theorem (p '&&' q) => p is ctaut
  proof
    let g;
    set v = VAL g;
A1: v.p = 1 or v.p = 0 by XBOOLEAN:def 3;
    thus v.((p '&&' q) => p) = v.(p '&&' q) => v.p by LTLAXIO1:def 15
    .= v.p '&' v.q => v.p by LTLAXIO1:31
    .= 1 by A1;
  end;
