Theorem sbn1 880

Description: Removal of negation from substitution.

Assertion

Ref	Expression
sbn1	|- ([y / x] -. ph -> -. [y / x]ph)

Proof of Theorem sbn1

Step	Hyp	Ref	Expression
1		sbequ2 864	. . . 4 |- (x = y -> ([y / x] -. ph -> -. ph))
2		sbequ2 864	. . . . 5 |- (x = y -> ([y / x]ph -> ph))
3	2	con3d 87	. . . 4 |- (x = y -> (-. ph -> -. [y / x]ph))
4	1, 3	syld 27	. . 3 |- (x = y -> ([y / x] -. ph -> -. [y / x]ph))
5	4	a4s 682	. 2 |- (A.x x = y -> ([y / x] -. ph -> -. [y / x]ph))
6		sb4 861	. . 3 |- (-. A.x x = y -> ([y / x] -. ph -> A.x(x = y -> -. ph)))
7		sb1 858	. . . . 5 |- ([y / x]ph -> E.x(x = y /\ ph))
8		eqs3 830	. . . . 5 |- (E.x(x = y /\ ph) <-> -. A.x(x = y -> -. ph))
9	7, 8	sylib 173	. . . 4 |- ([y / x]ph -> -. A.x(x = y -> -. ph))
10	9	con2i 89	. . 3 |- (A.x(x = y -> -. ph) -> -. [y / x]ph)
11	6, 10	syl6 23	. 2 |- (-. A.x x = y -> ([y / x] -. ph -> -. [y / x]ph))
12	5, 11	pm2.61i 110	1 |- ([y / x] -. ph -> -. [y / x]ph)

Syntax hints:  -. wn 1   -> wi 2   /\ wa 196  A.wal 672  E.wex 678   = weq 797  [wsb 852

This theorem is referenced by:  sbn 882

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853

metamath.org