theorem
  M,v |= (p => q) => ((p => r) => (p => q '&' r)) & M |= (p => q) => ((p
  => r) => (p => q '&' r))
proof
  now
    let v;
    now
      assume
A1:   M,v |= p => q;
      now
        assume M,v |= p => r;
        then M,v |= p implies M,v |= r & M,v |= q by A1,ZF_MODEL:18;
        then M,v |= p implies M,v |= q '&' r by ZF_MODEL:15;
        hence M,v |= p => q '&' r by ZF_MODEL:18;
      end;
      hence M,v |= (p => r) => (p => q '&' r) by ZF_MODEL:18;
    end;
    hence M,v |= (p => q) => ((p => r) => (p => q '&' r)) by ZF_MODEL:18;
  end;
  hence thesis;
end;
