WEAK PROBABILITY POLYADIC ALGEBRAS. S. Marinković, M. Rašković, R. D ord ević

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1 FACTA UNIVERSITATIS (NIŠ) Ser. Math. Inform. 16 (2001), 1 12 WEAK PROBABILITY POLYADIC ALGEBRAS S. Marinković, M. Rašković, R. D ord ević Abstract. We are motivated by the connection between Boolean algebra and propositional logic, cylindric algebra and predicate logic, Keisler s L(V,ν,m) logic and polyadic algebra [2], to introduce an algebra calling weak probability polyadic algebra which corresponds to the weak probability logic with infinitely predicates L(V,ν,m,R), introduced in [4]. Structure (B,+,,, 0, 1,C,C r,s τ ), where (B,+,,, 0, 1) is a Boolean algebra, C, C r and S τ are unary operations on B, for each sequence of ordinals from β of length α, α β, eachr R (R is as in [4]) and τ β β, is a weak probability polyadic algebra of dimension β, brieflywpp β, if the following postulates hold: (WPP 1 ) (B,+,,, 0, 1,C,S τ ) is a polyadic algebra of dimension β, (WPP 2 ) (i) C( ) r x = x, r > 0, (ii) C r 0=0, r > 0, (WPP 3 ) C 0 x =1, (WPP 4 ) C r x Cs x, r s, (WPP 5 ) C r Cs x = Cs x, r > 0, (WPP 6 ) (i) C r x C1 s y Cmin{1,r+s} (x + y), (ii) C r x Cs y C1 (x y) Cmin{1,r+s} (x + y), (WPP 7 ) (i) C r x C1 r x, (ii) C s x C1 r x, s > r, (iii) C 1 s x Cs+ x,wheres+ =min{r R r>s}, (WPP 8 ) S σ C r x = S τ C r x, if σ (β \ rang K) =τ (β \ rang K), (WPP 9 ) S σ Cσ r x = Cr S σx, if σ σ is 1 1 function, (WPP 10 ) (i) C r x C x, r > 0, (ii) C C(K r 1 ) x = Cr (K 1 ) x, rang K rang K 1 (iii) C r C (K 1 )x = C (K1 )x, r > 0, rang K rang K 1. Received November 5, Mathematics Subject Classification. Primary 03C70. 1

2 2 S. Marinković, M. Rašković, R. D ord ević Theorem 1. If A is WPP β algebra, then (1) C r 1=1, (2) If x y, then C r x Cr y, (3) C r x + Cr y Cr (x + y), (4) C r (x y) Cr x Cr y, (5) C 1 x = x iff C x = x, (6) S τ 1=1. The proof is similar to that one in [6]. Let V 1 be a set of new variables such that V 1 V =, V 1 P = and V 1 = V. Let be V = V V 1, τ 0 : V 1 V 1, G the set of all sentences in L, H = {Φ V f (Φ) V, Φ Form L } and Γ G. For each Φ, Ψ H we define Φ/Γ = {Ψ Γ L Ψ Φ, Ψ H}, H Γ = {Φ/Γ Φ H}, 0 Γ = /Γ, 1 Γ = /Γ and operations + Γ, Γ, Γ on H Γ as usual. It is well known the next result: Lemma 1. The set of sentences Γ is consistent in L iff (H Γ, + Γ, Γ, Γ, 1 Γ, 0 Γ ) is notrivial Boolean algebra. For each sequence x of different variables from V, r R, τ V V and Φ H we define: ( Γ x)(φ/γ )= ( ( x)φ ) /Γ, (P Γ x r)(φ/γ )= ( (Px r)φ ) /Γ, S Γ (τ)(φ/γ )= ( S f (τ)s f (τ0 )S(τ 0)Φ ) /Γ. Lemma 2. (i) If Φ is atomic formula and τ V V,then S Γ (τ)(φ/γ )= ( S f (τ)φ ) /Γ =(S(τ)Φ) /Γ. (ii) If τ (V \ V b (Φ)) V, then S Γ (τ)(φ/γ )= ( S f (τ)φ ) /Γ. (iii) If τ V V is 1 1, thens Γ (τ)(φ/γ )=(S(τ)Φ)/Γ. Lemma 3. If m = V + and Γ is consistent in L,then H Γ = ( H Γ, + Γ, Γ, Γ, 1 Γ, 0 Γ, Γ x, P Γ x r, S Γ (τ) ), where τ V V, r R and x is a sequence of different variables from V of length α, α<m, is a weak probability polyadic algebra of dimension V. Proof. Let be Φ Γ = Φ/Γ. It follows from Lemma 1 (H Γ, + Γ, Γ, Γ, 1 Γ, 0 Γ )is a Boolean algebra. The operations Γ x, P Γ x r and S Γ (τ) are well defined and the set H Γ is closed for them. It is easy to verify (by using the axioms and the rules of inference of L, see [4]) that all axioms (WPP 1 ) (WPP 10 )holdinh Γ. We shall only prove (WPP 10 ):

