Metamath Proof Explorer


Theorem s6cld

Description: A length 6 string is a word. (Contributed by Mario Carneiro, 27-Feb-2016)

Ref Expression
Hypotheses s2cld.1
|- ( ph -> A e. X )
s2cld.2
|- ( ph -> B e. X )
s3cld.3
|- ( ph -> C e. X )
s4cld.4
|- ( ph -> D e. X )
s5cld.5
|- ( ph -> E e. X )
s6cld.6
|- ( ph -> F e. X )
Assertion s6cld
|- ( ph -> <" A B C D E F "> e. Word X )

Proof

Step Hyp Ref Expression
1 s2cld.1
 |-  ( ph -> A e. X )
2 s2cld.2
 |-  ( ph -> B e. X )
3 s3cld.3
 |-  ( ph -> C e. X )
4 s4cld.4
 |-  ( ph -> D e. X )
5 s5cld.5
 |-  ( ph -> E e. X )
6 s6cld.6
 |-  ( ph -> F e. X )
7 df-s6
 |-  <" A B C D E F "> = ( <" A B C D E "> ++ <" F "> )
8 1 2 3 4 5 s5cld
 |-  ( ph -> <" A B C D E "> e. Word X )
9 7 8 6 cats1cld
 |-  ( ph -> <" A B C D E F "> e. Word X )