A New Constructive Proof of Graham's Theorem and More New Classes of Functionally Complete Functions

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1 A New Costructive Proof of Graham's Theorem ad More New Classes of Fuctioally Complete Fuctios Azhou Yag, Ph.D. Zhu-qi Lu, Ph.D. Abstract A -valued two-variable truth fuctio is called fuctioally complete, if all -valued fuctios of m variables ca be expressed i terms of the fiite compositios of this fuctio. R. L. Graham, usig a idirect proof, proved the existece of -valued ( > 3) fuctioally complete truth fuctios of two variables. He foud oe class of fuctioally complete fuctios. I this article, we will provide a costructive direct proof of Graham's Theorem. Usig the same procedure, we foud several more classes of fuctioally complete fuctios. The method ad strategy of our proof ca be widely used to fid more ew types of fuctioally complete fuctios. 1 Itroductio It is well kow that there exist two ad oly two fuctioally complete Sheffer stroke fuctios, ( P) ( Q),ad ( P) ( Q), of -valued prepositioal m calculus. (i.e. all two-valued truth fuctios of m variables ca be defied i terms of fiite compositios of ay oe of these two Sheffer stroke fuctios. For detailed discussio of this cocept, see [) R. L. Graham exteded the result to -valued truth fuctios. He proved the existece of fuctioally complete fuctios for. First, he proved all -valued fuctios of m variables ca be defied i terms of fiite compositios of -valued fuctios of two variables. The, he defied a class of -valued fuctios of two variables, ad usig a idirect proof, proved all -valued fuctios of two variables ca be defied i terms of fiite compositios of ay oe of these fuctios i the class. This implies that all -valued fuctios of m variables ca be defied i terms of fiite compositios of ay oe of these fuctios i the class. Therefore, all fuctios i this class are fuctioally complete (cf. [1).If T() deotes the umber of -valued fuctioally complete truth fuctios of two variables, it is kow that T() =, T(3) = 3774 (cf. [3, 4). For > 3, it is still ukow. Graham showed there are! fuctios i his class. For example, whe = 3, there are 16 such fuctios. Comparig with T(3) = 3774, there should be may other types of fuctioally complete fuctios. I order to fid more fuctioally complete fuctios, we eed to have a method to examie the completeess directly. I the ext sectio, we will give a direct costructive proof of the completeess of the Graham's theory. The direct proof is much easier to uderstad ad ca be widely used o proofs of other types of fuctioally complete fuctios. Joural of Mathematical Scieces & Mathematics Educatio 1

2 A New Costructive Proof of Graham's Theorem I order to compare our proof with Graham s proof, we will use most of the same otatio as i Graham s article (cf.[1). Here are some otatios: I = { 0, 1,, -1 }, where. C = { all -valued truth fuctios of two variables from I I ito I }. C has cardiality. If F C, P(F) deote the set of all truth fuctios of two variables which ca be defied i terms of fiite compositios of the -variable fuctio F(p,q). The otatio [ a1,a,...,a P(F) implies that the fuctio G (p,q) with the truth table [G(0, 0);G(0, 1),,G(0, - 1);G(1, 0) G( - 1; - 1) = [ a1,a,...,a belogs to P(F). Let ω be the bijectio from { 0, 1,, } ito I I such that ω ( 1) = (0; 0); ω () = (0; 1),, ω ( ) =(-1,-1). So, If ω (k) = (p, q), the a k = G (ω (k)) = G(p,q). A fuctio F is called fuctioally complete, if P(F) = C. I other words, F is fuctioal complete if ay truth table [ a1,a,...,a i C belogs to P(F). I order to defie a fuctioally complete fuctio, we itroduce two mappigs. Let σ ad π be arbitrary fixed mappigs of I ito I such that for all a I: ( i ) 0 < r < implies σ (r) (a) a ( where σ (r) (a) deotes the r th iterate ( (... (a)...)) 1 σ 443 σ σ, σ(0) (a) = a. Note that σ is just a permutatio of I, that r cosists of a sigle cycle. ( ii ) There exists r >0 such that π (r) (a) = 0. Now, we defie the fuctio from I I ito I: F σ, π σ (p), 0, (p,q) = π (p), ay value if if if p = q; p = 0 ad q 0; p 0 ad q = 0; otherwise; Abbreviate F σ,π (p,q) by F. Let T r = { t 1,t,,t r } be a subset of { 1,,, }. For b i I, the otatio B r = [b 1, b,, b r P(F) will idicate there exists [ a,a,...,a 1 P(F) such that t i Joural of Mathematical Scieces & Mathematics Educatio Tr a i = b for 1 i r.

