Theorem sbn2 881

Description: Introduction of negation into substitution.

Assertion

Ref	Expression
sbn2	|- (-. [y / x]ph -> [y / x] -. ph)

Proof of Theorem sbn2

Step	Hyp	Ref	Expression
1		sbequ1 863	. . . . 5 |- (x = y -> (ph -> [y / x]ph))
2	1	con3d 87	. . . 4 |- (x = y -> (-. [y / x]ph -> -. ph))
3	2	com12 13	. . 3 |- (-. [y / x]ph -> (x = y -> -. ph))
4		sb2 859	. . . . . 6 |- (A.x(x = y -> -. -. ph) -> [y / x] -. -. ph)
5		pm4.13 142	. . . . . . 7 |- (ph <-> -. -. ph)
6	5	bisb 855	. . . . . 6 |- ([y / x]ph <-> [y / x] -. -. ph)
7	4, 6	sylibr 175	. . . . 5 |- (A.x(x = y -> -. -. ph) -> [y / x]ph)
8	7	con3i 90	. . . 4 |- (-. [y / x]ph -> -. A.x(x = y -> -. -. ph))
9		eqs3 830	. . . 4 |- (E.x(x = y /\ -. ph) <-> -. A.x(x = y -> -. -. ph))
10	8, 9	sylibr 175	. . 3 |- (-. [y / x]ph -> E.x(x = y /\ -. ph))
11	3, 10	jca 236	. 2 |- (-. [y / x]ph -> ((x = y -> -. ph) /\ E.x(x = y /\ -. ph)))
12		df-sb 853	. 2 |- ([y / x] -. ph <-> ((x = y -> -. ph) /\ E.x(x = y /\ -. ph)))
13	11, 12	sylibr 175	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-gen 677  ax-9 799

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

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