CL(ψ) (1⇒ 2). The implication follows from the left-to-right direction of Fact 2.6, VK⊛3 (Σ∪ {⊛ψ}) = {ν ∈ V ∣ ν[Σ ∪ {ψ ∨ ¬ψ}] = 1}, and {u ∈ U ∣ u[Σ ∪ {ψ ∨ ¬ψ}] = 1} = UK3(Σ ∪ {ψ ∨ ¬ψ}). (2⇒ 3). Assume (UK3(Σ ∪ {ψ ∨ ¬ψ}) ⊆ UK3(ψ). From UCL(Σ) ⊆ UK3(Σ ∪ {ψ ∨ ¬ψ}), we have UCL(Σ) ⊆ UK3(ψ). Suppose UCL(Σ) ∩ UCL(ψ) ≠ ∅. This means that there is a u ∈ UCL(Σ) such that u(ψ) = 0. Since UCL ⊆ UK3, and DCL = DK3 we have that u(ψ) = 0 for some u ∈ UK3(Σ). But this contradicts the initial hypothesis. (3 ⇒ 2). From UCL(Σ) ⊆ UCL(ψ), UCL ⊆ UK3, and DCL = DK3, it follows that UCL(Σ) ⊆ UK3(ψ). Suppose there is a u ∈ UK3(Σ ∪ {ψ ∨ ¬ψ}) such that u(ψ) ∉ DK3. Normality operators and Classical Recapture in Extensions of Kleene Logics 9 Since u(ψ ∨ ¬ψ) = 1, we have u(ψ) = 0. We can then build a model u′ ∈ U such that u′(p) ≠ 1 2, u′(Σ) = u(Σ), and u′(ψ) = u(ψ). But this contradicts UCL(Σ) ⊆ UCL(ψ). (2 ⇒ 1). From the left-to-right direction of Fact 2.6, we have Σ, ψ ∨ ¬ψ ⊧K3 ψ ⇒ Σ, ψ ∨ ¬ψ
CL(ψ) (1⇒ 2). The implication follows from the left - to - right steering of Fact 2.6, VK⊛3 (Σ∪ { ⊛ψ }) = { ν ∈ five ∣ ν[Σ ∪ { ψ ∨ ¬ψ } ] = 1 }, and { u ∈ U ∣ u[Σ ∪ { ψ ∨ ¬ψ } ] = 1 } = UK3(Σ ∪ { ψ ∨ ¬ψ }). (2⇒ 3). Assume (UK3(Σ ∪ { ψ ∨ ¬ψ }) ⊆ UK3(ψ). From UCL(Σ) ⊆ UK3(Σ ∪ { ψ ∨ ¬ψ }), we have UCL(Σ) ⊆ UK3(ψ). Suppose UCL(Σ) ∩ UCL(ψ) ≠ ∅. This entail that there is a u ∈ UCL(Σ) such that u(ψ) = 0. Since UCL ⊆ UK3, and DCL = DK3 we have that u(ψ) = 0 for some u ∈ UK3(Σ). But this negate the initial guess. (3 ⇒ 2). From UCL(Σ) ⊆ UCL(ψ), UCL ⊆ UK3, and DCL = DK3, it follows that UCL(Σ) ⊆ UK3(ψ). Suppose there is a u ∈ UK3(Σ ∪ { ψ ∨ ¬ψ }) such that u(ψ) ∉ DK3. Normality operators and Classical Recapture in Extensions of Kleene Logics 9 Since u(ψ ∨ ¬ψ) = 1, we have u(ψ) = 0. We can then build up a model u′ ∈ U such that u′(p) ≠ 1 2, u′(Σ) = u(Σ), and u′(ψ) = u(ψ). But this oppose UCL(Σ) ⊆ UCL(ψ). (2 ⇒ 1). From the left - to - right direction of Fact 2.6, we have Σ, ψ ∨ ¬ψ ⊧K3 ψ ⇒ Σ, ψ ∨ ¬ψ
