theorem Th25:
  (root-tree [d1,d2])`1 = root-tree d1 & (root-tree [d1,d2])`2 = root-tree d2
proof reconsider r = {} as Node of root-tree [d1,d2] by TREES_1:22;
A1: dom (root-tree [d1,d2])`1 = dom root-tree [d1,d2] by Th24;
A2: dom (root-tree [d1,d2])`2 = dom root-tree [d1,d2] by Th24;
A3: (root-tree [d1,d2]).r = [d1,d2];
A4: [d1,d2]`1 = d1;
A5: [d1,d2]`2 = d2;
  thus (root-tree [d1,d2])`1 = root-tree ((root-tree [d1,d2])`1.r) by A1,Th5
    .= root-tree d1 by A3,A4,TREES_3:39;
  thus (root-tree [d1,d2])`2 = root-tree ((root-tree [d1,d2])`2.r) by A2,Th5
    .= root-tree d2 by A3,A5,TREES_3:39;
end;
