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1438 JOURNAL OF COMPUTERS, VOL. 8, NO. 6, JUNE 2013 (U20) ( ∀x)( φ → χ) →(( ∃x) φ → χ) (x is not free in χ ) (U21) ( ∀x)( φ ∨χ) →(( ∀x) φ∨ χ) (x is not free in χ ) Deduction rules of ∀ UL − h∈ (0, 1] are three rules. They are: Modus Ponens (MP): from φ, φ → ψ infer ψ ; Necessitation: from φ infer Δ φ ; Generalization: from φ infer ( ∀ x) φ . The meaning of “t substitutable for x in φ ( x) ” and “x is not free in χ ” in the above definition have the same meaning in the classical first-order predicate logic, moreover, we can define the concepts such as proof, theorem, theory, deduction from a theory T, T- consequence in the system ∀ UL − h∈ (0, 1] . T φ denotes that φ is provable in the theory T. φ denotes that φ is a theorem of system ∀ ULh ∈ (0 , 1] . Let − Thm( ∀ UL h∈ (0, 1] ) = { φ ∈ J | φ} , Ded( T) = { φ∈ J | T φ} . Being the axioms of propositional system UL − h∈ (0, 1] are in predicate system ∀ UL − h∈ (0, 1] , then the theorems in UL h∈ (0, 1] are theorems in ∀ UL − h∈ (0, 1] . According the similar proof in [3, 16, 17] we can get the following lemmas. Lemma 1 The hypothetical syllogism holds in ∀ , i.e. let Γ= { φ → ψψ , → χ} , then Γ φ → χ . UL − h∈ (0, 1] Lemma 2 ∀ UL − h∈ (0, 1] proves: (1) φ → φ ; (2) φ →( ψ → φ) ; (3) ( φ →ψ) →(( φ →γ) →( ψ → γ)) ; (4) ( φ & ( φ →ψ)) → ψ ; (5) Δφ ≡ Δφ& Δ φ . Lemma 3 If T = { φ → ψ, χ → γ} , then T ( φ & χ) → ( ψ & γ) . Let J is first-order predicate language, L is linearly ordered ŁΠG − algebra, M = ( M, ( rP ) P, ( mc) c) is called a L-evaluation for first-order predicate language J, which M is non-empty domain, according to each n-ary predicate P and object constant c, r P is L-fuzzy n-ary n relation: rP : M → L, m c is an element of M. Definition 7 Let J be predicate language, M is L- evaluation of J, x is object variable, P∈ J . (i) A mapping v: Term( J ) → M is called M- evaluation, if for each c ∈Const (J), vc () = mc ; (ii)Two M-evaluation vv′ , are called equal denoted by v ≡ x v′ if for each y∈ Var( J) \ x , there is vy ( ) = v′ ( y) . (iii) The value of a term given by M, v is defined by: x = v( x) ; c = mc . We define the truth value M, v M, v L M v φ , of a formula φ as following. Clearly, ∗,⇒,Δ denote the operations of L . L 1, 2,, n = M v P 1 ,, , M, v n M, v L L L → = ⇒ M, v M, v M, v L L L & = ∗ M, v M, v M, v L L Pt ( t t) r( t t ) φ ψ φ ψ φ ψ φ ψ 0 = 0; 1 = 1 M, v M, v L L φ M, v M, v L L φ M, v M, v L L φ φ M, v M, v′ L L φ φ M, v M, v′ Δ φ = Δ − φ =− ( ∀ x) = inf{ | v ≡ v ′ } ( ∃ x) = sup{ | v ≡ v ′ } In order to the above definitions are reasonable, the infimum/supremum should exist in the sense of L. So the structure M is L-safe if all the needed infima and suprema L exist, i.e. φ M , is defined for all φ, v . v Definition 8 Let φ ∈ J , M be a safe L-structure for J. (i) The truth value of φ in M is L L φ = inf{ φ ,v | v M − evaluation} . M M (ii) A formula φ of a language J is an L -tautology if φ L = 1 for each safe L-structure M. i.e. L M L φ 1 Mv , = for