Theorem sb5y 883

Description: Equivalence involving substitution with a variable not free.

Hypothesis

Ref	Expression
sb5y.1	|- (ph -> A.yph)

Assertion

Ref	Expression
sb5y	|- ([y / x]ph <-> E.x(x = y /\ ph))

Proof of Theorem sb5y

Step	Hyp	Ref	Expression
1		sb5y.1	. . . . . 6 |- (ph -> A.yph)
2	1	hbne 699	. . . . 5 |- (-. ph -> A.y -. ph)
3	2	sb6y 872	. . . 4 |- ([y / x] -. ph <-> A.x(x = y -> -. ph))
4		sbn 882	. . . 4 |- ([y / x] -. ph <-> -. [y / x]ph)
5	3, 4	bitr3 153	. . 3 |- (A.x(x = y -> -. ph) <-> -. [y / x]ph)
6	5	bicon2i 194	. 2 |- ([y / x]ph <-> -. A.x(x = y -> -. ph))
7		eqs3 830	. 2 |- (E.x(x = y /\ ph) <-> -. A.x(x = y -> -. ph))
8	6, 7	bitr4 154	1 |- ([y / x]ph <-> E.x(x = y /\ ph))

Syntax hints:  -. wn 1   -> wi 2   <-> wb 127   /\ wa 196  A.wal 672  E.wex 678   = weq 797  [wsb 852

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

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