theorem 'not' (p '&&' q) => (p => 'not' q)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;
A3: v.(p => 'not' q) = v.p => v.('not' q) by LTLAXIO1:def 15
    .= v.p => (v.q => v.tf) by LTLAXIO1:def 15;
A4: v.q = 1 or v.q = 0 by XBOOLEAN:def 3;
    v.('not' (p '&&' q)) = v.(p '&&' q) => v.tf by LTLAXIO1:def 15
    .= v.p '&' v.q => v.tf by LTLAXIO1:31;
    hence v.('not' (p '&&' q) => (p => 'not' q))
    = (v.p '&' v.q => v.tf) => (v.p =>(v.q => v.tf)) by LTLAXIO1:def 15,A3
    .= 1 by A2,A4,A1;
  end;