3 Weak Probability Polyadic Algebras 3 (i) (P Γ x r)φ Γ = ( (Px r)φ ) Γ ( ) Γ Γ (Px > 0)Φ for r>0, by A 10 ( ) Γ Γ ( x)φ by contraposition A 12 =( Γ x)φ Γ. (ii) ( Γ x)(p Γ y r)φ Γ = ( ( x)(py r)φ ) Γ Γ ( (Py r)φ ) Γ =(P Γ y r)φ Γ, by supposing rang x rang y. The above inequality follows from Γ L ( x)(py r)φ (Py r)φ that we obtain from Γ L ( x) ( (Py < r)φ (Py < r)φ ) by using A 2. (iii) ( Γ y)φ Γ = ( ( y)φ ) Γ ( ) Γ ( ) Γ (Px r)( y)φ for V f ( y)φ rang x =, byr3 and A 4 =(P Γ x r)( Γ y)φ Γ. Conversely, (P Γ x r)( Γ y)φ Γ = ( (Px r)( y)φ ) Γ ( ) Γ Γ (Px > 0)( y)φ for r>0, by A 10 ( ) Γ Γ ( x)( y)φ by A 12 = ( ( y)φ ) Γ from A 2, since rang x rang y =( Γ y)φ Γ. Let A =(A, R p,µ α ) p P,α<V + be a weak probability structure for L and V = β, Φ A = {a A β = A Φ[a V ]} and A = {Φ A Φ H}.

4 4 S. Marinković, M. Rašković, R. D ord ević We shall define for each sequence of ordinals from β of length α, α β, each r R and each σ β β : c (Φ A )= { a A β b (β \ rang ) = a (β \ rang ), for some b Φ A}, c r (ΦA )={a A β µ α {b b Φ A, b (β \ rang ) = a (β \ rang )} r}, { s σ (Φ A )= a A β a σ ( S f (τ0 )S(τ 0)Φ ) } A, where a σ = ( a σ(α) )α<β for a =(a α) α<β. It is easy to verify that the above operations are well defined and that the sets from definition of c r are measurable. We may show that A is closed for these operations. c (Φ A )= { a A β b (β \ rang ) = a (β \ rang ), for some b Φ A} = {a A β b (β \ rang ) = a (β \ rang ), for some b such that = A Φ[b V ]} = {a A β = A ( x)φ[a V ]} =(( x) Φ) A A. c r (ΦA )={a A β µ α {b K b Φ A, b (β \ rang ) = a (β \ rang )} r} = {a A β µ α {b K = A Φ[b V ], b (β \ rang ) = a (β \ rang )} r} = { a A β = A (Px r)φ[a V ] } =((Px r)φ) A A. { s σ (Φ A )= a A β a σ ( S f (τ0 )S(τ 0)Φ ) } A = { a A β = A S f (τ0 )S(τ 0)Φ[a (σ V )] } = { a A β = A S f (σ)s f (τ0 )S(τ 0)Φ[a V ] } = ( S f (σ)s f (τ0 )S(τ 0)Φ ) A A. The algebra (A,,,,,A β,c,c r,s σ) will be called weak probability polyadic set algebra.