3 Graham s Theorem: The above fuctio F σ,π (p,q) is fuctioally complete. ( i.e. P(F σ,π ) = C or all truth tables [ b1,b,...,b i C belog to P(F). ) Before proceedig to the proof of the theorem, we prove some lemmas. Lemma 1: Assume = [t 1, t,, t k { 1,,, }is oempty,. If [b 1, b,, b k P(F), the [σ (m) (b 1 ), σ (m) (b ),,σ (m) (b k ) P(F) for all m. Tk Proof: Whe m = 0, [σ (0) (b 1 ), σ (0) (b ),,σ (0) (b k ) = [b T 1, b,, b k k Tk P(F). If [σ (m) (b 1 ), σ (m) (b ),,σ (m) (b k ) P(F), the F([σ (m) (b 1 ), σ (m) (b ),,σ (m) (b k ) T, [σ (m) (b 1 ), σ (m) (b ),,σ (m) (b k ) k ) = [σ (m1) (b 1 ), σ (m1) (b ),,σ (m1) (b k ) By iductio, the lemma is true. Let A k (1 k ) be the statemet: For ay = [t 1, t,, t k { 1,,, }, all truth tables B = [b 1, b,, b k P(F), for arbitrary b T i I, 1 i k. k If we prove that A is true, i.e. all truth tables B = [b 1, b,, b P(F), for ay bi I, 1 i, the F is fuctioally complete. The proof of the Graham s theorem will be complete. Here is the idea of our proof: First, prove A 1 is true, the prove A is true, ad fially, by iductio, prove A 3,, A are true. First, prove the statemet A 1 : For ay t 1 I, let T 1 = { t 1 }. There exists a [a T 1 P(F), for example, a = F (p, q), where ω (t 1 )= (p,q). By Lemma 1, [σ(a) T 1 P(F), [σ () (a) T 1 P(F),, [σ (-1) (a) T 1 Sice the fuctio σ is a sigle cycle permutatio, we proved for ay b i I, [b i T A 1 1 has bee proved. Before we prove the statemet A, we eed to prove two lemmas. Lemma : For ay = { t 1, t } { 1,,, }, t 1 t, there exist two umbers a ad b i I, such that a b ad [ a, b Proof: For ay p I, ad q I, cosider these two fuctios: K( p, q ) = F( p, p ) = σ(p) P(F), H( p, q ) = F ( q, q ) = σ(q). These two fuctios are defied i terms of F. Therefore, they belog to P(F). Assume ω (t 1 )= ( m 1, 1 ) ad ω (t )= ( m, ). Sice t 1 t, we have either m 1 m or 1. Joural of Mathematical Scieces & Mathematics Educatio 3

