theorem
  L is subst-correct & x = x0 in X.s & y = y0 in X.s implies
  A\and\not(A/(x0,y0))\imp\not(x '=' (y,L)) in G1
  proof
    assume that
A0: L is subst-correct and
A1: x = x0 in X.s & y = y0 in X.s;
    s in dom X = the carrier of J by A1,FUNCT_1:def 2,PARTFUN1:def 2;
    then X.s c= (the Sorts of T).s = (the Sorts of L).s
    by Th16,PBOOLE:def 2,def 18;
    then reconsider t1 = x, t2 = y as Element of L,s by A1;
    A\and (x '=' (y,L))\imp(A/(x0,y0)) in G1 &
    (A\and(t1 '=' (t2,L))\imp(A/(x0,y0)))\imp
    (A\imp((t1 '=' (t2,L))\imp(A/(x0,y0)))) in G1 by A0,A1,ThTwo,Th47;
    then
    A\imp((t1 '=' (t2,L))\imp(A/(x0,y0))) in G1 &
   ((t1 '=' (t2,L))\imp(A/(x0,y0)))\imp(\not(A/(x0,y0))\imp\not(t1 '=' (t2,L)))
   in G1 by Def38,Th57;
   then
   A\imp(\not(A/(x0,y0))\imp\not(t1 '=' (t2,L))) in G1 &
   A\imp(\not(A/(x0,y0))\imp\not(t1 '=' (t2,L)))\imp
   (A\and\not(A/(x0,y0))\imp\not(t1 '=' (t2,L))) in G1 by Th45,Th48;
   hence thesis by Def38;
  end;
