let n be Ordinal; for i being Nat
for b, b1, b2 being bag of n st i > 1 & i < len (decomp b) & (decomp b) /. i = <*b1,b2*> holds
( b1 <> EmptyBag n & b2 <> EmptyBag n )
let i be Nat; for b, b1, b2 being bag of n st i > 1 & i < len (decomp b) & (decomp b) /. i = <*b1,b2*> holds
( b1 <> EmptyBag n & b2 <> EmptyBag n )
let b, b1, b2 be bag of n; ( i > 1 & i < len (decomp b) & (decomp b) /. i = <*b1,b2*> implies ( b1 <> EmptyBag n & b2 <> EmptyBag n ) )
assume that
A1:
( i > 1 & i < len (decomp b) )
and
A2:
(decomp b) /. i = <*b1,b2*>
; ( b1 <> EmptyBag n & b2 <> EmptyBag n )
reconsider p = (divisors b) /. i as bag of n ;
A3:
i in dom (decomp b)
by A1, FINSEQ_3:25;
then A4:
(decomp b) /. i = <*p,(b -' p)*>
by Def16;
then A5:
b2 = b -' p
by A2, FINSEQ_1:77;
A6:
dom (decomp b) = dom (divisors b)
by Def16;
then A7:
len (decomp b) = len (divisors b)
by FINSEQ_3:29;
b1 = p
by A2, A4, FINSEQ_1:77;
hence
b1 <> EmptyBag n
by A1, A7, Th64; b2 <> EmptyBag n
assume
b2 = EmptyBag n
; contradiction
then
p = b
by A6, A3, A5, Th55, Th62;
hence
contradiction
by A1, A7, Th64; verum