Metamath Proof Explorer


Theorem wrddrin

Description: A word whose alphabet is intersection of two classes is also a word in each of those alphabets. (Contributed by Ender Ting, 24-Jul-2026)

Ref Expression
Assertion wrddrin
|- ( A e. Word ( B i^i C ) -> ( A e. Word B /\ A e. Word C ) )

Proof

Step Hyp Ref Expression
1 inss1
 |-  ( B i^i C ) C_ B
2 sswrd
 |-  ( ( B i^i C ) C_ B -> Word ( B i^i C ) C_ Word B )
3 1 2 ax-mp
 |-  Word ( B i^i C ) C_ Word B
4 3 sseli
 |-  ( A e. Word ( B i^i C ) -> A e. Word B )
5 inss2
 |-  ( B i^i C ) C_ C
6 sswrd
 |-  ( ( B i^i C ) C_ C -> Word ( B i^i C ) C_ Word C )
7 5 6 ax-mp
 |-  Word ( B i^i C ) C_ Word C
8 7 sseli
 |-  ( A e. Word ( B i^i C ) -> A e. Word C )
9 4 8 jca
 |-  ( A e. Word ( B i^i C ) -> ( A e. Word B /\ A e. Word C ) )