 p. 276) claimed that the derivation of ⊥ from the assumption ∃!a where a = {x|¬x ∈ x} enters a looping reduction and said, 'Russell's [paradox] remains an intrinsically troublesome case of paradox.' Later, from the result of Proposition 4.1.1, Tennant (2016, Sec. 3) claims that Russell's is not a genuine paradox because the derivation in question does not enter into loops. He diagnoses that the standard form of the elimination rule for set-abstraction creates an artefact feature of the looping reduction sequence. Unfortunately, as we have seen in Σ6 of Proposition 4.1.2, the generalized form of the elimination rule either creates a reduction loop. The real issue is not which form of elimination rules we choose, but which set of reduction procedures we accept. Moreover, even when we use standard elimination rules, it is readily proved that there is a normal derivation of ¬∃y(y = {x|¬x ∈ x}). His assessment of the genuineness of Russell's paradox was wrong in the first place. 142 Then, we have an open derivation Σ8 of Φ from [Φ ↔¬T (⌜Φ⌝)]. [Φ ↔¬T (⌜Φ⌝)]6 ********************************************** de f (Φ →¬T (⌜Φ⌝))∧ (¬T (⌜Φ⌝)→ Φ) [¬T (⌜Φ⌝→ Φ)]5 [T (⌜Φ⌝)]1 D13 ⊥ ¬I,1¬T (⌜Φ⌝) [Φ]7 → E,7 Φ ∧E,5 Φ With Σ7 and Σ8, we finally have a closed derivation 9 of ¬(Φ ↔¬T (⌜Φ⌝)). [Φ ↔¬T (⌜Φ⌝)]6 ********************************************** de f (Φ →¬T (⌜Φ⌝))∧ (¬T (⌜Φ⌝)→ Φ) [Φ →¬T (⌜Φ⌝)]7 [Φ ↔¬
p. 276) claimed that the derivation of ⊥ from the assumption ∃!a where a = { x|¬x ∈ x } enters a loop decrease and said,' Russell's [ paradox ] remains an intrinsically troublesome subject of paradox.' by and by, from the result of Proposition 4.1.1, Tennant (2016, Sec. 3) claims that Russell's is not a actual paradox because the derivation in question does not enter into closed circuit. He diagnoses that the standard form of the elimination principle for set - abstraction produce an artefact feature of speech of the looping reduction succession. Unfortunately, as we have seen in Σ6 of Proposition 4.1.2, the generalized mannequin of the elimination convention either creates a reduction loop. The veridical issue is not which form of elimination rules we choose, but which stage set of reduction procedures we accept. Moreover, even when we use standard elimination convention, it is readily proved that there is a normal derivation of ¬∃y(y = { x|¬x ∈ x }). His assessment of the authenticity of Russell's paradox was incorrect in the first place. 142 Then, we have an open derivation Σ8 of Φ from [ Φ ↔¬T (⌜ Φ ⌝) ]. [ Φ ↔¬T (⌜ Φ ⌝ )]6 * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * de f (Φ →¬T (⌜ Φ ⌝ ))∧ (¬T (⌜ Φ ⌝ )→ Φ) [ ¬T (⌜ Φ ⌝ → Φ)]5 [ T (⌜ Φ ⌝ )]1 D13 ⊥ ¬I,1¬T (⌜ Φ ⌝) [ Φ]7 → E,7 Φ ∧E,5 Φ With Σ7 and Σ8, we finally have a closed deriving 9 of ¬(Φ ↔¬T (⌜ Φ ⌝) ). [ Φ ↔¬T (⌜ Φ ⌝ )]6 * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * de f (Φ →¬T (⌜ Φ ⌝ ))∧ (¬T (⌜ Φ ⌝ )→ Φ) [ Φ →¬T (⌜ Φ ⌝ )]7 [ Φ ↔¬
