theorem
not l2 in rng p implies (l2,u) ReassignIn I-TermEval.p= I-TermEval.p
proof
set tt=p, II=U-InterpretersOf S, I2=(l2,u) ReassignIn I, f2=l2.-->({}.-->u);
tt null {} is ({}\/rng tt)-valued FinSequence; then
tt is FinSequence of (rng tt) by FOMODEL0:26; then
reconsider ttt=tt as Element of (rng tt)*;
I2-TermEval|((rng tt)*).ttt \+\ I2-TermEval.ttt={} &
I-TermEval|((rng tt)*).ttt \+\ I-TermEval.ttt={}; then
A1: I2-TermEval|((rng tt)*).tt=I2-TermEval.tt &
I-TermEval|((rng tt)*).tt=I-TermEval.tt by FOMODEL0:29;
assume not l2 in rng tt; then
{l2} misses rng tt by ZFMISC_1:50; then dom f2 misses rng tt; then I2|(rng tt)=
I|(rng tt) by FUNCT_4:72;
hence thesis by A1, Lm41;
end;