4 If m 1 m, the the truth table of K( p, q ) = F( p, p ) o will be [σ(m 1 ), σ(m ) T 1 Sice m 1 m, we have σ(m 1 ) σ(m ). If 1, the the truth table of H( p, q ) = F( q, q ) o will be [σ( 1 ), σ( ) T 1 Sice 1, we have σ( 1 ) σ( ). Lemma is proved. Lemma 3: For ay ={ t 1, t } { 1,,, }, t 1 t, There exists c 0 i I, ad d 0 i I such that [ c, 0 (ii) [ 0, d (iii) [ 0, 0 Proof: Assume [ a, b P(F) ad a b. There exist s ad t such that σ (s) (a) = 0, ad σ (t) (b) = 0. Sice a b, σ (s) (b) 0, σ (t) (a) 0. By Lemma 1, [ σ (s) (a), σ (s) (b) = [ 0, σ (s) (b) P(F), ad [ σ (t) (a), σ (t) (b) = [ σ (t) (a), 0 P(F). Let c = σ (t) (a), ad d = σ (s) (b). ad (ii) are proved. Sice there exist c 0ad d 0. [ c, 0 P(F) ad [ 0, d F([ 0, d, [ c, 0 ) = [0, π(d) If π(d) = 0, the we have [ 0, 0 If π(d) 0, F([0, π(d), [c, 0 ) = [0, π () (d) If π () (d) = 0, the [ 0, 0 P(F), otherwise, we will cotiue the above procedure to get F([0, π () (d), [ c, 0 ) = [0, π (3) (d) P(F), ad so o. Evetually, accordig to the property of π, there is a umber r such that π (r) (d) = 0, ad [0, π (r) (d) = [ 0, 0 (iii) is proved. Therefore Lemma 3 is true. Now, we prove A is true: For ay = { t 1, t } { 1,,, }, by Lemma 3 (iii), [ 0, 0 By Lemma 1, we kow that [ σ σ(0) [ σ () σ () (0) P(F),, [ σ (-1) σ (-1) (0) Joural of Mathematical Scieces & Mathematics Educatio 4

5 By Lemma 3, there exists c I, such that c 0 ad [ c, 0 By Lemma 3 (iii), [ 0, 0 Therefore, F( [ 0, 0, [ c, 0 F([ 0, σ(0), [ σ σ(0), [ σ () σ () (0) ) = [ 0, σ(0) F([ 0, σ () (0) ; F([ 0, σ (-) (0), [ σ (-) σ (-) (0) P(F); ) = [ 0, σ () (0) ) = [ 0, σ (3) (0) ) = [ 0, σ (-1) (0) P(F); P(F); Accordig to the property of σ, we kow that for all b I, [ 0, b By Lemma 1, for ay s I, ad ay b I, [ σ (s) σ (s) (b) That is: for ay s I, for ay b I, [ σ (s) b P(F). This implies for ay a I ad b I, [ a, b A has bee proved. Now, we prove A k by iductio, for all k where 3 k : Assume A m is true for all m k, where k, we are goig to prove that A k1 is true. We abbreviate truth table [ b 1, b,, b k by B K. ad deote [ σ (s) (b 1 ), σ (s) (b ),,σ (s) (b k ) By σ (s) (B k ). We will prove that for ay B k o ay set = { t 1, t,, t k } 1 { 1,,, }, ad ay β I, the truth table [ B k, β We eed to prove some lemmas: Lemma 4: For ay sets 1 = { t 1, t,, t k1 } { 1,,, }, ad ay B k-1 = [ b 1, b,, b k-1 -, where -1 = { t 1, t,, t k-1 }, there exists a iteger s, such 1 that [ σ (s) (B k-1 ), 0, 0 T Proof: Sice A k is true, there exist α I ad β 0 I, such that [ B k-1, α, 0 T P(F) ad [ B, 0, β k-1 0 T If α = 0 or β = 0, the Lemma 4 is 0 true, with s = 0. If ot, the α 0 ad β 0 0. Cosider F( [ B k-1, α, 0 T, [ B, 0, β k-1 0 T ) = [ σ( B ), π( α ), 0 k-1 T If π( α ) = 0, the Lemma 4 is proved with s =1. Otherwise there exists a umber β 1 I, such that [ σ( B k-1 ), 0, β 1 T If β = 0, the Lemma 4 is 1 true, with s = 1. Otherwise, if β 1 0, the Joural of Mathematical Scieces & Mathematics Educatio 5

