reserve A,B,p,q,r,s for Element of LTLB_WFF,
  n for Element of NAT,
  X for Subset of LTLB_WFF,
  g for Function of LTLB_WFF,BOOLEAN,
  x,y for set;
reserve P,Q,P1,R for PNPair;

theorem
  for P be consistent PNPair st A in rng P & B in rng P & A => B in rng P
  holds (A => B in rng P`1 iff (A in rng P`2 or B in rng P`1))
  proof
    let P be consistent PNPair;
    assume that
A1: A in rng P and
A2: B in rng P and
A3: A => B in rng P;
    set p = P^,pa = p => A, pna = p => 'not' A,pb = p => B,pnb = p => 'not' B,
    pab = p => (A => B),pnab = p => 'not' (A => B), np = 'not' p;
    hereby
      pa => (pnb => (pab => np)) is ctaut
      proof
        let g;
        set v = VAL g,vf = v.TFALSUM;
A4:     v.A = 1 or v.A = 0 by XBOOLEAN:def 3;
A5:     v.B = 1 or v.B = 0 by XBOOLEAN:def 3;
A6:     v.p = 1 or v.p = 0 by XBOOLEAN:def 3;
A7:     vf = 0 & v.pa = v.p => v.A by LTLAXIO1:def 15;
A8:     v.pab = v.p => v.(A => B) by LTLAXIO1:def 15
        .= v.p => (v.A => v.B) by LTLAXIO1:def 15;
A9:     v.pnb = v.p => v.('not' B) by LTLAXIO1:def 15
        .= v.p => (v.B => vf) by LTLAXIO1:def 15;
        thus v.(pa => (pnb => (pab => np))) = v.pa => v.(pnb => (pab => np))
        by LTLAXIO1: def 15
        .= v.pa => (v.pnb => v.(pab => np)) by LTLAXIO1:def 15
        .= v.pa => (v.pnb => (v.pab => v.np)) by LTLAXIO1:def 15
        .= 1 by A4,A5,A6,A7,A8, LTLAXIO1:def 15,A9;
      end;
      then pa => (pnb => (pab => np)) in LTL_axioms by LTLAXIO1:def 17;
      then A10: {} l |- pa => (pnb => (pab => np)) by LTLAXIO1:42;
      assume A => B in rng P`1;
      then A11: {} l |- pab by Th28;
      assume that
A12:  not A in rng P`2 and
A13:  not B in rng P`1;
      B in rng P`2 by A13,A2,XBOOLE_0:def 3;
      then A14: {} l |- pnb by Th29;
      A in rng P`1 by A12,A1,XBOOLE_0:def 3;
      then {} l |- pa by Th28;
      then {} l |- pnb => (pab => np) by A10,LTLAXIO1:43;
      then {} l |- pab => np by LTLAXIO1:43,A14;
      hence contradiction by LTLAXIO1:43,A11,Def10;
    end;
    assume
A15: A in rng P`2 or B in rng P`1;
     assume not A => B in rng P`1;
     then A => B in rng P`2 by XBOOLE_0:def 3,A3;
     then A16: {} l |- pnab by Th29;
     per cases by A15;
     suppose
A17:   A in rng P`2;
       pna => (pnab => np) is ctaut
       proof
         let g;
         set v = VAL g,vf = v.TFALSUM;
A18:     vf = 0 by LTLAXIO1:def 15;
A19:     v.A = 1 or v.A = 0 by XBOOLEAN:def 3;
A20:     v.p = 1 or v.p = 0 by XBOOLEAN:def 3;
A21:     v.pna = v.p => v.('not' A) by LTLAXIO1:def 15
         .= v.p => (v.A => vf) by LTLAXIO1:def 15;
A22:     v.pnab = v.p => v.('not' (A => B)) by LTLAXIO1:def 15
         .= v.p => (v.(A => B) => vf) by LTLAXIO1:def 15
         .= v.p => (v.A => v.B => vf) by LTLAXIO1:def 15;
         thus v.(pna => (pnab => np)) = v.pna => v.(pnab => np)
         by LTLAXIO1:def 15
         .= v.pna => (v.pnab => v.np) by LTLAXIO1:def 15
         .= 1 by A19,A20,A18,LTLAXIO1:def 15,A21,A22;
       end;
       then pna => (pnab => np) in LTL_axioms by LTLAXIO1:def 17;
       then A23: {} l |- pna => (pnab => np) by LTLAXIO1:42;
       {} l |- pna by A17,Th29;
       then {} l |- pnab => np by A23,LTLAXIO1:43;
       hence contradiction by LTLAXIO1:43,A16,Def10;
     end;
     suppose
A24:   B in rng P`1;
       pb => (pnab => np) is ctaut
       proof
         let g;
         set v = VAL g,vf = v.TFALSUM;
A25:     v.B = 1 or v.B = 0 by XBOOLEAN:def 3;
A26:     v.p = 1 or v.p = 0 by XBOOLEAN:def 3;
A27:     vf = 0 & v.pb = v.p => v.B by LTLAXIO1:def 15;
A28:     v.pnab = v.p => v.('not' (A => B)) by LTLAXIO1:def 15
         .= v.p => (v.(A => B) => vf) by LTLAXIO1:def 15
         .= v.p => (v.A => v.B => vf) by LTLAXIO1:def 15;
         thus v.(pb => (pnab => np)) = v.pb => v.(pnab => np)
         by LTLAXIO1:def 15
         .= v.pb => (v.pnab => v.np) by LTLAXIO1:def 15
         .= 1 by A25,A26,A27,LTLAXIO1:def 15,A28;
       end;
       then pb => (pnab => np) in LTL_axioms by LTLAXIO1:def 17;
       then A29: {} l |- pb => (pnab => np) by LTLAXIO1:42;
       {} l |- pb by A24,Th28;
       then {} l |- pnab => np by A29,LTLAXIO1:43;
       hence contradiction by LTLAXIO1:43,A16,Def10;
     end;
   end;
