Theorem sbeqi 33968

Description: Equality deduction for substitution. (Contributed by Giovanni Mascellani, 10-Apr-2018.)

Assertion

Ref	Expression
sbeqi	|- ( ( x = y /\ A. z ( ph <-> ps ) ) -> ( [ x / z ] ph <-> [ y / z ] ps ) )

Proof of Theorem sbeqi

Step	Hyp	Ref	Expression
1		spsbbi 2402	. 2 |- ( A. z ( ph <-> ps ) -> ( [ x / z ] ph <-> [ x / z ] ps ) )
2		sbequ 2376	. 2 |- ( x = y -> ( [ x / z ] ps <-> [ y / z ] ps ) )
3	1, 2	sylan9bbr 737	1 |- ( ( x = y /\ A. z ( ph <-> ps ) ) -> ( [ x / z ] ph <-> [ y / z ] ps ) )

Syntax hints:   -> wi 4   <-> wb 196   /\ wa 384   A.wal 1481   = wceq 1483   [wsb 1880

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-10 2019  ax-12 2047  ax-13 2246

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-ex 1705  df-nf 1710  df-sb 1881

This theorem is referenced by: (None)