let p1, p2, p3 be set ; ( p1 <> p3 & p2 <> p3 implies p3 .. <*p1,p2,p3*> = 3 )
assume that
A1:
p1 <> p3
and
A2:
p2 <> p3
; p3 .. <*p1,p2,p3*> = 3
A6:
now for i being Nat st 1 <= i & i < 2 + 1 holds
<*p1,p2,p3*> . i <> <*p1,p2,p3*> . 3let i be
Nat;
( 1 <= i & i < 2 + 1 implies <*p1,p2,p3*> . i <> <*p1,p2,p3*> . 3 )assume
1
<= i
;
( i < 2 + 1 implies <*p1,p2,p3*> . i <> <*p1,p2,p3*> . 3 )then A7:
i <> 0
;
assume
i < 2
+ 1
;
<*p1,p2,p3*> . i <> <*p1,p2,p3*> . 3then
i <= 2
by NAT_1:13;
then
not not
i = 0 & ... & not
i = 2
;
hence
<*p1,p2,p3*> . i <> <*p1,p2,p3*> . 3
by A1, A2, A7;
verum end;
3 <= len <*p1,p2,p3*>
by FINSEQ_1:45;
then
3 in dom <*p1,p2,p3*>
by FINSEQ_3:25;
hence
p3 .. <*p1,p2,p3*> = 3
by A6, Th2; verum