
theorem Th23:
for A,B be non empty Interval, p,q,r,s be R_eal st
 A = ].p,q.[ & B = ].r,s.[ & A misses B holds not A \/ B is Interval
proof
    let A,B be non empty Interval, p,q,r,s be R_eal;
    assume that
A1:  A = ].p,q.[ and
A2:  B = ].r,s.[ and
A3:  A misses B;
A4: p < q & r < s by A1,A2,XXREAL_1:28; then
A5: inf A = p & sup A = q & inf B = r & sup B = s
      by A1,A2,MEASURE6:8,12;
    per cases by A1,A2,A3,Th13;
    suppose A6: q <= r; then
A7:  inf A < inf B & sup A < sup B by A5,A1,A2,XXREAL_1:28,XXREAL_0:2;
     not q in A & not q in B by A1,A2,A6,XXREAL_1:4; then
A8:  not q in A \/ B by XBOOLE_0:def 3;
     now assume
A9:   A \/ B is Interval;
      inf(A \/ B) = min(inf A,inf B)
    & sup(A \/ B) = max(sup A,sup B) by XXREAL_2:9,10; then
      inf(A \/ B) = inf A & sup(A \/ B) = sup B
        by A6,A4,A5,XXREAL_0:2,def 9,def 10;
      hence contradiction by A8,A5,A7,A4,A9,XXREAL_2:83;
     end;
     hence not A \/ B is Interval;
    end;
    suppose A10: s <= p;
     not s in A & not s in B by A1,A2,A10,XXREAL_1:4; then
A11:  not s in A \/ B by XBOOLE_0:def 3;

A12: inf B < s & s < sup A by A5,A10,A1,A2,XXREAL_1:28,XXREAL_0:2;
     now assume
A13:   A \/ B is Interval;
      inf(A \/ B) = min(inf A,inf B)
    & sup(A \/ B) = max(sup A,sup B) by XXREAL_2:9,10; then
      inf(A \/ B) = inf B & sup(A \/ B) = sup A
        by A10,A4,A5,XXREAL_0:2,def 9,def 10;
      hence contradiction by A11,A12,A13,XXREAL_2:83;
     end;
     hence not A \/ B is Interval;
    end;
end;
