theorem
  M,v |= (p => q) '&' (q => r) => (p => r) & M |= (p => q) '&' (q => r)
  => (p => r)
proof
  now
    let v;
    now
      assume M,v |= (p => q) '&' (q => r);
      then M,v |= (p => q) & M,v |= (q => r) by ZF_MODEL:15;
      hence M,v |= p => r by Th102;
    end;
    hence M,v |= (p => q) '&' (q => r) => (p => r) by ZF_MODEL:18;
  end;
  hence thesis;
end;