5 Weak Probability Polyadic Algebras 5 Lemma 4. Let f : H Γ Abe defined by f(φ Γ )=Φ A. Then, if = A Γ then f is homomorphism H Γ onto A and A is a weak probability polyadic algebra. Theorem 2. Let B =(B,+,,, 0, 1,C,C r,s σ) be a weak probability polyadic algebra of infinite dimension β. Let be V = β, P = {p p B}, m = β + and ν(p) =β for each p P. Then there exists a set of sentences Γ in logic L = L(V,ν,m,R) such that H Γ = B. Proof. We shall put V = β, P = B, ν(p) =β for each p P and x β β 1 1 and onto. Let Γ be the next set of sentences: ( x)(( p)(x) (p(x))), ( x)((p + q)(x) p(x) q(x)), ( x)((p q)(x) p(x) q(x)), ( x)((s τ p)(x) S(τ)p(x)), ( x) ( (C p)(x) ( y)p(x) ), ) ( x) ((C r p)(x) (Py r)p(x), where p, q P, τ V V, r R and =y is a sequence { of different variables of length α, α<m. First, we shall show that H Γ = (p(x)) Γ p P. } For each formula Φ H there exists formula Ψ = S f (τ0 )S(τ 0)Φ H such that Φ Γ = Ψ Γ, V b (Ψ) V 1 and V f (Ψ) V. Really, Ψ Γ = S f (id)s f (τ 0 )S(τ 0 )Φ Γ = S Γ (id)φ Γ = Φ Γ. We shall prove that for each formula Φ from H, with free variables in V and bound in V 1, there exists formula Ψ F such that Φ Γ = Ψ Γ. Let be τ 1 : V 1 1 V. Then na Γ L S(τ 1 )Φ S f (τ 0 )S(τ 0)S(τ 1 )Φ. By (R 3 ), we have Γ L S(τ 1 )S(τ 1)Φ S(τ 1 )S f (τ 0 )S(τ 0)S(τ 1 )Φ. Since V b (S f (τ 0 )S(τ 0)S(τ 1 )Φ) V 1, it follows S(τ 1 )S f (τ 0 )S(τ 0)S(τ 1 )Φ = S f (τ 1 )S f (τ 0 )S(τ 0)S(τ 1 )Φ = S f (τ 0 )S(τ 0)S f (τ 1 )S(τ 1)Φ.

6 6 S. Marinković, M. Rašković, R. D ord ević So, Γ L Φ S f (τ0 )S(τ 0)S f (τ1 )S(τ 1)Φ. Putting Ψ = S f (τ1 )S(τ 1)Φ, wehave Γ L Φ Ψ, Ψ F. It follows that it is enough to prove that for each formula Φ F there exists p P such that Φ Γ = p(x) Γ. We can prove it by induction of complexity of Φ. Now, we shall define a function g : B H Γ on the following way: g(p) =(p(x)) Γ. It is easy to verify that g is well defined, onto and homomorphism. To prove that it is one to one we shall define h : F B such that, for Φ F holds: if Γ L Φ then h(φ) =1. For formulas from F = {Φ F V b (Φ) V f (Φ) = } we define the mapping h by induction: h (p(x)) = p, h (p(τ y)) = S τ p, h( Φ) = h(φ), h(φ Ψ) =h(φ)+h(ψ), h (( y)φ) =C h(φ), h ((Py r)φ) =C r h(φ). For Φ F \ F, let be h(φ) =S (τ 1 V f (Ψ)) V h(ψ), where Ψ = S(τ 1 )S f (τ 0 )S(τ 0)Φ and τ 1 is a fixed injection from V onto V. First, we shall prove that for each function σ : V f (Φ) V \ V b (Φ) and every formula Φ F we have (*) h(s f (σ)φ) =S σ V h(φ). For Φ F the proof is by induction on the complexity of formula. If Φ = p(x), then h(s f (σ)φ) =h(p(σ x)) = S σ p = S σ V h(φ).

