Theorem spim 1975

Description: Specialization, using implicit substitution. Compare Lemma 14 of [Tarski] p. 70. The spim 1975 series of theorems requires that only one direction of the substitution hypothesis hold. (Contributed by NM, 5-Aug-1993.) (Revised by Mario Carneiro, 3-Oct-2016.)

Hypotheses

Ref	Expression
spim.1	|- F/xps
spim.2	|- (x = y -> (ph -> ps))

Assertion

Ref	Expression
spim	|- (A.xph -> ps)

Proof of Theorem spim

Step	Hyp	Ref	Expression
1		spim.1	. 2 |- F/xps
2		spim.2	. . 3 |- (x = y -> (ph -> ps))
3	2	ax-gen 1546	. 2 |- A.x(x = y -> (ph -> ps))
4		spimt 1974	. 2 |- (( F/xps /\ A.x(x = y -> (ph -> ps))) -> (A.xph -> ps))
5	1, 3, 4	mp2an 653	1 |- (A.xph -> ps)

Syntax hints:   -> wi 4  A.wal 1540   F/wnf 1544

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:  spime  1976  chvar  1986  spimv  1990