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  6616  tfr1onlemsucaccv  6617  tfr1onlembxssdm  6619  tfr1onlembfn  6620  tfr1onlemres  6625  tfri1dALT  6627  tfrcllemsucfn  6629  tfrcllemsucaccv  6630  tfrcllembxssdm  6632  tfrcllembfn  6633  tfrcllemres  6638  findcard2  7199  findcard2s  7200  bj-nalset  17143