7 Weak Probability Polyadic Algebras 7 If Φ = p(y), y = τ x, τ V V,wehave h (S f (σ)p(y)) = h (p(σ τ x)) by definition S f = S σ τ p by definition h = S σ V S τ p by (P 9 ) (see [1]) = S σ V h (p(y)) by definition h. If Φ = Ψ or Φ = Ψ Θ it is easy to verify. If Φ =(Py r)ψ, then h (S f (σ)φ) =h (S f (σ)(py r)ψ) = h((py r)s f (σ 1 )Ψ where σ 1 = σ V f (Ψ) = C r S σ 1 V h(ψ) = S σ1 V C r σ h(ψ) 1 = S σ V C r h(ψ) since σ =id = S σ V h((py r)ψ). To show that ( ) holds for each formula from F \ F, we shall prove, first, that (**) h(s(σ)φ) =S (σ Vf (Φ)) V h(φ), for each formula Φ F and each σ : V (Φ) 1 1 V. First, let be Φ F. If Φ = p(x), then h(s(σ)p(x)) = h(p(σ x)) = S σ p = S σ Vf (Φ)h(Φ). If Φ =( y)ψ, wehave h(s(σ)φ) =h(( σ y)s(σ)ψ) = C σ h(s(σ)ψ) = C σ S (σ Vf (Ψ)) V h(ψ) by induction hypothesis = S (σ Vf (Ψ)) V C h(ψ) by (P 12 ) = S (σ Vf (Φ)) V C h(ψ) by (P 11 ) = S (σ Vf (Φ)) V h(( y)ψ) =S (σ Vf (Φ)) V h(φ).

8 8 S. Marinković, M. Rašković, R. D ord ević If Φ F \ F and σ : V (Φ) 1 V, let be Θ = S(τ 1 )S f (τ0 )S(τ 0)S(σ)Φ, Ψ = S(τ 1 )S f (τ0 )S(τ 0)Φ and σ 1 : V (Ψ) V defined by σ 1 V b (Ψ) = τ 1 τ 0 σ τ0 τ1 V b (Ψ) andσ 1 V f (Ψ) =τ 1 σ τ1 V f (Ψ). Then h(s(σ)φ) =S (τ 1 V f (Θ)) V h(θ) =S (τ 1 V f (Θ)) V h(s(τ 1)S f (τ0 )S(τ 0)S(σ)Φ) 1 V f (Θ)) V h(s(σ 1)S(τ 1 )S f (τ0 )S(τ 0)Φ) 1 V f (Θ)) V h(s(σ 1)Ψ 1 V f (Θ)) V S (σ 1 V f (Ψ)) V h(ψ) 1 V f (Θ)) V S (τ 1 σ τ 1 V f (Ψ)) V h(ψ) 1 V f (Θ)) V (τ 1 σ τ 1 V f (Ψ)) V h(ψ) = S (σ Vf (Φ)) V S τ 1 V f (Ψ)) V h(ψ) =S (σ V f (Φ)) V h(φ). By using ( ) it is easy to verify that for each formula Φ F holds h(φ) =S (τ 1 V f (Ψ)) V h(ψ), where Ψ = S(τ 1 )S f (τ0 )S(τ o)φ. Now, we shall prove that ( ) is valid for each formula Φ from F \F.Letbe Θ = S(τ 1 )S f (τ0 )S(τ 0)S f (σ)φ and Ψ = S(τ 1 )S f (τ0 )S(τ 0)Φ. Then h(s f (σ)φ) =S (τ 1 V f (Θ)) V h(θ) 1 V f Θ)) V h(s f (τ 1 σ τ1 )S(τ 1)S f (τ0 )S(τ 0))Φ) 1 V f (Θ)) V S (τ 1 σ τ 1 V f (Ψ)) V h(ψ) = S σ V S (τ 1 V f (Ψ)) V h(ψ) =S σ V h(φ). Now, we shall prove that Γ L Φ implies h(φ) = 1. It follows from Γ F, Φ F, that Γ L Φ iff Γ L Φ. So, it is enough to show that from Γ L Φ follows h(φ) = 1. At the beginning, we shall prove that each axiom maps to 1. First, let the axiom be from F. (A 1 ) It is sufficient to show that h(φ) 1 implies Φ is not tautology. Let I be maximal ideal of Boolean algebra (B,+,,, 0, 1) which contains h(φ) and π natural homomorphism from B onto two-elements Boolean algebra B/I. Fromh(Φ) 1 it follows π h(φ) = 0, i.e. Φ is not tautology.

