Metamath Proof Explorer


Theorem als1d

Description: Deduction rule: Given "all some" applied to a top-level inference, you can extract the "for all" part. (Contributed by David A. Wheeler, 20-Oct-2018)

Ref Expression
Hypothesis als1d.1
|- ( ph -> AE x ( ps -> ch ) )
Assertion als1d
|- ( ph -> A. x ( ps -> ch ) )

Proof

Step Hyp Ref Expression
1 als1d.1
 |-  ( ph -> AE x ( ps -> ch ) )
2 df-als
 |-  ( AE x ( ps -> ch ) <-> ( A. x ( ps -> ch ) /\ E. x ps ) )
3 1 2 sylib
 |-  ( ph -> ( A. x ( ps -> ch ) /\ E. x ps ) )
4 3 simpld
 |-  ( ph -> A. x ( ps -> ch ) )