theorem Th91:
  Subformulae (F => G) = Subformulae F \/ Subformulae G \/ {
  'not' G, F '&' 'not' G, F => G }
proof
  thus Subformulae (F => G) = Subformulae (F '&' 'not' G) \/ { F => G } by Th88
    .= Subformulae F \/ Subformulae 'not' G \/ {F '&' 'not' G} \/ {F => G}
  by Th89
    .= Subformulae F \/ (Subformulae G \/ {'not' G}) \/ {F '&' 'not' G} \/ {
  F => G } by Th88
    .= Subformulae F \/ Subformulae G \/ {'not' G} \/ {F '&' 'not' G} \/ { F
  => G } by XBOOLE_1:4
    .= Subformulae F \/ Subformulae G \/ {'not' G} \/ ({F '&' 'not' G} \/ {
  F => G }) by XBOOLE_1:4
    .= Subformulae F \/ Subformulae G \/ ({'not' G} \/ ({F '&' 'not' G} \/ {
  F => G })) by XBOOLE_1:4
    .= Subformulae F \/ Subformulae G \/ ({'not' G} \/ { F '&' 'not' G,F =>
  G }) by ENUMSET1:1
    .= Subformulae F \/ Subformulae G \/ { 'not' G,F '&' 'not' G,F => G } by
ENUMSET1:2;
end;
