theorem Th43:
  for U0 being non-empty OSAlgebra of S1 for x,y being Element of
  OSSub(U0) holds (OSAlg_meet(U0)).(x,y) = (MSAlg_meet(U0)).(x,y)
proof
  let U0 be non-empty OSAlgebra of S1;
  let x,y be Element of OSSub(U0);
  x is strict OSSubAlgebra of U0 & y is strict OSSubAlgebra of U0 by Def14;
  then consider U1,U2 being strict OSSubAlgebra of U0 such that
A1: x = U1 & y = U2;
  (OSAlg_meet(U0)).(x,y) = U1 /\ U2 by A1,Def16
    .= (MSAlg_meet(U0)).(x,y) by A1,MSUALG_2:def 21;
  hence thesis;
end;
