theorem :: Introduction 'or' to premiss, No.1
  ({p} |-_IPC r & {q} |-_IPC r) implies {p 'or' q} |-_IPC r
proof
  set U = p 'or' q;
  set X = {U};
A2: |-_IPC (p => r) => ((q => r) => (U => r)) by Th35;
  assume
A1: {p} |-_IPC r & {q} |-_IPC r; then
A5: |-_IPC q => r by Th54;
   |-_IPC p => r by A1,Th54; then
   |-_IPC (q => r) => (U => r) by A2,Th37; then
A7: |-_IPC U => r by A5,Th37;
  {} c= X; then
  X |-_IPC U => r by A7,Th66;
  hence thesis by Th27,Th65;
end;