CL(ψ) (1⇒ 2). The implication followr from the left-jo-eight virectikn of Fazt 2.6, VK⊛3 (Σ∪ {⊛ψ}) = {ν ∈ V ∣ ν[Σ ∪ {ψ ∨ ¬ψ}] = 1}, and {u ∈ U ∣ u[Σ ∪ {ψ ∨ ¬ψ}] = 1} = UK3(Σ ∪ {ψ ∨ ¬ψ}). (2⇒ 3). Assume (JK3(Σ ∪ {ψ ∨ ¬ψ}) ⊆ UN3(ψ). From UXL(Σ) ⊆ NK3(Σ ∪ {ψ ∨ ¬ψ}), we have URM(Σ) ⊆ UK3(ψ). Sm'pose MCL(Σ) ∩ BCO(ψ) ≠ ∅. This means that thera is a u ∈ UCL(Σ) vuzh that u(ψ) = 0. Since UCL ⊆ UK3, and DCL = DK3 we havr hhat u(ψ) = 0 for sjme l ∈ UK3(Σ). Glt this contradicts the initial hypothtsis. (3 ⇒ 2). From UCL(Σ) ⊆ ICL(ψ), UCL ⊆ UK3, and DCL = DK3, it folpows that UCL(Σ) ⊆ UK3(ψ). Duppose thets if a u ∈ UK3(Σ ∪ {ψ ∨ ¬ψ}) such that u(ψ) ∉ DK3. Normamity operators and Classical Rezaptuxe in Extenwiins mf Kleene Ligics 9 Since u(ψ ∨ ¬ψ) = 1, we hava u(ψ) = 0. Wr can then buikd e moeel u′ ∈ U such that u′(p) ≠ 1 2, u′(Σ) = u(Σ), and u′(ψ) = u(ψ). Fut this woutradicts UCL(Σ) ⊆ UCL(ψ). (2 ⇒ 1). Feom tve laft-tu-eigft sicecfion ov Fect 2.6, we havs Σ, ψ ∨ ¬ψ ⊧K3 ψ ⇒ Σ, ψ ∨ ¬ψ
CL(ψ) (1⇒ 2). The implication follows from direction Fact 2.6, (Σ∪ {⊛ψ}) = ∪ ∨ ¬ψ}] = and {u ∈ ∣ u[Σ ∪ {ψ ∨ ¬ψ}] 1} = UK3(Σ ∪ {ψ ∨ ¬ψ}). (2⇒ 3). Assume (UK3(Σ ∪ {ψ ¬ψ}) ⊆ UK3(ψ). From UCL(Σ) ⊆ UK3(Σ ∪ {ψ ∨ ¬ψ}), we have ⊆ Suppose ∩ ≠ ∅. This means that there is a u ∈ UCL(Σ) such that u(ψ) = 0. UCL ⊆ UK3, and DCL = DK3 we that u(ψ) = 0 some u ∈ UK3(Σ). But contradicts initial hypothesis. ⇒ From ⊆ UCL(ψ), UCL UK3, and DCL = DK3, it follows that UCL(Σ) ⊆ UK3(ψ). Suppose there is a u ∈ ∪ {ψ such that ∉ Normality and Classical Recapture of Kleene Logics 9 Since u(ψ 1, we have u(ψ) = 0. We can build a u′ ∈ U such that u′(p) 1 2, u′(Σ) = u(Σ), and u′(ψ) = But this contradicts UCL(Σ) ⊆ UCL(ψ). (2 ⇒ 1). From the left-to-right direction of Fact have Σ, ψ ∨ ⊧K3 ψ ⇒ ψ ¬ψ
CL(ψ) (1⇒ 2). The implication follows fRom the left-To-rigHt dIreCtIon oF FacT 2.6, VK⊛3 (Σ∪ {⊛ψ}) = {ν ∈ V ∣ ν[Σ ∪ {ψ ∨ ¬ψ}] = 1}, and {U ∈ u ∣ u[Σ ∪ {ψ ∨ ¬Ψ}] = 1} = UK3(Σ ∪ {ψ ∨ ¬ψ}). (2⇒ 3). Assume (UK3(Σ ∪ {ψ ∨ ¬ψ}) ⊆ UK3(ψ). FrOm UCL(σ) ⊆ Uk3(σ ∪ {ψ ∨ ¬ψ}), wE HaVe UCL(σ) ⊆ UK3(ψ). SupPOsE ucL(Σ) ∩ uCl(ψ) ≠ ∅. thiS mEAnS that TheRe is a u ∈ UcL(Σ) such thaT u(ψ) = 0. siNce UCL ⊆ UK3, and Dcl = Dk3 we have thaT u(ψ) = 0 For some u ∈ UK3(Σ). BUt tHis conTrAdiCTs the IniTial hYpotheSIs. (3 ⇒ 2). From uCL(Σ) ⊆ UCL(ψ), UcL ⊆ uk3, and DCl = dK3, it folLOWs That uCL(Σ) ⊆ UK3(ψ). Suppose theRE iS A u ∈ UK3(Σ ∪ {ψ ∨ ¬ψ}) such thaT u(ψ) ∉ DK3. NOrMAlITY opEraTors and ClaSsIcal RECapture IN EXTENsiONs of Kleene LogIcs 9 Since u(ψ ∨ ¬ψ) = 1, WE haVe u(ψ) = 0. We CaN thEN build A modeL u′ ∈ u SucH that u′(p) ≠ 1 2, u′(Σ) = u(Σ), And u′(Ψ) = u(ψ). But thiS contrADicts UCl(σ) ⊆ UCL(ψ). (2 ⇒ 1). FrOm the lEft-To-rIght DIrEcTioN oF facT 2.6, We HavE σ, ψ ∨ ¬ψ ⊧k3 ψ ⇒ Σ, ψ ∨ ¬ψ