6 F( [ σ( B k-1 ), π( α ), 0, [ σ( B ), 0, β ) = [ σ () ( B ), π () ( α ), 0 k-1 1 k-1 If π () ( α ) = 0, the Lemma 4 is proved with s =. Otherwise cotiue the above procedure. Evetually, there exists a iteger s such that either β s = 0, or π (s) ( α ) = 0. Therefore, [ σ (s) ( B k-1 ), 0, 0 P(F) for some atural umber s. Lemma 5: For ay set 1 = { t 1, t,, t k1 } { 1,,, }, ad ay B k-1 = [ b 1, b,, b k-1, where -1 = { t 1, t,, t k-1 } 1, if [ σ (s) (B k-1 ), 0, 0-1 P(F), the [ σ (s1) (B ), 0, 0 k-1 Proof: Let s discuss the followig two cases: Case 1: There exist α 0, ad β 0, such that [ σ (s) ( B k-1 ), α, β T Case : If Case 1 is ot true, the for ay α ad β, [ σ (s) ( B k-1 ), α, β T P(F) implies that either α = 0 or β = 0. Proof of Case 1: If there exist α 0, ad β 0 such that [ σ (s) ( B k-1 ), α, β T P(F), the F( [ σ (s) (B), 0, 0 T, [ σ (s) ( B ), α, β k-1 T ) = [ σ (s1) ( B ), 0, 0 k-1 T P(F). Proof of Case : For ay α 0, there exists a umber β I, such that [ σ (s) ( B k-1 ), α, β T Sice α 0, β must be 0. Therefore, for all α 0, [ σ (s) ( B k- 1 ), α, 0 Similarly, we have that for all β 0 [ σ (s) ( B ), 0, β k-1 By choosig some α ad β, such that α 0, π( α ) = 0 ad β 0, we have that F( [ σ (s) ( B k-1 ), α, 0 T, [ σ (s) ( B ), 0, β k-1 T ) = [ σ (s1) ( B ), 0, 0 k-1 T P(F). Lemma 6: For ay B k-1 = [ b 1, b,, b k-1-1,where -1 = { t 1, t,, t k-1 } 1, if [ σ (s) (B k-1 ), 0, 0 T P(F), for some s I, the [ B, α, α k-1 T P(F), for all α I.. Proof: Sice [ σ (s) (B k-1 ), 0, 0 P(F), by Lemma 5, we have [ σ (s1) (B ), 0, k-1 0 Cotiuously usig Lemma 5, we have that [ σ (s) (B ), 0, 0 k-1 Joural of Mathematical Scieces & Mathematics Educatio 6

7 [ σ (s3) (B k-1 ), 0, 0 P(F),, [ σ (s-1) (B k-1 ), 0, 0 P(F). This meas [ σ (r) (B k-1 ), 0, 0 P(F) for all r. For ay α = σ (t) ( 0 ), for some t, by the result above, let r = -t, [σ (-t) (B k-1 ), 0, 0 By Lemma 1, [ σ () (B), σ (t) ( 0 ), σ (t) (0) Therefore, [ B k-1, α, α By Lemma 4, ad Lemma 6, we will get Lemma 7 immediately: Lemma 7: For ay B k-1 = [ b 1, b,, b k-1 α, α -1, ad ay α I, we have [ B k-1, This implies that if b i = b j for some i j, the B k1 = [ b 1, b, b i,... b j,, b k1 Now, we assume that A k is true, where k, ad we goig to prove for ay B k = [ b 1, b,, b k, ad ay β I, [ B k, β This implies that A k1 is true. Proof: For ay B k = [ b 1, b,, b k, ad ay β = σ (t) ( 0 ) I for some t, If If all b s are i b i =b j, for some i j, the by Lemma 7, [ B k, β distict, there exists a umber γ, such that [ B k, γ T Assume that σ (s) ( γ ) = 0, for some s. Sice all b i s are distict, there exists some b i γ. Without loosig of geerality, assume that b k γ. We have B k = [ B k, b k, ad [ B k-1, b k, γ T P). Sice σ (s) ( γ ) = 0, by Lemma 1, we have that [σ (s) ( B k-1 ), σ (s) ( b k ), σ (s) ( γ ) T = [σ (s) (B ), σ (s) ( b ), 0 k-1 k T By Lemma 7, we have [σ (s) ( B k-1 ), σ (s) ( b k ), σ (s) ( b k ) T Sice b γ, ad σ (s) ( γ ) = k 0, so, σ (s) ( b k ) 0. F([σ (s) ( B k-1 ), σ (s) ( b k ), 0 T, [σ (s) ( B ), σ (s) ( b ), σ (s) ( 0 ) k-1 k T ) = [σ (s1) ( B k-1 ), σ (s1) ( b k ), 0 T F([σ (s1) ( B k-1 ), σ (s1) ( b k ), 0 T, [σ (s1) ( B ), σ (s1) ( b ), σ( 0 ) k-1 k T ) = [σ (s) ( B k-1 ), σ (s) ( b k ), 0 T F([σ (s) ( B k-1 ), σ (s) ( b k ), 0 T,[σ (s) ( B ), σ (s) ( b ), σ( 0 ) k-1 k T ) = [σ (s3) ( B k-1 ), σ (s3) ( α ), 0 T, Joural of Mathematical Scieces & Mathematics Educatio 7

