 1} ⊆ ULP(ψ), we have UCL(Σ) ⊆ ULP(ψ). From this, it follows that UCL(Σ) ⊆ UCL(ψ)-otherwise, we would have {u ∈ US(Σ) ∣ u[Σ] = 1} ∩ {u ∈ US ∣ u(ψ) = 0}, which contradicts {u ∈ US(Σ) ∣ u[Σ] = 1} ⊆ ULP(ψ). (3 ⇒ 2). Assume UCL(Σ) ⊆ UCL(ψ), and suppose that there is a u ∈ U such that u[Σ] = 1 and u(ψ) = 0. We can build a corresponding classical model u′ ∈ UCL such that u′[Σ] = u[Σ], u′(ψ) = u(ψ), and u(p) ≠ 1 2 for every p ∈ var. We have u′[Σ] = 1 and u′(ψ) = 0, which contradicts the initial assumption. (2⇒ 1). The implication follows from the fact that, for every φ ∈ L2 and ν ∈ V, we can build a valuation u ∈ U such that u(p) = ν(p). Since ν ∈ V(Σ∪Σ⊛) = {ν ∈ V ∣ ν[Σ] = 1}, this implies that, if {u ∈ US(Σ) ∣ u[Σ] = 1} ⊆ US(ψ), then VS(Σ ∪Σ⊛) ⊆ VS(ψ). This proves 1⇔ 3, which suffices to prove Σ,Σ⊛ ⊧LP⊛ ψ ⇔ Σ ⊧CL ψ. We go to K⊛3. We prove that, if Σ ∪ {ψ} ⊆ L2, the following are equivalent: 1. VK⊛3 (Σ ∪ {⊛ψ}) ⊆ VK⊛3 (ψ) 2. (UK3(Σ ∪ {ψ ∨ ¬ψ}) ⊆ UK3(ψ) 3. UCL(Σ) ⊆ U
1 } ⊆ ULP(ψ), we have UCL(Σ) ⊆ ULP(ψ). From this, it follows that UCL(Σ) ⊆ UCL(ψ)-otherwise, we would have { u ∈ US(Σ) ∣ u[Σ ] = 1 } ∩ { u ∈ US ∣ u(ψ) = 0 }, which contradicts { u ∈ US(Σ) ∣ u[Σ ] = 1 } ⊆ ULP(ψ). (3 ⇒ 2). Assume UCL(Σ) ⊆ UCL(ψ), and presuppose that there be a u ∈ U such that u[Σ ] = 1 and u(ψ) = 0. We can build a corresponding authoritative model u′ ∈ UCL such that u′[Σ ] = u[Σ ], u′(ψ) = u(ψ), and u(p) ≠ 1 2 for every p ∈ var. We hold u′[Σ ] = 1 and u′(ψ) = 0, which contradicts the initial assumption. (2⇒ 1). The deduction follows from the fact that, for every φ ∈ L2 and ν ∈ V, we can build a evaluation u ∈ U such that u(p) = ν(p). Since ν ∈ V(Σ∪Σ⊛) = { ν ∈ V ∣ ν[Σ ] = 1 }, this implies that, if { u ∈ US(Σ) ∣ u[Σ ] = 1 } ⊆ US(ψ), then VS(Σ ∪Σ⊛) ⊆ VS(ψ). This raise 1⇔ 3, which suffices to prove Σ, Σ⊛ ⊧LP⊛ ψ ⇔ Σ ⊧CL ψ. We go to K⊛3. We prove that, if Σ ∪ { ψ } ⊆ L2, the following are equivalent: 1. VK⊛3 (Σ ∪ { ⊛ψ }) ⊆ VK⊛3 (ψ) 2. (UK3(Σ ∪ { ψ ∨ ¬ψ }) ⊆ UK3(ψ) 3. UCL(Σ) ⊆ uracil
