Theorem spv 1913

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 1864	1 |- ( A. x ph -> ps )

This proof depends on syntax axioms:   -> wi 4   <-> wb 105   A.wal 1400

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1500  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587

This proof depends on definitions:  df-bi 117  df-nf 1514

This theorem is used by:  spvv  1963  cbvalvw  1975  chvarv  1997  ru  3050  nalset  4265  tfisi  4736  tfr1onlemsucfn  6615  tfr1onlemsucaccv  6616  tfr1onlembxssdm  6618  tfr1onlembfn  6619  tfr1onlemres  6624  tfri1dALT  6626  tfrcllemsucfn  6628  tfrcllemsucaccv  6629  tfrcllembxssdm  6631  tfrcllembfn  6632  tfrcllemres  6637  findcard2  7197  findcard2s  7198  bj-nalset  17126