8 Cotiue this, we will have [σ (sj) ( B k-1 ), σ (sj) ( α ), 0 P(F), for ay umber j. Sice β = σ (t) ( 0 ), choose j such that s j = t. We have [σ (-t) ( B k- 1 ), σ (-t) ( b k ), 0 T P(F), By Lemma 1, we have [σ () ( B k-1 ), σ () ( b k ), σ (t) ( 0 ) = [ B, b, β = [ B, β P(F) k-1 k k A k1 is proved. Therefore, A k is true, for all k. Especially, A is true. This implies that all truth tables [ b1,b,...,b i C belog to P(F). The Graham s theorem is proved. 3 More New Classes of Fuctioally Complete Fuctios The followig fuctios are fuctioally complete. The proofs of completeess of these fuctios are similar to the above proof of the Graham s Theorem, therefore, we oly give the proof of the first fuctio. 1. (i σ F 1(p,q) = σ 1) (max{i,j}) if σ (0) = p = q 0; (j) if p = σ (0) ad q = σ i j, 0 i, j -1;. If we chage the max { i, j } of the fuctio F 1 to mi { i, j }, we will have aother fuctioally complete fuctio: (i σ F (p,q) = σ 1) (mi{i,j}) if σ (0) = p = q 0; (j) if p = σ (0) ad q = σ i j, 0 i, j -1; 3. If we chage the max { i, j } of the fuctio F 1 to ij, we will have aother fuctioally complete fuctio: F (p,q) 3 (i σ (i σ 1) = j) if p = σ if σ (0) ad q = σ (0) = p = q 0; (j) i j, 0 i, j -1; 4. If is prime, ad 3, the followig fuctio is fuctioally complete. Joural of Mathematical Scieces & Mathematics Educatio 8