each safe L-structure M and each M-valuation of object variables. Remark For each h∈ (0, 1] , k∈ (0, 1) , ([0, 1] ,∧ hk , ,⇒ hk , , min, max, 0, 1 ,Δ,−) is a ŁΠG − -algebra. So the predicate system ∀ UL − h∈ (0, 1] can be considered the axiomatization for 1-level universal AND operator. III. SOUNDNESS OF SYSTEM ∀ h (0 1] x x UL − ∈ , Definition 9 A logic system is soundness if for its each theorem φ , we can get φ is a tautology. Theorem 5 (Soundness of axioms) The axioms of ∀ are L-tautologies for each linearly ordered UL − h∈ (0, 1] G − ŁΠ -algebra L. Proof. The axioms of (U1)- (U16) are L-tautologies can be get as in propositional calculus. We verify (U17)- (U21) To verify (U17), (U18), let y is substitutable for x to φ , when v′′ ≡ x v and v′′ ( x) = v( y) , there is L L L L φ( y) = φ ( x) So, ( ∀ x) φ( x) = inf ( ) M, v M, v′′ M v v′≡ v φ x , M, v′ ≤ L L L φ( y) ≤ sup ( ) ( ) ( ) M v v φ x = ∃ ′ x φ x , then , ′′ M, v′ M, v ( ∀ x ) φ( x ) → φ( y ) = ( ∀ x ) φ( x ) → φ( y ) = 1. M, v M, v M, v For (U19), let x not free in χ , then for each M- valuation w, when w≡ x v , we have v = φ( x) . M, w M, v We have to prove L L L L inf ( v ⇒ φ ) ≤( v ⇒ inf φ ) . w M, w M, w M, v w M, w L L © 2013 ACADEMY PUBLISHER
JOURNAL OF COMPUTERS, VOL. 8, NO. 6, JUNE 2013 1439 L L Let v = v = a , φ M, v M, w M , = b w w , thus we must prove inf w( a⇒bw) ≤( a⇒ inf wbw) . On the one hand, inf w b w ≤ b w , thus a⇒bw ≥( a⇒ inf wbw) for each w, thus inf w( a⇒bw) ≥( a⇒ inf wbw) . On the other hand if z ≤( a⇒ b w ) for each w, then z∗a ≤ bw for each w, z∗a ≤ inf w b w , z ≤( a⇒ inf wbw) . Thus ( a⇒ inf wbw) is the infimum of all ( a⇒ b w ). So (U19) holds. For (U20), we need to verify inf w( aw ⇒ bw) = (sup w aw ⇒ b) . Indeed, sup w ≥ a w , thus (sup w aw ⇒b) ≤( aw ⇒ b) , hence (sup w aw ⇒b) ≤inf w( aw ⇒ b) , If z ≤ aw ⇒ b for all w , then aw ≤( z ⇒ b) for all w, then sup w aw ≤( z ⇒ b) , z ≤(sup waw ⇒ b) , so sup w a w ⇒ b is the infimum. So (U20) holds. Finally we verify (U21), we need to verify inf w( a∨ bw) = a∨ infwbw . Indeed, a ≤ a∨ bw , thus a ≤inf w( a∨ bw) ; similarly, infwbw ≤inf w( a∨ bw) , thus a∨infwbw ≤inf w( a∨ b) . Conversely, let z ≤ a∨ bw for all w , we prove z ≤ a∨ inf w b w . Case 1: Let a ≤ inf w b w . Then z ≤ bw for each w , z ≤ inf w b w and z ≤ a∨ inf w b w . Case 2: Let a ≥ inf w b w . Then for some w 0 , a ≥ b w0 , thus z ≤ a and z ≤ a∨ inf w b w . So we prove the soundness of axioms. Theorem 6 (Soundness of deduction rules) (1) For arbitrary formulas φ, ψ , safe-structure M and evaluation L L L v , ψ ≥ φ ∗ φ → ψ . In particular, if M, v M, v M, v L L L = → = 1 M, v M, v L then ψ 1 M , = v L . L L L (2) Consequently, ψ φ φ ψ M M M φ φ ψ ≥ ∗ → , thus if φ, φ → ψ are then ψ is 1 L -true in M. (3) If φ is 1 L -true in M then Δ φ is 1 L -true in M . (4) If φ is 1 L -true in M then ( ∀ x) φ is 1 L -true in M. Proof. (1) is just as in propositional calculus. To prove (2) put φ = aw, ψ = bw, inf w w w aw = a. We have to prove inf w( aw ⇒bw) ≤infw aw ⇒ infwbw . Observe the following: inf( aw ⇒bw) ≤( aw ⇒bw) ≤( a⇒ bw) , thus inf w( aw ⇒bw) ≤inf w( a⇒ bw) . It remains to prove inf w( a⇒bw) ≤ a⇒ infwbw, this is holds from Theorem 5. L (3) If φ is 1 L -true in M then φ M = 1L , so L L Δ φ =Δ φ = . Then (3) holds. 