Theorem spv 1906

Description: Specialization, using implicit substitition. (Contributed by NM, 30-Aug-1993.)

Hypothesis

Ref	Expression
spv.1	|- ( x = y -> ( ph <-> ps ) )

Assertion

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

Distinct variable group:   ps, x

Allowed substitution hints:   ph( x, y)   ps( y)

Proof of Theorem spv

Step	Hyp	Ref	Expression
1		spv.1	. . 3 |- ( x = y -> ( ph <-> ps ) )
2	1	biimpd 144	. 2 |- ( x = y -> ( ph -> ps ) )
3	2	spimv 1857	1 |- ( A. x ph -> ps )

Syntax hints:   -> wi 4   <-> wb 105   A.wal 1393

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1493  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580

This theorem depends on definitions:  df-bi 117  df-nf 1507

This theorem is referenced by:  spvv  1954  cbvalvw  1966  chvarv  1988  ru  3027  nalset  4214  tfisi  4679  tfr1onlemsucfn  6486  tfr1onlemsucaccv  6487  tfr1onlembxssdm  6489  tfr1onlembfn  6490  tfr1onlemres  6495  tfri1dALT  6497  tfrcllemsucfn  6499  tfrcllemsucaccv  6500  tfrcllembxssdm  6502  tfrcllembfn  6503  tfrcllemres  6508  findcard2  7051  findcard2s  7052  bj-nalset  16258