9 (p,q) F 4 σ (p), σ (q), = σ (p), ay value, if p = q; if p = 0 ad q 0; if p 0 ad q = 0; otherwise; 5. (i 1) σ π ( σ (0)), (i 1) σ F5 (p,q) = σ σ ay value, if σ if p = σ (0) = p = q 0; if p = 0 ad q = σ (0) 0 ad q = 0; (0) 0; if p = σ (0) ad p q ad q = σ if q = σ (0) ad p q ad p = σ otherwise; (0) 0; (0) 0; 6. F (p,q) 6 (i 1) σ π ( σ (0)), (i σ (max{ i, j}) σ = 1) if p = σ if p = σ if p = 0 ad q = σ (0) if σ 0, ad q (0) = p = q; (0) 0 ad q = 0; = σ (j) (0) 0; (0) 0, ad p q; As a example, we oly prove the first fuctio: Theorem. The fuctio F 1 is fuctioally complete, where (i σ F 1(p,q) = σ 1) (max{i,j}) if p = σ if σ (0) ad q = σ (0) = p = q 0; (j) i j, 0 i, j -1; We will follow the same procedure as the proof of the Graham s Theorem: Lemma 1, A 1, Lemma, Lemma 3, A, Lemma 4, Lemma 5, Lemma 6, Lemma 7 ad A k1. All lemmas state exactly the same as lemmas i the last sectio. If the proof is also the same, we will ot repeat. Abbreviate the fuctio F 1 (p,q) by F. Lemma 1: Assume = [t 1, t,, t k { 1,,, }is oempty,. If [b 1, b,, b k P(F), the [σ (m) (b 1 ), σ (m) (b ),,σ (m) (b k ) P(F) for all m. Tk Proof: The same as the proof of Lemma 1 i the last sectio. Joural of Mathematical Scieces & Mathematics Educatio 9

10 Let A k (1 k ) be the statemet: For ay = [t 1, t,, t k { 1,,, }, all truth tables B = [b 1, b,, b k P(F), for arbitrary b T i I, 1 i k. k Proof of A 1 : The same as the proof of A 1 i the last sectio. Lemma : For ay = { t 1, t } { 1,,, }, t 1 t, there exist two umbers a ad b i I, such that a b ad [ a, b Proof: The same as the proof of Lemma i the last sectio. Lemma 3: For ay ={ t 1, t } { 1,,, }, t 1 t, There exists c 0 i I, ad d 0 i I such that [ c, 0 (ii) [ 0, d (iii) [ 0, 0 Proof: The proof of case ad (iii) are the same as the proof of case ad (ii) of Lemma 3 i the last sectio. We ow prove case (iii): By case ad (ii) of Lemma 3 ad Lemma 1, there exist s -1 ad t -1 such that [ σ (-1) σ (s) (0) F([σ (-1) σ (s) (0), [ σ (t) σ (-1) (0) By Lemma 1, [ 0, 0 Proof of A : P(F), ad [ σ (t) σ (-1) (0) P(F); ) = [σ (-1) σ (-1) (0) P(F); Proof: By Lemma 3, there exists s such that [ σ (s) 0 Lemma 3 (iii) ad Lemma 1, we have [ σ (r) σ (r) (0) 1,. F( [σ (s) 0, [σ σ(0) ) = [σ (s), σ(0) F( [σ (s), σ(0), [ σ () σ () (0) ) = [σ (s) σ () (0) ; F([σ (s) 0, [ σ (s-1) σ (s-1) (0) ) = [σ (s) σ (s-1) (0) Joural of Mathematical Scieces & Mathematics Educatio 10 By P(F), for all 0 r P(F); P(F); Sice [ σ (s) 0 P(F), by Lemma 1, [ 0, σ (-s) (0) F( [0, σ (-s) (0), [ σ σ(0) ) = [σ σ (-s) (0) By Lemma 1, [σ (s) σ (-1) (0) F( [0, σ (-s) (0), [ σ () σ () (0) ) = [σ () σ (-s) (0)

