Metamath Proof Explorer


Theorem ralseu2d

Description: Deduction rule: Given "all some one" applied to a class, you can extract the "exactly one" part. Note that the witness must satisfy the antecedent ps , not merely be a member of A . (Contributed by David A. Wheeler, 21-Jul-2026)

Ref Expression
Hypothesis ralseu2d.1
|- ( ph -> AE! x e. A ( ps -> ch ) )
Assertion ralseu2d
|- ( ph -> E! x e. A ps )

Proof

Step Hyp Ref Expression
1 ralseu2d.1
 |-  ( ph -> AE! x e. A ( ps -> ch ) )
2 df-ralseu
 |-  ( AE! x e. A ( ps -> ch ) <-> ( A. x e. A ( ps -> ch ) /\ E! x e. A ps ) )
3 1 2 sylib
 |-  ( ph -> ( A. x e. A ( ps -> ch ) /\ E! x e. A ps ) )
4 3 simprd
 |-  ( ph -> E! x e. A ps )