 ( ) n 3 (1 ) n 1 8 n n 3/ n 8 Γ( 1/) = 3n πn 3 We are now ready to tackle Y(x), and state the main theorem of the paper 7

8 Theorem 33 Let y n be number of rows with the value false in the truth tables of all the bracketed m-implications, case(i), with n distinct variables Then ( 10 ) 10 3n y n 10 πn 3 Proof Recall that Y(x) = 1 8x 3 4x 1 8x Wefindthatr = 1 8, andf(x) = 1 8x SinceY(1/8) = ( 5)/ 0, we need a function which vanishes at Y(1/8), thus we let A(x) = Y(x) Y(1/8) A(x) lim x 1/8 f(x) = lim x 1/8 Let v = 1 8x Then v v 4v+5+ 5 γ = lim v 0 v = x 3 4x 1 8x x = 10 10, 0 1 = lim (v 4)(v 4v +5) 1 v 0 where we have used l Hôpital s Rule in the penultimate line Finally, y n 10 ( )( ) ( 10 n 3 n 1 10 ) 10 3n 0 n 8 10, πn 3 and the proof is finished The importance of the constant = lies in the following fact: Corollary 34 Letg n be the total number of rows in alltruth tablesfor bracketed m-implications, case(i), with n distinct variables, and y n the number of rows with the value false Then lim n y n /g n =

9 The table below illustrates the convergence Corollary 35 Let n y n g n y n /g n then we have the following inequality Where f n is defined in Theorem 11 P(y n ) = y n g n and P(f n ) = f n g n P(y n ) P(f n ) Corollary 36 Let d n be the number of rows with the value true in the truth tables of all bracketed formulae with n distinct variables connected by the binary connective of m-implication, case(i) Then d n = g n y n, with t 0 = 0, and for large n, d n ( ) 3n 5 πn 3 Using this Corollary 36, it is straightforward to calculate the values of d n The table below illustrates this up to n = 10 n d n

10 4 Parity For brevity, we represent the set of even counting numbers by the capital letter E, the set of odd counting numbers by the capital letter O, and the set of natural numbers, {1,,3,4,}, by N We begin by determining the parity of Catalan number C n, which has the following recurrence relation n 1 C n = C i C n 1, with C 0 = 0,C 1 = 1 (8) i=1 From the Segner s recurrence relation, C n can be expressed as a piecewise function, with respect to the parity of n, (see [7, page 39]) (C 1 C n 1 +C C n ++Cn 1Cn+1) if n O, C n = (C 1 C n 1 +C C n ++Cn Cn+)+C n if n E Lemma 41 (Parity of C n ) [8] Proof C n O n = i, where i N For n, C n O C n Note that C 1 = 1 O O Cn O n = i i N By using Proposition 14, we get the following triangular table Where the left hand side column represents the sum of the corresponding row y : 1 y 3 : 3 3 y 4 : y 5 : y 6 : Theorem 4 (Parity of y n ) The sequence {y n } n 1 preserves the parity of C n 10

11 Proof If an additive partition of y n, (which is determined by the recurrence relation (4)), is odd, then it comes as a pair; ie ( i C i f i )( n i C n i y n i ) O y i,y n i ( ) Hence, ( i C i y i )( n i C n i y n i )+( n i C n i y n i )( i C i y i ) E Thus, y n canbeexpressed asapiecewise functiondepending ontheparity of n: n 1 i=1 ((i C i y i )( n i C n i y n i )) if n O, y n = ) ( n i=1 ((i C i y i )( n i C n i y n i )) +( n Cn y n ) if n E Finally, y n O ( n C n y n ) O yn O n = i, i N Note that y 1 = 1 O Proposition 43 (Parity of d n ) The sequence {d n } n 1 preserves the parity of C n Proof Since d n = g n y n = n C n y n, with n 1 The sequence {g n } n 1 is always even, and the sequence {y n } n 1 preserves the parity of C n by Theorem 4 Therefore the sequence {d n } n 1 preserves the parity of C n 5 A fruitful tree We begin by recalling following definitions: Definition 51 [8], The nth Catalan tree, A n, is a combinatorical object, characterized by one root, (n 1) main-branches, and C n sub-branches Where each main-branch gives rise to a number of sub-branches, and the number of these sub-branches is determined by the additive partition of the corresponding Catalan number, as determined by the recurrence relation (8) 11

12 Definition 5 [8], The Catalan tree A n is fruitful iff each sub-branch of A n has fruits We denote this new tree by A n (µ i ), where {µ i } i 1 is the corresponding fruit sequence Example 53 Let {y n } n 1 be the corresponding fruit sequence for the Catalan tree A n Then A n (y n ) has the following symbolic representation, (( 1 C 1 f 1 )( n 1 C n 1 y n 1 ),,( n 1 C n 1 y n 1 )( 1 C 1 f 1 )) (C 1 C n 1,C C n,,c n C,C n 1 C 1 ) (1,1,,1,1) (1) Example 54 Let {d n } n 1 be the corresponding fruit sequence for the Catalan tree A n Then A n (d n ) has the following symbolic representation, (( n ( 1 C 1 f 1 )( n 1 C n 1 y n 1 )),,( n ( n 1 C n 1 y n 1 )( 1 C 1 f 1 ))) (C 1 C n 1,C C n,,c n C,C n 1 C 1 ) (1,1,,1,1) Proposition 55 For n > 1, let a n (y n ) and a n (d n ) be the total number of components of the fruitful trees A n (y n ) and A n (d n ) respectively Then (1) a n (y n ) = y n +C n +n, and a n (d n ) = d n +C n +n Using Proposition 55, it is straightforward to calculate the values of a n (y n ), and a n (d n ) The table below illustrates this up to n = 10 n a n (y n ) a n (d n ) Corollary 56 For n > 1, a n (y n ), and a n (d n ) are odd iff n O Proof Since, a n = (C n +n) O n O or n = i, and y n,d n O n = i Therefore, a n (y n ),a n (d n ) O n O 1

13 References [1] E A Bender and S G Williamson, Foundations of Applied Combinatorics, Addison-Wesley Publishing Company, Reading, MA, 1991 [] P J Cameron, Combinatorics: Topics, Techniques, Algorithms, Cambridge University Press, Cambridge, 1994 [3] P J Cameron, Sets, Logic and Categories, Springer, London, 1998 [4] P J Cameron and V Yildiz, Counting false entries in truth tables of bracketed formulae connected by implication, Preprint, (arxivorg/abs/ ) [5] T Koshy, Catalan Numbers with Applications, Oxford University Press, New York, 009 [6] D Makinson, Sets, Logic and Maths for Computing, Springer, London, 009 [7] N J A Sloane, The Online Encyclopedia of Integer Sequences, [8] V Yildiz, Catalan tree & Parity of some Sequences which are related to Catalan numbers Preprint, ( Onlar ki kurtulamaz ikiyüzlülükten Canı ayırmaya kalkarlar bedenden; Horoz gibi tepemde testere olsa Aklımın kafasını keser atarım ben Ö Hayyam 13