now

let Seqts1, Seqts2 be Subset of (S -sequents); :: thesis: ( Seqts1 c= Seqts2 implies (R3e S) . Seqts1 c= (R3e S) . Seqts2 )
set X = Seqts1;
set Y = Seqts2;
assume Seqts1 c= Seqts2 ; :: thesis: (R3e S) . Seqts1 c= (R3e S) . Seqts2
set R = R3e S;
set Q = S -sequents ;

now

let x be set ; :: thesis: ( x in (R3e S) . Seqts1 implies x in (R3e S) . Seqts2 )
assume CC0: x in (R3e S) . Seqts1 ; :: thesis: x in (R3e S) . Seqts2
reconsider seqt = x as Element of S -sequents by CC0;
[Seqts1,seqt] in P3e S by CC0, Lm1e;
then seqt Rule3e Seqts1 by DefP3e;
then ex s being relational Element of S ex T, U being abs (ar b₁) -element Element of (AllTermsOf S) * st
( seqt `1 = {(s -compound T)} \/ { ((<*(TheEqSymbOf S)*> ^ (TT . j)) ^ (UU . j)) where j is Element of Seg (abs (ar s)), TT, UU is Function of (Seg (abs (ar s))),(((AllSymbolsOf S) *) \ {{}}) : ( TT = T & UU = U ) } & seqt `2 = s -compound U ) by Def3e;
then seqt Rule3e Seqts2 by Def3e;
then [Seqts2,seqt] in P3e S by DefP3e;
hence x in (R3e S) . Seqts2 by Lm1; :: thesis: verum

end;

hence (R3e S) . Seqts1 c= (R3e S) . Seqts2 by TARSKI:def 3; :: thesis: verum

end;

hence for b₁ being Rule of S st b₁ = R3e S holds
b₁ is isotone by DefMonotonic1; :: thesis: verum