11 By Lemma 1, [σ (s) σ (-) (0) ; F( [0, σ (-s) (0), [ σ (-s-1) σ (-s-1) (0) ) = [σ (-s-1) σ (-s) (0) By Lemma 1, [σ (s) σ (s1) (0) Also, we have [σ (s) σ (s) (0) Therefore, for ay b I, we have [σ (s) b By Lemma 1, for ay r, [σ (r) ( σ (s) (0)), σ (r) (b) That is, for ay a, b i I, [a, b A is proved. Now, we prove A k for all k by iductio, where 3 k : Assume A m is true for all m k, we are goig to prove that A k1 is true. Lemma 4: For ay sets 1 = { t 1, t,, t k1 } { 1,,, }, ad ay B k-1 = [ b 1, b,, b k-1 -, where T 1 k-1 = { t 1, t,, t k-1 }, there exists a iteger s, such that [ σ (s) (B k-1 ), 0, 0 T Proof: Sice A k is true, there exist α I ad β I, such that [ B k-1, α, σ (-1) (0) T P(F) ad [ B, σ (-1) β k-1 T If α = σ (-1) (0) or β = σ (-1) the by Lemma 1, [ σ( B k-1 ), 0, 0 T Lemma 4 is true, ad s = 1. If ot, the α σ (-1) (0) ad β σ (-1) (0). Cosider F( [ B k-1, α, σ (-1) (0) T, [ B, σ (-1) β k-1 T ) = [ σ( B ), σ (-1) σ (- k-1 1) (0) T By Lemma 1, [ σ () ( B k-1 ), 0, 0 Lemma 4 is true, ad s =. Lemma 5: For ay set 1 = { t 1, t,, t k1 } { 1,,, }, ad ay B k-1 = [ b 1, b,, b k-1, where -1 = { t 1, t,, t k-1 } 1, if [ σ (s) (B k-1 ), 0, 0-1 P(F), the [ σ (s1) (B ), 0, 0 k-1 Proof: Sice [ σ (s) (B k-1 ), 0, 0 σ (-1) (0) P(F), by Lemma 1, [ σ (s-1) (B k-1 ), σ (-1) Case 1: There exist α 0, ad β 0, such that [ σ (s) ( B k-1 ), α, β Joural of Mathematical Scieces & Mathematics Educatio 11

12 Case : If Case 1 is ot true, the for ay α ad β, [ σ (s) ( B k-1 ), α, β T P(F) implies that either α = 0 or β = 0. Proof of Case 1: If there exist α 0, ad β 0 such that [ σ (s) ( B k-1 ), α, β T P(F), the [ σ (s-1) ( B k-1 ), σ (-1) ( α ), σ (-1) ( β ) T P(F) Sice α 0, ad β 0, so σ (-1) ( α ) σ (-1) ( 0 ), ad σ (-1) ( β ) σ (-1) ( 0 ), F([ σ (s-1) ( B k-1 ), σ (-1) ( 0 ), σ (-1) ( 0 ) T, [ σ (s-1) ( B ), σ (-1) ( α ), σ (-1) ( β k-1 ) T ) = [ σ (s) ( B k-1 ), σ (-1) ( 0 ), σ (-1) ( 0 ) T By Lemma 1, [ σ (s1) (B k-1 ), 0, 0 T Proof of Case : For ay α 0, ad β 0, we have that [ σ (s) ( B k-1 ), α, 0 T P(F), ad [ σ (s) ( B k-1 ), 0, β T By Lemma 1, [ σ (s-1) ( B k-1 ), σ (-1) ( α ), σ (-1) ( 0 ) T P(F), ad [ σ (s-1) ( B ), k-1 σ (-1) ( 0 ), σ (-1) ( β ) T Sice α 0, ad β 0, so σ (-1) ( α ) σ (-1) ( 0 ), ad σ (-1) ( β ) σ (-1) ( 0 ), F([ σ (s-1) ( B k-1 ), σ (-1) ( α ), σ (-1) ( 0 ) T, [ σ (s-1) ( B ), σ (-1) ( 0 ), σ (-1) ( β k-1 ) T ) = [ σ (s) ( B k-1 ), σ (-1) ( 0 ), σ (-1) ( 0 ) T By Lemma 1, [ σ (s1) (B k-1 ), 0, 0 Lemma 6: For ay B k-1 = [ b 1, b,, b k-1-1,where -1 = { t 1, t,, t k-1 } 1, if [ σ (s) (B k-1 ), 0, 0 T P(F), for some s I, the [ B, α, α k-1 T P(F), for all α I.. Proof: The same as the proof of Lemma 6 i the last sectio. Lemma 7: For ay B k-1 = [ b 1, b,, b k-1-1, ad ay α I, we have [ B k-1, α, α T This implies that if b i = b j for some i j, the B k1 = [ b 1, b, b i,... b j,, b k1 Proof: The same as the proof of Lemma 7 i the last sectio. Joural of Mathematical Scieces & Mathematics Educatio 1

