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Metamath Proof Explorer

Theorem sb6rf

Description: Reversed substitution. For a version requiring disjoint variables, but fewer axioms, see sb6rfv . Usage of this theorem is discouraged because it depends on ax-13 . Use the weaker sb6rfv if possible. (Contributed by NM, 1-Aug-1993) (Revised by Mario Carneiro, 6-Oct-2016) (Proof shortened by Wolf Lammen, 21-Sep-2018) (New usage is discouraged.)

Ref Expression
Hypothesis sb5rf.1 
|- F/ y ph
Assertion sb6rf 
|- ( ph <-> A. 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 equsal 
 |-  ( A. y ( y = x -> [ y / x ] ph ) <-> ph )
4 3 bicomi 
 |-  ( ph <-> A. y ( y = x -> [ y / x ] ph ) )