 p. 276) claimed that the derivauion of ⊥ from the assumpvion ∃!a shere a = {x|¬x ∈ x} enters a looping redurtiob and said, 'Russell's [paradox] remains wn intribsicelly troublesome case of paradkw.' Latzr, from the resukt of Propmsition 4.1.1, Tennatt (2016, Dec. 3) claims that Russell's is not a denuine pwradox because the qeribation in question does not enter jnto lomps. He diagnoxes that the standard form of hhe elimination rupe for set-avstrwxtion creater an artefact feature kf the looping reduction sequenze. Unyortunately, aw wf have seen mn Σ6 of Proposition 4.1.2, the geteralizrd form of the elmminqtion rule either creetes a reduction look. The real iasue is not which firm ox elhminxriov rhlxs se cholse, but which set of redyction procedures wt asbrpt. Moreover, even rhqn we use standard elimination rules, it is readily proved that thwre is a normal derivwtion of ¬∃r(y = {x|¬x ∈ x}). His assessment of the genuineness of Ruvsell'a paxqdox dqs wrong in the first place. 142 Then, we have an opqh cegivation Σ8 of Φ froi [Φ ↔¬T (⌜Φ⌝)]. [Φ ↔¬T (⌜Φ⌝)]6 ********************************************** ce f (Φ →¬T (⌜Φ⌝))∧ (¬T (⌜Φ⌝)→ Φ) [¬T (⌜Φ⌝→ Φ)]5 [T (⌜Φ⌝)]1 D13 ⊥ ¬I,1¬C (⌜Φ⌝) [Φ]7 → E,7 Φ ∧E,5 Φ With Σ7 and Σ8, we finwlly yave a cljsed derivation 9 of ¬(Φ ↔¬T (⌜Φ⌝)). [Φ ↔¬T (⌜Φ⌝)]6 ********************************************** de f (Φ →¬T (⌜Φ⌝))∧ (¬T (⌜Φ⌝)→ Φ) [Φ →¬N (⌜Φ⌝)]7 [Φ ↔¬
p. 276) claimed that the derivation of the ∃!a where = {x|¬x ∈ and 'Russell's [paradox] remains intrinsically troublesome case paradox.' Later, from the result of 4.1.1, Tennant (2016, Sec. 3) claims that Russell's is not a genuine paradox the derivation in question does not enter into loops. He diagnoses that the form the rule set-abstraction creates an artefact feature of the looping reduction sequence. Unfortunately, as we have seen in of Proposition 4.1.2, the generalized form of the rule either creates a loop. The real issue is which of elimination we but set of reduction we accept. Moreover, even when we use standard elimination rules, it is readily proved that there is normal derivation = {x|¬x x}). assessment the genuineness of was wrong in the first place. have an open derivation Σ8 of Φ from ↔¬T (⌜Φ⌝)]. ↔¬T (⌜Φ⌝)]6 ********************************************** de f (Φ (⌜Φ⌝))∧ (¬T (⌜Φ⌝)→ Φ) [¬T (⌜Φ⌝→ Φ)]5 [T D13 ⊥ ¬I,1¬T (⌜Φ⌝) [Φ]7 → E,7 Φ ∧E,5 Φ With Σ7 and Σ8, we a closed derivation 9 ¬(Φ ↔¬T (⌜Φ⌝)). ↔¬T ********************************************** f →¬T (⌜Φ⌝))∧ (⌜Φ⌝)→ Φ) [Φ →¬T (⌜Φ⌝)]7 [Φ ↔¬
 p. 276) claimed that the derivation Of ⊥ from the aSsumpTioN ∃!a wHeRe a = {x|¬X ∈ x} enTers a looping reDUctiOn and said, 'Russell's [paradOx] remAiNS an iNTrInsicAlly troUBlESOme CaSe Of pArADoX.' LateR, frOm the reSult of PropOsiTiOn 4.1.1, Tennant (2016, Sec. 3) CLaIms that RusSelL's is not a genuIne ParadoX bEcaUSe the DerIvatiOn in quEStion dOes not entEr INto looPS. He diagNOSeS thaT the standard form oF ThE Elimination rulE for seT-aBStRACtiOn cReates an arTeFact fEAture of THe LOOPinG Reduction sequEnce. UnfortuNAteLy, as we HaVe sEEn in Σ6 oF PropOsITioN 4.1.2, the generalIzed Form of the EliminATion rulE Either cReates A reDucTion LOoP. THe rEaL IssUE iS noT WhiCh form of ElImInatiOn ruLES WE choOse, But wHich sEt of reduction ProCeduREs wE accePt. MorEoveR, eVen whEn we usE stanDaRd elimination ruLes, iT is readilY prOvEd tHaT therE Is a norMal DerIvation Of ¬∃y(y = {x|¬x ∈ x}). hIs aSsESSMeNt of the genuineness Of rUSsEll's paraDox was WRoNg IN the firsT pLacE. 142 TheN, WE have An opEN dErivatioN Σ8 of Φ fROm [φ ↔¬T (⌜φ⌝)]. [Φ ↔¬T (⌜Φ⌝)]6 ********************************************** de f (φ →¬T (⌜φ⌝))∧ (¬T (⌜Φ⌝)→ Φ) [¬T (⌜Φ⌝→ φ)]5 [T (⌜φ⌝)]1 D13 ⊥ ¬I,1¬t (⌜Φ⌝) [Φ]7 → e,7 Φ ∧E,5 Φ WITh Σ7 aNd Σ8, we fInally haVe a clOSed derivation 9 oF ¬(φ ↔¬T (⌜Φ⌝)). [Φ ↔¬T (⌜Φ⌝)]6 ********************************************** de f (Φ →¬T (⌜Φ⌝))∧ (¬T (⌜φ⌝)→ φ) [Φ →¬t (⌜φ⌝)]7 [φ ↔¬
