theorem
  M,v |= (p => q) '&' (p => 'not' q) => 'not' p & M |= (p => q) '&' (p
  => 'not' q) => 'not' p
proof
  now
    let v;
    now
      assume M,v |= (p => q) '&' (p => 'not' q);
      then M,v |= p => q & M,v |= p => 'not' q by ZF_MODEL:15;
      then M,v |= p implies M,v |= q & M,v |= 'not' q by ZF_MODEL:18;
      hence M,v |= 'not' p by ZF_MODEL:14;
    end;
    hence M,v |= (p => q) '&' (p => 'not' q) => 'not' p by ZF_MODEL:18;
  end;
  hence thesis;
end;
