Metamath Proof Explorer


Theorem cbv3v2

Description: Version of cbv3 with two disjoint variable conditions, which does not require ax-11 nor ax-13 . (Contributed by BJ, 24-Jun-2019) (Proof shortened by Wolf Lammen, 30-Aug-2021)

Ref Expression
Hypotheses cbv3v2.nf
|- F/ x ps
cbv3v2.1
|- ( x = y -> ( ph -> ps ) )
Assertion cbv3v2
|- ( A. x ph -> A. y ps )

Proof

Step Hyp Ref Expression
1 cbv3v2.nf
 |-  F/ x ps
2 cbv3v2.1
 |-  ( x = y -> ( ph -> ps ) )
3 1 2 spimfv
 |-  ( A. x ph -> ps )
4 3 alrimiv
 |-  ( A. x ph -> A. y ps )