Metamath Proof Explorer


Theorem sb6rfv

Description: Reversed substitution. Version of sb6rf requiring disjoint variables, but fewer axioms. (Contributed by NM, 1-Aug-1993) (Revised by Wolf Lammen, 7-Feb-2023)

Ref Expression
Hypothesis sb6rfv.nf
|- F/ y ph
Assertion sb6rfv
|- ( ph <-> A. y ( y = x -> [ y / x ] ph ) )

Proof

Step Hyp Ref Expression
1 sb6rfv.nf
 |-  F/ y ph
2 sbequ12r
 |-  ( y = x -> ( [ y / x ] ph <-> ph ) )
3 1 2 equsalv
 |-  ( A. y ( y = x -> [ y / x ] ph ) <-> ph )
4 3 bicomi
 |-  ( ph <-> A. y ( y = x -> [ y / x ] ph ) )