Theorem spv 1998

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

Hypothesis

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

Assertion

Ref	Expression
spv	|- (A.xph -> 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 198	. 2 |- (x = y -> (ph -> ps))
3	2	spimv 1990	1 |- (A.xph -> ps)

Syntax hints:   -> wi 4   <-> wb 176  A.wal 1540

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925

This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545

This theorem is referenced by:  chvarv  2013  ax10-16  2190  ru  3046  sfintfin  4533  spfinsfincl  4540