 1} ⊆ KLP(ψ), we have UCL(Σ) ⊆ ULP(ψ). Froo this, it follocw that UCL(Σ) ⊆ HCL(ψ)-otherdise, we would have {u ∈ US(Σ) ∣ u[Σ] = 1} ∩ {u ∈ UW ∣ u(ψ) = 0}, which contradicgs {u ∈ US(Σ) ∣ u[Σ] = 1} ⊆ ULP(ψ). (3 ⇒ 2). Qssume UCL(Σ) ⊆ UCL(ψ), and suppkde tkav there is a u ∈ U such thdt u[Σ] = 1 and u(ψ) = 0. Wd ean build a corresponding classical iodel u′ ∈ KCL such that o′[Σ] = u[Σ], l′(ψ) = u(ψ), ahd u(p) ≠ 1 2 for every p ∈ var. We have h′[Σ] = 1 and u′(ψ) = 0, which comtradicts the initial assulptiln. (2⇒ 1). The implicatiln follows drom rhe fact thag, for everj φ ∈ L2 and ν ∈ V, we can build a valuation u ∈ J suck that u(p) = ν(k). Aijwe ν ∈ V(Σ∪Σ⊛) = {ν ∈ V ∣ ν[Σ] = 1}, nhis implies bnat, if {u ∈ US(Σ) ∣ u[Σ] = 1} ⊆ US(ψ), then YS(Σ ∪Σ⊛) ⊆ TS(ψ). Tyis proves 1⇔ 3, which suhfices to prove Σ,Σ⊛ ⊧LP⊛ ψ ⇔ Σ ⊧CL ψ. We gk to K⊛3. We prove tyar, if Σ ∪ {ψ} ⊆ N2, thd foulosiig zre eqkivelent: 1. VK⊛3 (Σ ∪ {⊛ψ}) ⊆ VK⊛3 (ψ) 2. (UK3(Σ ∪ {ψ ∨ ¬ψ}) ⊆ UK3(ψ) 3. UCL(Σ) ⊆ U
1} ⊆ ULP(ψ), we have UCL(Σ) ⊆ this, follows that ⊆ UCL(ψ)-otherwise, we ∣ = 1} ∩ ∈ US ∣ = 0}, which contradicts {u ∈ ∣ u[Σ] = 1} ⊆ ULP(ψ). (3 ⇒ 2). Assume UCL(Σ) ⊆ UCL(ψ), suppose that there is a u ∈ U such that u[Σ] = 1 u(ψ) 0. can a corresponding classical model u′ ∈ UCL such that u′[Σ] = u[Σ], u′(ψ) = u(ψ), and ≠ 1 2 for every p ∈ var. have u′[Σ] = 1 u′(ψ) = 0, which contradicts initial (2⇒ 1). implication from fact that, for φ ∈ L2 and ν ∈ V, we can build a valuation u ∈ U such that = ν(p). ∈ V(Σ∪Σ⊛) {ν V ν[Σ] = 1}, that, if {u ∈ US(Σ) ∣ ⊆ US(ψ), then VS(Σ ∪Σ⊛) ⊆ VS(ψ). This 1⇔ 3, suffices to prove Σ,Σ⊛ ⊧LP⊛ ψ Σ ⊧CL ψ. We go to K⊛3. We that, if Σ ∪ {ψ} ⊆ L2, the following are equivalent: 1. VK⊛3 (Σ ∪ VK⊛3 (ψ) 2. (UK3(Σ {ψ ∨ ¬ψ}) UK3(ψ) UCL(Σ) U
 1} ⊆ ULP(ψ), we have UCL(Σ) ⊆ ULP(ψ). From thiS, it follows That UcL(Σ) ⊆ uCL(Ψ)-oTherWise, We would have {u ∈ US(σ) ∣ U[Σ] = 1} ∩ {u ∈ Us ∣ u(ψ) = 0}, which contradicts {u ∈ US(σ) ∣ u[Σ] = 1} ⊆ ULp(ψ). (3 ⇒ 2). aSsumE uCl(Σ) ⊆ UCL(Ψ), and supPOsE THat ThErE is A u ∈ u SuCh thaT u[Σ] = 1 And u(ψ) = 0. We Can build a cOrrEsPonding classICaL model u′ ∈ UCL SucH that u′[Σ] = u[Σ], u′(ψ) = u(Ψ), anD u(p) ≠ 1 2 for EvEry P ∈ Var. We HavE u′[Σ] = 1 anD u′(ψ) = 0, whiCH contrAdicts the InITial asSUmption. (2⇒ 1). tHE iMpliCation follows from THe FAct that, for everY φ ∈ L2 and Ν ∈ V, WE cAN BuiLd a Valuation u ∈ u sUch thAT u(p) = ν(p). SiNCe Ν ∈ v(σ∪σ⊛) = {ν ∈ V ∣ Ν[σ] = 1}, this implies tHat, if {u ∈ US(Σ) ∣ u[Σ] = 1} ⊆ us(ψ), tHen VS(Σ ∪σ⊛) ⊆ Vs(ψ). THIs provEs 1⇔ 3, whiCh SUffIces to prove σ,Σ⊛ ⊧LP⊛ Ψ ⇔ Σ ⊧CL ψ. We go To K⊛3. We pROve that, IF Σ ∪ {ψ} ⊆ L2, the FollowIng Are EquiVAlEnT: 1. VK⊛3 (σ ∪ {⊛ψ}) ⊆ vk⊛3 (ψ) 2. (Uk3(σ ∪ {ψ ∨ ¬Ψ}) ⊆ UK3(Ψ) 3. uCL(σ) ⊆ U
