Metamath Proof Explorer


Theorem lswcl

Description: Closure of the last symbol: the last symbol of a not empty word belongs to the alphabet for the word. (Contributed by AV, 2-Aug-2018) (Proof shortened by AV, 29-Apr-2020)

Ref Expression
Assertion lswcl
|- ( ( W e. Word V /\ W =/= (/) ) -> ( lastS ` W ) e. V )

Proof

Step Hyp Ref Expression
1 lsw
 |-  ( W e. Word V -> ( lastS ` W ) = ( W ` ( ( # ` W ) - 1 ) ) )
2 1 adantr
 |-  ( ( W e. Word V /\ W =/= (/) ) -> ( lastS ` W ) = ( W ` ( ( # ` W ) - 1 ) ) )
3 lennncl
 |-  ( ( W e. Word V /\ W =/= (/) ) -> ( # ` W ) e. NN )
4 fzo0end
 |-  ( ( # ` W ) e. NN -> ( ( # ` W ) - 1 ) e. ( 0 ..^ ( # ` W ) ) )
5 3 4 syl
 |-  ( ( W e. Word V /\ W =/= (/) ) -> ( ( # ` W ) - 1 ) e. ( 0 ..^ ( # ` W ) ) )
6 wrdsymbcl
 |-  ( ( W e. Word V /\ ( ( # ` W ) - 1 ) e. ( 0 ..^ ( # ` W ) ) ) -> ( W ` ( ( # ` W ) - 1 ) ) e. V )
7 5 6 syldan
 |-  ( ( W e. Word V /\ W =/= (/) ) -> ( W ` ( ( # ` W ) - 1 ) ) e. V )
8 2 7 eqeltrd
 |-  ( ( W e. Word V /\ W =/= (/) ) -> ( lastS ` W ) e. V )