Theorem sbel2x 995

Description: Elimination of double substitution.

Assertion

Ref	Expression
sbel2x	|- (ph <-> E.xE.y((x = z /\ y = w) /\ [y / w][x / z]ph))

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

Proof of Theorem sbel2x

Step	Hyp	Ref	Expression
1		sbelx 994	. . . . 5 |- ([x / z]ph <-> E.y(y = w /\ [y / w][x / z]ph))
2	1	anbi2i 367	. . . 4 |- ((x = z /\ [x / z]ph) <-> (x = z /\ E.y(y = w /\ [y / w][x / z]ph)))
3	2	biex 733	. . 3 |- (E.x(x = z /\ [x / z]ph) <-> E.x(x = z /\ E.y(y = w /\ [y / w][x / z]ph)))
4		sbelx 994	. . 3 |- (ph <-> E.x(x = z /\ [x / z]ph))
5		exdistr 967	. . 3 |- (E.xE.y(x = z /\ (y = w /\ [y / w][x / z]ph)) <-> E.x(x = z /\ E.y(y = w /\ [y / w][x / z]ph)))
6	3, 4, 5	3bitr4 158	. 2 |- (ph <-> E.xE.y(x = z /\ (y = w /\ [y / w][x / z]ph)))
7		anass 336	. . 3 |- (((x = z /\ y = w) /\ [y / w][x / z]ph) <-> (x = z /\ (y = w /\ [y / w][x / z]ph)))
8	7	bi2ex 734	. 2 |- (E.xE.y((x = z /\ y = w) /\ [y / w][x / z]ph) <-> E.xE.y(x = z /\ (y = w /\ [y / w][x / z]ph)))
9	6, 8	bitr4 154	1 |- (ph <-> E.xE.y((x = z /\ y = w) /\ [y / w][x / z]ph))

Syntax hints:   <-> wb 127   /\ wa 196  E.wex 678   = weq 797  [wsb 852

This theorem is referenced by:  opabid 2099

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  ax-17 925

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

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