theorem Th26: p => ('not' 'not' p) is ctaut
  proof
    let g;
    set v = VAL g;
A1: v.tf = 0 by LTLAXIO1:def 15;
A2: v.p = 1 or v.p = 0 by XBOOLEAN:def 3;
    thus v.(p => ('not' 'not' p)) = v.p => v.('not' 'not' p) by LTLAXIO1:def 15
    .= v.p => (v.('not' p) => v.tf) by LTLAXIO1:def 15
    .= v.p => (v.p => v.tf => v.tf) by LTLAXIO1:def 15
    .= 1 by A2,A1;
  end;