1 M, v M, v L L M φ ′ ′ , } L M L M (4) Being φ = inf{ φ ,v | v M − evaluation} ≤ inf{ Mv| v ≡ v = ( ∀ x) φ } , So (4) holds. So we can get the following soundness theorem. L Theorem 7 (Soundness) Let L is linearly ordered ŁΠG − -algebra and φ is a formula in J, if φ , then φ is L L-tautology, i.e. φ = 1 . M L Similarly, we can get the following strong soundness theorem. Definition 10 Let T be a theory, L be a linearly ordered ŁΠG − -algebra and M a safe L-structure for the language of T. M is an L-model of T if all axioms of T are 1 L -true in M, i.e. φ = 1 L in each φ ∈ T . Theorem 8 (Strong Soundness) Let T be a theory, L is linearly ordered ŁΠG − -algebra and φ is a formula in J, L if T φ ( φ is provable in T), then φ M = 1L for each linearly ordered ŁΠG − -algebra L and each L-model M of T. Proof. In fact, from the proof of Theorem 5, for each L-model M of T, the axioms are true, and the formulas in T are true, from the proof of Theorem 6, the deduction rules preserve true. So the theorem holds. Theorem 9 (Deduction Theorem) Let T be a theory, φ, ψ are closed formulas. Then ( T ∪φ) ψ iff T Δφ →ψ . Proof. Sufficiency: Let T Δφ →ψ , from φ ( φ∈ ( T ∪ φ) ), then Δ φ by necessitation, so we can get ψ by MP rules. Necessity: Let m is the proof length from T ∪ φ} to ψ , we prove by induction for the length m. When m = 1 , ψ ∈T ∪φ∪Axm(C ∀ ) , if ψ = φ , The result holds. If ψ ∈ T or ψ is axiom, from Lemma2 (2), we have ψ →( Δφ → ψ) , then by ψψ , →( Δφ→ ψ) , we get Δφ → ψ , thus T Δφ →ψ . Assume that the result holds when m≤ k , i.e. we get γ at k step, then T Δφ → γ . Now Let m= k + 1. If ψ is obtained from MP rule by the above results γγ , → ψ in the proof sequence, then by induction hypothesis, we get T Δφ → γ, T Δφ →( γ →ψ) . From Lemma 3, we can get T ( Δφ& Δφ) →( γ &( γ →ψ)) . Being T ( Δφ) &( Δφ) ≡ Δφ , so T Δφ →( γ &( γ →ψ)) . From lemma 2 (4) we have ( γ & ( γ →ψ) → ψ , so we get T Δφ →ψ by the hypothetical syllogism. If ψ is obtained from necessitation rule by the above step γ in the proof sequence, i.e. Δ γ = ψ , then by induction hypothesis, we get T Δφ →γ . T Δ( Δφ →γ) , from (U16) we can get T ΔΔφ → Δγ , from (U15) we can get Δφ →ΔΔ φ , thus by the hypothetical syllogism we can get T Δφ →Δγ , i.e. T Δφ →ψ . If ψ is obtained from generalization rule by the above step γ in the proof sequence, i.e. ( ∀ x) γ = ψ , then by © 2013 ACADEMY PUBLISHER
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Journal of Computers ISSN 1796-203X
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Corn Moisture Measurement using a C
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Call for Papers and Special Issues
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(Contents Continued from Back Cover
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