13 Proof of A k1 : Proof: We eed to prove that for ay B k = [ b 1,, b i,... b j,, b k, ad ay β = σ (t) (0) I, [ B k, β T By Lemma 7, if there exist i j, ad b = b, i j where 1 i, j k, the [ B k, β T Now, we discuss the case: All b i s i B k are distict. There exists s such that [ B k, σ (s) (0) T By Lemma 1, [ σ (-1-s) ( B ), σ (-1) ( 0 ) k T Sice all b i s i B k are distict, there exists a b i i B k, such that σ (-1-s) ( b i ) σ (- 1) ( 0 ). By Lemma 7, [ σ (-1-s) ( B k ), σ (-1-s) ( b i ) T F([ σ (-1-s) ( B k ), σ (-1) ( 0 ) T, [ σ (-1-s) ( B ), σ (-1-s) ( b ) k i T ) = [ σ (-s) ( B k ), σ (-1) ( 0 ) Sice all b i s i B k are distict, there exists a b i i B k, such that σ (-s) ( b i ) σ (-1) ( 0 ). By Lemma 7, [ σ (-s) ( B k ), σ (-s) ( b i ) T F([ σ (-s) ( B k ), σ (-1) ( 0 ) T, [ σ (-s) ( B ), σ (-s) ( b ) k i T ) = [ σ (-s1) ( B k ), σ (-1) ( 0 ) T Cotiue doig this, we have for ay r, [ σ (r) ( B k ), σ (-1) ( 0 ) T We kow that β = σ (t) (0). Let r = -1-t. We have that [ σ (-1-t) ( B k ), σ (-1) ( 0 ) T By Lemma 1, we have [ B, σ (t) (0) k T = [ B k, β A is proved. k1 Therefore, by iductio, A k is true for all k. Especially, fuctio F 1 is fuctioally complete. 4 Summary A is true. The By similar methods, we ca prove that other classes of fuctios F, F 3,, F 6 are fuctioally complete. The completeess of may classes of fuctios with the permutatio fuctio σ ca be examied i this way: A 1, A, the by iductio A. If a fuctio F( p, q ) is fuctioally complete, the either this fuctio satisfies the property F ( p, p ) = σ (p) for some permutatio σ, or some compositio of fuctio F ( p, q ) satisfies this property. Therefore, fidig more classes of fuctioally complete fuctios with fuctio σ is very importat i the research of fidig all fuctioally complete fuctios i the space C. Joural of Mathematical Scieces & Mathematics Educatio 13

14 Azhou Yag, Ph.D., Beijig Uiversity of Techology, Chia Zhu-qi Lu, Ph.D., Uiversity of Maie at Presque Isle, USA Refereces [1 Graham, R.I. O -valued fuctioally complete truth fuctios, Joural of Symbolic Logic. Vol 3 (1967) pp [ Rosser, J. B. May-valued Logic, North-Hollad, Amsterdam, 195 [3 Foxley, E. The determiatio of all Sheffer fuctios i 3-valued logic, usig a logical computer, Notre Dame Joural of Formal Logic. Vol 3 (196) pp [4 Marti, N. M. The Sheffer fuctios of 3-valued logic, Joural of Symbolic Logic. Vol 19 (1954) pp Joural of Mathematical Scieces & Mathematics Educatio 14

Section Mathematical Induction and Section Strong Induction and Well-Ordering

Section Mathematical Induction and Section Strong Induction and Well-Ordering Sectio 4.1 - Mathematical Iductio ad Sectio 4. - Strog Iductio ad Well-Orderig A very special rule of iferece! Defiitio: A set S is well ordered if every subset has a least elemet. Note: [0, 1] is ot well