9 Weak Probability Polyadic Algebras 9 (A 2 ) h (( y)(φ Ψ) (Φ ( y)ψ)) = h ( ( y) (Φ Ψ) (Φ ( y) Ψ)) = C ( h( Φ Ψ)) + h(φ)+ C ( h(ψ)) = C (h(φ) h(ψ)) + h(φ)+ C ( h(ψ)) = C h(φ) C ( h(ψ)) + h(φ)+ C ( h(ψ)) h(φ) C ( h(ψ)) + h(φ)+ C ( h(ψ)) = ( h(φ) C ( h(ψ)) ) h(φ) C ( h(ψ)) = ( h(φ) h(φ) C ( h(ψ)) + C ( h(ψ)) h(φ) C ( h(ψ)) ) = (0 + 0) = 1, with condition V f (Φ) rang y =. (A 3 ) h (( y)φ S f (τ)φ), where τ :rang y V \ V b (Φ) = h (( y) Φ S f (τ)φ) = C ( h(φ)) + S τ h(φ), where τ = τ V = S τ C ( h(φ)) + S τ h(φ) by (P 6 ) since τ (V \ rang y) =id (V \ rang y) ( = S τ C ( h(φ)) + h(φ) ) S τ ( h(φ)+h(φ)) = S τ 1=1. (A 4 ) h (( y)φ ( τ y)) Φ = C h(φ)+ C h(φ) =1 by (P 13 ) supposing τ : y 1 1 y. onto (A 5 ) h ((Py 0)Φ) =C 0 h(φ) =1 from(wpp 2). (A 6 ) h ((Py r)φ (Py s)φ) = C r h(φ)+cs h(φ) C r h(φ)+cr h(φ) =1, for r s, by (WPP 4).

10 10 S. Marinković, M. Rašković, R. D ord ević (A 7 ) h ((Py r)φ (Py s)ψ (Py 0)(Φ Ψ) (Py min{1,r+ s})(φ Ψ)) ) = (C r h(φ) Cs h(ψ) C1 (h(φ) h(ψ)) (A 8 ) + C min{1,r+s} (h(φ)+h(ψ)) C min{1,r+s} (h(φ)+h(ψ)) + C min{1,r+s} (h(φ)+h(ψ)) = 1. h ((Py r)φ (Py < s)ψ (Py < max{0,r+ s 1})(Φ Ψ)) ( ) = C 1 r h(φ) Cs h(ψ) + C max{0,r+s} (h(φ) h(ψ)) C max{0,r+s} (h(φ) h(ψ)) + C max{0,r+s} (h(φ) h(ψ)) = 1. (A 9 ) (A 10 ) (A 11 ) (A 12 ) h ((Py < s)φ (Py s)φ) = C s h(φ)+c1 s h(φ) C s h(φ)+ Cs h(φ) =1, by (WPP 8). h ((Py s)φ (Py > r)φ) = C s h(φ)+cr h(φ) C s h(φ)+cs h(φ) =1, by (WPP 8), for s>r. h ( (Py > s)φ (Py s + )Φ ) = C 1 s h(φ)+cs+ h(φ) C s+ h(φ)+cs+ h(φ) =1, by (WPP 8). h (( y)φ (Py 1)Φ) =C h(φ)+c 1 h(φ) C h(φ)+ C 0+ h(φ) C h(φ)+ C h(φ) =1. by (WPP 7) (iii) from (WPP 10 ) (ii)