 1} ⊆ ULP(ψ), we have UCL( Σ) ⊆ ULP(ψ ). Fr omthi s,  itfoll ows that UCL(Σ )  ⊆ U CL(ψ)-otherwise, we wo uld h av e  {u∈  U S(Σ)∣ u[Σ]=  1 } ∩ { u∈US∣u (ψ ) = 0 },which c ontradicts  {u  ∈  US(Σ) ∣ u[Σ ]  =  1} ⊆ ULP( ψ).  (3 ⇒ 2). As sum e UCL( Σ)  ⊆U CL(ψ) , a nd su pposet hat th ere is au∈  U suc h  that u [ Σ ]= 1and u(ψ) = 0. Wec an build a corres pondin gc la s s ica l m odel u′ ∈UC L suc h  that u ′ [Σ ] =  u[ Σ ], u′(ψ) = u( ψ), and u(p )  ≠1 2 fo reve r y p ∈var.We hav e u′[Σ] = 1  and  u′(ψ) =0, whi c h contr a dicts t he ini tia l a ssum p ti on . ( 2⇒ 1). Th e i m pli cation f ol lo ws fr om t h e f acttha t, f or ev ery φ ∈ L2 an d ν  ∈ V ,  we  canbuild  a v al uatio n u ∈U suc hthat u(p) = ν(p ). S ince ν ∈V(Σ ∪Σ ⊛)={ν ∈V  ∣ ν[Σ ] =  1} , thisimplies tha t, i f  { u ∈ US(Σ) ∣ u[Σ] =  1 } ⊆US(ψ), t hen VS ( Σ∪Σ ⊛ ) ⊆ VS(ψ ).  Th is p r o ves 1 ⇔ 3, wh ich suff ices t o  p ro ve Σ,Σ⊛  ⊧ LP⊛ ψ⇔Σ ⊧ CLψ. We go t o K⊛3.  We prov e tha t , if Σ ∪ {ψ} ⊆ L2, the follo w in g ar e  equ iva lent: 1. VK ⊛3 ( Σ  ∪ { ⊛ψ}) ⊆VK⊛ 3  (ψ)2. (U K3 ( Σ∪  {ψ ∨ ¬ψ}) ⊆ UK3(ψ)  3 . UCL( Σ) ⊆U
 1}_⊆ ULP(ψ),_we have UCL(Σ) ⊆_ULP(ψ). From_this,_it follows_that_UCL(Σ) ⊆ UCL(ψ)-otherwise,_we would have_{u ∈ US(Σ) ∣_u[Σ] = 1}_∩_{u ∈ US ∣ u(ψ) = 0}, which contradicts {u ∈ US(Σ) ∣ u[Σ]_=_1} ⊆_ULP(ψ)._(3_⇒ 2). Assume UCL(Σ) ⊆_UCL(ψ), and suppose that there_is a_u ∈ U such that u[Σ] = 1_and_u(ψ) = 0._We can build a corresponding classical model u′ ∈_UCL such that u′[Σ] = u[Σ],_u′(ψ) = u(ψ),_and_u(p)_≠ 1 2 for_every p ∈ var. We have_u′[Σ] = 1 and u′(ψ) =_0, which contradicts the initial assumption. (2⇒_1). The implication follows from the_fact that, for every φ_∈ L2_and ν ∈ V, we_can build a_valuation u_∈ U such_that u(p) = ν(p). Since ν_∈ V(Σ∪Σ⊛) =_{ν ∈ V ∣ ν[Σ] =_1},_this implies that,_if_{u_∈ US(Σ)_∣ u[Σ] =_1}_⊆ US(ψ),_then_VS(Σ ∪Σ⊛) ⊆ VS(ψ). This proves_1⇔_3, which suffices to prove Σ,Σ⊛ ⊧LP⊛_ψ ⇔ Σ ⊧CL_ψ._We go to K⊛3._We prove that, if Σ_∪ {ψ} ⊆ L2, the following_are equivalent:_1. VK⊛3_(Σ ∪ {⊛ψ}) ⊆ VK⊛3 (ψ) 2. (UK3(Σ ∪ {ψ ∨_¬ψ}) ⊆ UK3(ψ) 3. UCL(Σ) ⊆_U