reserve Y for RealNormSpace;

theorem LM519B1:
for a,b,z be Real, p,q,x be Point of REAL-NS 1
   st p= <*a*> & q= <*b*> & x = <*z*> holds
    ( z in ].a,b.[ implies x in ].p,q.[ )
  & ( x in ].p,q.[ implies
        a <> b
      & (a < b implies z in ].a,b.[)
      & (a > b implies z in ].b,a.[) )
proof
   let a,b,z be Real,p,q,x be Point of REAL-NS 1;
   reconsider I= proj(1,1) qua Function" as Function of REAL,REAL 1
            by PDIFF_1:2;
   reconsider J= proj(1,1) as Function of REAL-NS 1,REAL by Lm1;
   assume
P2: p = <*a*> & q = <*b*> & x = <*z*>;
A2:I.a = p & I.b = q & I.z = x & J.p = a & J.q = b & J.x = z by P2,PDIFF_1:1;
   hereby assume z in ].a,b.[; then
A4: a < z & z < b by XXREAL_1:4; then
A5: a < b by XXREAL_0:2;
    reconsider r=(z-a)/(b-a) as Real;
A6: 0 < z-a & 0 < b-a & z-a < b-a by A4,A5,XREAL_1:14,50; then
C1: 0 < r & r < 1 by XREAL_1:139,191;
C3: (1-r)*a + r*b = a + (z-a)/(b-a)*(b-a)
     .= a + (z-a) by A6,XCMPLX_1:87;
    q-p = I.(b-a) by A2,PDIFF_1:3; then
    r*(q-p) = I.(r*(b-a)) by PDIFF_1:3; then
    p + r*(q-p) = x by A2,C3,PDIFF_1:3; then
    x in { p+t*(q-p) where t is Real : 0 < t & t < 1} by C1;
    hence x in ].p,q.[ by A4,P2,FINSEQ_1:76,NDIFF_5:16;
   end;
   assume x in ].p,q.[; then
X1:x in [.p,q.] & not x in {p,q} by XBOOLE_0:def 5; then
   x in {(1-r)*p + r*q where r is Real : 0 <= r & r <= 1}
     by RLTOPSP1:def 2; then
   consider r be Real such that
B1: x = (1-r)*p+r*q & 0 <= r & r <= 1;
B2:J.x = J.((1-r)*p) + J.(r*q) by B1,PDIFF_1:4
      .= (1-r)*J.p + J.(r*q) by PDIFF_1:4
      .=(1-r)*J.p + r*J.q by PDIFF_1:4;
   hence a <> b by A2,X1,TARSKI:def 2;
X3:now assume r=0; then
    x = p + (0 qua Real)*q by B1,RLVECT_1:def 8
     .= p + 0.(REAL-NS 1) by RLVECT_1:10;
    hence contradiction by X1,TARSKI:def 2;
   end;
   now assume r=1; then
    x = (0 qua Real)*p + q by B1,RLVECT_1:def 8
     .= 0.(REAL-NS 1) + q by RLVECT_1:10;
    hence contradiction by X1,TARSKI:def 2;
   end; then
X4:r < 1 by B1,XXREAL_0:1;
   set s = 1-r;
X5:r = 1-s & 0<s & s<1 by X3,B1,X4,XREAL_1:44,50;
   hereby assume a < b; then
    a < z & z < b by A2,X3,B1,X4,XREAL_1:177,178,B2;
    hence z in ].a,b.[;
   end;
   assume a > b; then
   b < z & z < a by A2,X5,XREAL_1:177,178,B2;
   hence z in ].b,a.[;
end;