 p. 276) claimed that thederivation  of ⊥  fr omth e as sump tion ∃!a where a ={x|¬x ∈ x} enters a lo oping  r e duct i on  andsaid, ' R us s e ll' s[p ara do x ]remai nsan intr insicallytro ub lesome caseo fparadox.'Lat er, from the  re sult o fPro p ositi on4.1.1 , Tenn a nt (20 16, Sec.3) claims that Ru s s el l'sis not a genuinep ar a dox because th e deri va t io n inque stion does  n ot en t er into lo o p s . H e  diagnoses th at the stan d ard  formof  th e  elimi natio nr ule  for set-ab stra ction cre ates a n  artefa c t featu re ofthe  lo opin g  r ed uct io n  se q ue nce .  Un fortunat el y,  as w e ha v e s eeninΣ6 o f Pro position 4.1. 2,theg ene raliz ed fo rm o fthe e limina tionru le either creat es a  reductio n l oo p.Th e rea l  issue  is  no t which  form o f  el im i n a ti on rules we choose ,b u twhich se t of r e du ct i on proce du res  wea c cept.  Mor e ov er, even  whenw eus e stand ar d elim in ati onrules ,  itis rea dily pro ved t h at there is an ormal derivat i on o f¬ ∃y(y  ={x|¬x ∈ x}) . Hi s  ass essm e nt  of the g enuin en e ss of Russell's parado xwas wr ong i n the first p lace. 142T h e n, we ha ve a n  o p en derivationΣ8 of  Φ from [Φ ↔¬T (⌜Φ⌝ )]. [ Φ ↔¬T (⌜ Φ⌝)]6 *** * * ******** *** *** *** *** * * ** ************* * * ** d ef (Φ →¬ T ( ⌜Φ⌝))∧(¬T  (⌜ Φ⌝) → Φ )[¬T (⌜Φ⌝→  Φ)]5 [T  ( ⌜Φ ⌝) ]1  D1 3 ⊥ ¬ I ,1¬T (⌜Φ ⌝)  [Φ ]7  →E,7 Φ ∧E,5 Φ  With  Σ7an dΣ 8,we fina l ly h aveacl osed  de ri vatio n 9 of¬(Φ ↔¬T  (⌜Φ⌝)).[Φ↔ ¬T ( ⌜Φ ⌝) ]6 **** ************* ** ********** ** *** ****** * * **** def (Φ →¬T (⌜Φ⌝))∧ (¬T (⌜ Φ ⌝)→ Φ)[Φ→¬T ( ⌜Φ⌝) ]7 [Φ ↔¬
 p._276) claimed_that the derivation of_⊥ from_the_assumption ∃!a_where_a = {x|¬x_∈ x} enters_a looping reduction and_said, 'Russell's [paradox]_remains_an intrinsically troublesome case of paradox.' Later, from the result of Proposition 4.1.1, Tennant_(2016,_Sec. 3)_claims_that_Russell's is not a genuine_paradox because the derivation in_question does_not enter into loops. He diagnoses that the_standard_form of the_elimination rule for set-abstraction creates an artefact feature of_the looping reduction sequence. Unfortunately, as_we have seen_in_Σ6_of Proposition 4.1.2, the_generalized form of the elimination rule_either creates a reduction loop. The_real issue is not which form of_elimination rules we choose, but which_set of reduction procedures we_accept. Moreover,_even when we use standard_elimination rules, it_is readily_proved that there_is a normal derivation of ¬∃y(y_= {x|¬x ∈_x}). His assessment of the genuineness_of_Russell's paradox was_wrong_in_the first_place. 142 Then,_we_have an_open_derivation Σ8 of Φ from [Φ_↔¬T_(⌜Φ⌝)]. [Φ ↔¬T (⌜Φ⌝)]6 ********************************************** de f_(Φ →¬T (⌜Φ⌝))∧ (¬T_(⌜Φ⌝)→_Φ) [¬T (⌜Φ⌝→ Φ)]5_[T (⌜Φ⌝)]1 D13 ⊥ ¬I,1¬T_(⌜Φ⌝) [Φ]7 → E,7 Φ ∧E,5_Φ With_Σ7 and_Σ8, we finally have a closed derivation 9 of ¬(Φ ↔¬T_(⌜Φ⌝)). [Φ ↔¬T (⌜Φ⌝)]6 ********************************************** de_f (Φ →¬T (⌜Φ⌝))∧_(¬T (⌜Φ⌝)→_Φ)_[Φ →¬T (⌜Φ⌝)]7_[Φ_↔¬