Metamath Proof Explorer


Theorem alseuals

Description: "All some one" implies "all some": requiring exactly one witness is stronger than requiring at least one. Any consequence of an allsome statement is therefore a consequence of the corresponding "all some one" statement, which is how alseu-no-surprise is proved. (Contributed by David A. Wheeler, 21-Jul-2026)

Ref Expression
Assertion alseuals
|- ( AE! x ( ph -> ps ) -> AE x ( ph -> ps ) )

Proof

Step Hyp Ref Expression
1 euex
 |-  ( E! x ph -> E. x ph )
2 1 anim2i
 |-  ( ( A. x ( ph -> ps ) /\ E! x ph ) -> ( A. x ( ph -> ps ) /\ E. x ph ) )
3 df-alseu
 |-  ( AE! x ( ph -> ps ) <-> ( A. x ( ph -> ps ) /\ E! x ph ) )
4 df-als
 |-  ( AE x ( ph -> ps ) <-> ( A. x ( ph -> ps ) /\ E. x ph ) )
5 2 3 4 3imtr4i
 |-  ( AE! x ( ph -> ps ) -> AE x ( ph -> ps ) )