Metamath Proof Explorer


Theorem sb5rf

Description: Reversed substitution. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 3-Feb-2005) (Revised by Mario Carneiro, 6-Oct-2016) (Proof shortened by Wolf Lammen, 20-Sep-2018) (New usage is discouraged.)

Ref Expression
Hypothesis sb5rf.1
|- F/ y ph
Assertion sb5rf
|- ( ph <-> E. y ( y = x /\ [ y / x ] ph ) )

Proof

Step Hyp Ref Expression
1 sb5rf.1
 |-  F/ y ph
2 sbequ12r
 |-  ( y = x -> ( [ y / x ] ph <-> ph ) )
3 1 2 equsex
 |-  ( E. y ( y = x /\ [ y / x ] ph ) <-> ph )
4 3 bicomi
 |-  ( ph <-> E. y ( y = x /\ [ y / x ] ph ) )