theorem
  L is subst-eq-correct & x0 in X.s & t1 '=' (t2,L) in G1
  implies (t1/(x0,t) '=' (t2/(x0,t),L)) in G1
  proof
    assume A1: L is subst-eq-correct;
    set Y = X extended_by({}, the carrier of S1);
    assume A2: x0 in X.s;
    then
A3: s in dom X = the carrier of J & dom Y = the carrier of S1
    by FUNCT_1:def 2,PARTFUN1:def 2;
    then reconsider x = x0 as Element of Union X by A2,CARD_5:2;
A4: Y.s = X.s by A3,Th1;
    assume t1 '=' (t2,L) in G1;
    then \for(x,t1 '=' (t2,L)) in G1 &
    \for(x,t1 '='(t2,L))\imp((t1 '=' (t2,L))/(x0,t)) in G1 by A3,A4,A2,Def39;
    then (t1 '=' (t2,L))/(x0,t) in G1 by Def38;
    hence (t1/(x0,t) '=' (t2/(x0,t),L)) in G1 by A1,A2;
  end;
