let a, b be set ; ( a <> b implies Terminals (IterGrammar a,b) = {b} )
set T = IterGrammar a,b;
assume A1:
a <> b
; Terminals (IterGrammar a,b) = {b}
A2:
the carrier of (IterGrammar a,b) = {a,b}
by Def9;
then reconsider x = a, y = b as Symbol of (IterGrammar a,b) by TARSKI:def 2;
A3:
the Rules of (IterGrammar a,b) = {[a,<*b,a*>],[a,{} ]}
by Def9;
thus
Terminals (IterGrammar a,b) c= {b}
XBOOLE_0:def 10 {b} c= Terminals (IterGrammar a,b)proof
let c be
set ;
TARSKI:def 3 ( not c in Terminals (IterGrammar a,b) or c in {b} )
assume
c in Terminals (IterGrammar a,b)
;
c in {b}
then consider s being
Symbol of
(IterGrammar a,b) such that A4:
c = s
and A5:
for
n being
FinSequence holds not
s ==> n
;
[a,<*b,a*>] in the
Rules of
(IterGrammar a,b)
by A3, TARSKI:def 2;
then
x ==> <*y,x*>
by Def1;
then
s <> x
by A5;
then
c = b
by A2, A4, TARSKI:def 2;
hence
c in {b}
by TARSKI:def 1;
verum
end;
let c be set ; TARSKI:def 3 ( not c in {b} or c in Terminals (IterGrammar a,b) )
assume
c in {b}
; c in Terminals (IterGrammar a,b)
then A6:
b = c
by TARSKI:def 1;
assume
not c in Terminals (IterGrammar a,b)
; contradiction
then consider n being FinSequence such that
A7:
y ==> n
by A6;
A8:
[a,{} ] <> [b,n]
by A1, ZFMISC_1:33;
[a,<*b,a*>] <> [b,n]
by A1, ZFMISC_1:33;
then
not [b,n] in {[a,<*b,a*>],[a,{} ]}
by A8, TARSKI:def 2;
hence
contradiction
by A3, A7, Def1; verum