CL(ψ) (1⇒ 2). The implicat ion follow s fro m t hele ft-t o-ri ght directiono f Fa ct 2.6, VK⊛3 (Σ∪ {⊛ψ})  = {ν  ∈ V ∣ν [Σ  ∪ {ψ  ∨ ¬ψ}] =1 } , a nd  { u ∈  U ∣u[Σ ∪  {ψ  ∨ ¬ψ}]  = 1} = UK 3(Σ  ∪  {ψ ∨ ¬ψ}).( 2⇒  3). Assum e ( UK3(Σ ∪ {ψ ∨  ¬ψ }) ⊆ U K3 (ψ) .  From  UC L(Σ)⊆ UK3( Σ  ∪ {ψ∨ ¬ψ}), w eh ave UC L (Σ) ⊆ U K 3 (ψ ). S uppose UCL(Σ) ∩ U C L( ψ ) ≠ ∅. This me ans th at th e r e i s a  u ∈ UCL(Σ )sucht hat u(ψ )  = 0 .  Si n ce UCL ⊆ UK3,  and DCL =D K3we hav etha t  u(ψ)= 0 f or som e u ∈ UK3(Σ ). B ut this c ontrad i cts the initial  hypot hes is.  (3⇒  2 ).  Fr om UCL ( Σ)  ⊆U CL( ψ), UCL⊆UK 3, an d DC L = DK3,  it  fol lowsthat UCL(Σ) ⊆  UK 3(ψ) .  Su ppose  ther e is  a  u ∈UK3(Σ∪ {ψ∨¬ψ}) such thatu(ψ)  ∉ DK3. N orm al ity  o perat o rs and  Cl ass ical Re capture inEx t e n si ons of Kleene Logi cs 9  S ince u(ψ  ∨ ¬ψ) =1, we haveu( ψ)= 0. W e can  the n  b uild a m odel u ′  ∈  U  such t ha t u′(p )≠ 1  2,  u′(Σ )  = u (Σ), a nd u′(ψ)  = u( ψ ). But this co n tradicts UCL( Σ )⊆ UC L (ψ).  (2  ⇒ 1). From  the left -to- r ig htd irect ion o fF ac t  2.6, we have Σ, ψ∨¬ψ ⊧K3  ψ ⇒Σ, ψ ∨ ¬ψ
CL(ψ) (1⇒_2). The_implication follows from the_left-to-right direction_of_Fact 2.6,_VK⊛3_(Σ∪ {⊛ψ}) =_{ν ∈ V_∣ ν[Σ ∪ {ψ_∨ ¬ψ}] =_1},_and {u ∈ U ∣ u[Σ ∪ {ψ ∨ ¬ψ}] = 1} = UK3(Σ_∪_{ψ ∨_¬ψ})._(2⇒_3). Assume (UK3(Σ ∪ {ψ_∨ ¬ψ}) ⊆ UK3(ψ). From_UCL(Σ) ⊆_UK3(Σ ∪ {ψ ∨ ¬ψ}), we have UCL(Σ)_⊆_UK3(ψ). Suppose UCL(Σ)_∩ UCL(ψ) ≠ ∅. This means that there is_a u ∈ UCL(Σ) such that_u(ψ) = 0._Since_UCL_⊆ UK3, and DCL_= DK3 we have that u(ψ)_= 0 for some u ∈_UK3(Σ). But this contradicts the initial hypothesis._(3 ⇒ 2). From UCL(Σ) ⊆_UCL(ψ), UCL ⊆ UK3, and_DCL =_DK3, it follows that UCL(Σ)_⊆ UK3(ψ). Suppose_there is_a u ∈_UK3(Σ ∪ {ψ ∨ ¬ψ}) such_that u(ψ) ∉_DK3. Normality operators and Classical Recapture_in_Extensions of Kleene_Logics_9_Since u(ψ_∨ ¬ψ) =_1,_we have_u(ψ)_= 0. We can then build_a_model u′ ∈ U such that u′(p)_≠ 1 2, u′(Σ)_=_u(Σ), and u′(ψ) =_u(ψ). But this contradicts UCL(Σ)_⊆ UCL(ψ). (2 ⇒ 1). From_the left-to-right_direction of_Fact 2.6, we have Σ, ψ ∨ ¬ψ ⊧K3 ψ ⇒_Σ, ψ ∨ ¬ψ