11 Weak Probability Polyadic Algebras 11 Now, let Φ be axiom from F \ F. Then Ψ = S(τ 1 )S f (τ0 )S(τ 0)Φ is an axiom from F. It follows h(ψ) =1. Then h(φ) =S (τ 1 V f (Ψ)) V h(ψ) =S (τ 1 V f (Ψ)) V 1=1. It is easy to verify that all formulas from Γ map to 1. Now, we shall show that from formula mapped to 1, by any rule of inference, we get the formula which maps to 1, i.e. that all theorems map to 1. It is only nontrivial for (R 3 )and(r 4 ). (R 3 )LetbeΦ L S(σ)Φ, where σ : V (Φ) 1 1 V, h(φ) =1. Then h (S(σ)Φ) =S (σ Vf (Φ)) V h(φ) =S (σ Vf (Φ)) V 1=1. (R 4 )LetbeS f (σ)φ L Φ, where σ : V f (Φ) 1 1 V \V b (Φ), h (S f (σ)φ) =1. If Φ F,wehaveΦ = S f (σ )S f (σ)φ. It follows h(φ) =S σ V h(s f (σ)φ) =S σ V 1=1. If Φ F \ F, let be Ψ = S(τ 1 )S f (τ 0 )S(τ 0)Φ and σ 1 : V (S f (σ)φ) V defined by σ 1 V b (S f (σ)φ) =τ 1 τ 0 and σ 1 V f (S f (σ)φ) =τ 1 σ. It is clear that σ 1 is 1 1 function. Then h(φ) =S (τ 1 V f (Ψ)) V h(ψ) =S (τ 1 V f (Ψ)) V h(s(σ 1)S f (σ)φ) 1 V f (Ψ)) V S (σ 1 V f (S f (σ)φ)) V h(s f (σ)ψ) 1 V f (Ψ)) V S (σ 1 V f (S f (σ)φ)) V 1=1. Finally, we shall show that g : B H Γ, g(p) =(p(x)) Γ, is 1 1. If g(p) =g(q), then Γ L p(x) q(x). It follows h(p(x) q(x)) = 1, and hence p = q. Theorem 3. If B is a weak probability polyadic algebra with infinite dimension β, then there exists homomorphism of B to some weak probability polyadic set algebra. Proof. By the previous theorem there exists the set of sentences Γ in logic L = L(β,ν,β +,R) such that H Γ = B. Since B>1, it follows that Γ is consistent in L. By the completeness theorem, Γ has a weak model A. By Lemma 4 A =(A,,,,A β,,c,c r,s τ ) is a weak probability polyadic

12 12 S. Marinković, M. Rašković, R. D ord ević set algebra and there exists homomorphism from H Γ onto A. It means that there exists homomorphism from B onto A. REFERENCES 1. L. Henkin, D. Monk and A. Tarski: Cylindric Algebras. North Holland, Amsterdam, H. J. Keisler: Acomplete first order logic with infinitary predicates. Fund. Math. 52 (1963), H. J. Keisler: Probability quantifiers. In: Model Theoretic Logics (J. Barwise and S. Feferman, eds.). Springer Verlag, Berlin, 1985, pp S. Marinković, M. Rašković and R. Djordjević: Weak probability logic with infinitary predicates. Publ. Inst. Math. Belgrade 65 (79) (1999), M. Rašković, R. Djordjević and M. Bradić: Weak cylindric probability algebras. Publ. Math. Inst. Belgrade 61 (75) (1997), M. Rašković and R. Djordjević: Cylindric probability algebras. (to appear) Department of Mathematics Faculty of Science University of Kragujevac P. O. Box 60, Kragujevac Serbia Yugoslavia