reserve A,B,p,q,r,s for Element of LTLB_WFF,
  i,j,k,n for Element of NAT,
  X for Subset of LTLB_WFF,
  f,f1 for FinSequence of LTLB_WFF,
  g for Function of LTLB_WFF,BOOLEAN;

theorem (for i be Nat st i in dom f holds {}LTLB_WFF |- (f/.i) => p) implies
   {}LTLB_WFF |- ('not' (con nega f)/.(len con nega f)) => p
   proof
     set nt = 'not' TVERUM;
     assume
A1:  for i be Nat st i in dom f holds {}l |- (f/.i) => p;
     per cases;
     suppose
A2:    len f = 0;
       nt => p is ctaut by Th23;
       then A3: nt => p in LTL_axioms by LTLAXIO1:def 17;
       len nega f = 0 by A2,Def4;
       then nega f = {};
       hence thesis by A3,Th10,LTLAXIO1:42;
     end;
     suppose
A4:    len f > 0;
       defpred P3[Nat] means $1 <= len f implies
       {}l |- ('not' ((con nega f)/.$1)) => p;
A5:    now
         let k being non zero Nat such that
A6:      P3[k];
         thus P3[k+1]
         proof
         set a = ('not' (con nega f)/.(k+1)), b = f/.(k+1),c = (con nega f)/.k,
         d = (nega f)/.(k+1),nc = 'not' c, nd = 'not' d;
         a => (nc 'or' nd) => (nd => b => (a => (nc 'or' b))) is ctaut by Th48;
         then a => (nc 'or' nd) => (nd => b => (a => (nc 'or' b)))
         in LTL_axioms by LTLAXIO1:def 17;then
A7:      {}l |- a => (nc 'or' nd) => (nd => b => (a => (nc 'or' b)))
         by LTLAXIO1:42;
         assume
A8:      k+1 <= len f;
         then k < len f by NAT_1:13;
         then 1 <= k & k < len nega f by NAT_1:25,Def4;
         then A9: a = 'not' (c '&&' d) by Th7;
         nc => p => (b => p => ((nc 'or' b) => p)) is ctaut by Th49;then
         nc => p => (b => p => ((nc 'or' b) => p)) in LTL_axioms
         by LTLAXIO1:def 17;then
         {}l |- nc => p => (b => p => ((nc 'or' b) => p)) by LTLAXIO1:42;
         then A10: {}l |- b => p => ((nc 'or' b) => p)
         by LTLAXIO1:43, A6, A8,NAT_1:13;
A11:     1 <= k+1 by NAT_1:25;
         then {}l |- b => p by FINSEQ_3:25,A8,A1;
         then A12: {}l |- (nc 'or' b) => p by A10,LTLAXIO1:43;
         k+1 in dom f by A11,FINSEQ_3:25,A8;
         then nd = 'not' 'not' b by Th8;
         then nd => b is ctaut by Th25;
         then nd => b in LTL_axioms by LTLAXIO1:def 17;
         then A13: {}l |- nd => b by LTLAXIO1:42;
         'not' (c '&&' d) => (nc 'or' nd) is ctaut by Th35;then
         'not' (c '&&' d) => (nc 'or' nd) in LTL_axioms by LTLAXIO1: def 17;
         then {}l |- a => (nc 'or' nd) by LTLAXIO1:42,A9;
         then {}l |- nd => b => (a => (nc 'or' b)) by A7,LTLAXIO1:43;
         then {}l |- a => (nc 'or' b) by LTLAXIO1:43,A13;
         hence {}l |- a => p by A12,LTLAXIO1:47;
       end;
     end;
A14: len nega f > 0 by A4,Def4;
A15: P3[1]
     proof
       set nnf = 'not' ('not' (f/.1));
       assume
A16:   1 <= len f;
       then A17: 1 in dom f by FINSEQ_3:25;
       nnf => f/.1 is ctaut by Th25;
       then nnf => f/.1 in LTL_axioms by LTLAXIO1:def 17;
       then A18: {}l |- nnf => f/.1 by LTLAXIO1:42;
A19:   {}l |- f/.1 => p by A16,FINSEQ_3:25,A1;
       'not' ((con nega f)/.1) = 'not' ((nega f)/.1) by A14,Th6
       .= nnf by Th8,A17;
       hence thesis by A18,A19,LTLAXIO1:47;
     end;
A20: for k being non zero Nat holds P3[k] from NAT_1:sch 10(A15,A5);
     len f = len nega f by Def4
     .= len con nega f by A14,Def2;
     hence {}l |- alt(f) => p by A20, A4;
   end;
 end;
