Theorem sbex 998

Description: Move existential quantifier in and out of substitution.

Assertion

Ref	Expression
sbex	|- ([z / y]E.xph <-> E.x[z / y]ph)

Distinct variable group(s):   x,y   x,z

Proof of Theorem sbex

Step	Hyp	Ref	Expression
1		sbn 882	. . 3 |- ([z / y] -. A.x -. ph <-> -. [z / y]A.x -. ph)
2		sbal 997	. . . . 5 |- ([z / y]A.x -. ph <-> A.x[z / y] -. ph)
3		sbn 882	. . . . . 6 |- ([z / y] -. ph <-> -. [z / y]ph)
4	3	bial 695	. . . . 5 |- (A.x[z / y] -. ph <-> A.x -. [z / y]ph)
5	2, 4	bitr 151	. . . 4 |- ([z / y]A.x -. ph <-> A.x -. [z / y]ph)
6	5	negbii 162	. . 3 |- (-. [z / y]A.x -. ph <-> -. A.x -. [z / y]ph)
7	1, 6	bitr 151	. 2 |- ([z / y] -. A.x -. ph <-> -. A.x -. [z / y]ph)
8		df-ex 679	. . 3 |- (E.xph <-> -. A.x -. ph)
9	8	bisb 855	. 2 |- ([z / y]E.xph <-> [z / y] -. A.x -. ph)
10		df-ex 679	. 2 |- (E.x[z / y]ph <-> -. A.x -. [z / y]ph)
11	7, 9, 10	3bitr4 158	1 |- ([z / y]E.xph <-> E.x[z / y]ph)

Syntax hints:  -. wn 1   <-> wb 127  A.wal 672  E.wex 678  [wsb 852

This theorem is referenced by:  sbabel 1189  sbcex 1465

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  ax-16 922

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853

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