Theorem spd 42425

Description: Specialization deduction, using implicit substitution. Based on the proof of spimed 2255. (Contributed by Emmett Weisz, 17-Jan-2020.)

Hypotheses

Ref	Expression
spd.1	|- ( ch -> F/ x ps )
spd.2	|- ( x = y -> ( ph <-> ps ) )

Assertion

Ref	Expression
spd	|- ( ch -> ( A. x ph -> ps ) )

Proof of Theorem spd

Step	Hyp	Ref	Expression
1		ax6e 2250	. . . 4 |- E. x x = y
2		spd.2	. . . . 5 |- ( x = y -> ( ph <-> ps ) )
3	2	biimpd 219	. . . 4 |- ( x = y -> ( ph -> ps ) )
4	1, 3	eximii 1764	. . 3 |- E. x ( ph -> ps )
5	4	19.35i 1806	. 2 |- ( A. x ph -> E. x ps )
6		spd.1	. . 3 |- ( ch -> F/ x ps )
7	6	19.9d 2070	. 2 |- ( ch -> ( E. x ps -> ps ) )
8	5, 7	syl5 34	1 |- ( ch -> ( A. x ph -> ps ) )

Syntax hints:   -> wi 4   <-> wb 196   A.wal 1481   E.wex 1704   F/wnf 1708

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-12 2047  ax-13 2246

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

This theorem is referenced by: (None)