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  4263  tfisi  4734  tfr1onlemsucfn  6611  tfr1onlemsucaccv  6612  tfr1onlembxssdm  6614  tfr1onlembfn  6615  tfr1onlemres  6620  tfri1dALT  6622  tfrcllemsucfn  6624  tfrcllemsucaccv  6625  tfrcllembxssdm  6627  tfrcllembfn  6628  tfrcllemres  6633  findcard2  7193  findcard2s  7194  bj-nalset  16921