Theorem sbel2x 1345

Description: Elimination of double substitution.

Assertion

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

Distinct variable groups:   x,y,z   y,w   ph,x,y

Proof of Theorem sbel2x

Step	Hyp	Ref	Expression
1		sbelx 1344	. . . . 5 |- ([x / z]ph <-> E.y(y = w /\ [y / w][x / z]ph))
2	1	anbi2i 480	. . . 4 |- ((x = z /\ [x / z]ph) <-> (x = z /\ E.y(y = w /\ [y / w][x / z]ph)))
3	2	exbii 1051	. . 3 |- (E.x(x = z /\ [x / z]ph) <-> E.x(x = z /\ E.y(y = w /\ [y / w][x / z]ph)))
4		sbelx 1344	. . 3 |- (ph <-> E.x(x = z /\ [x / z]ph))
5		exdistr 1309	. . 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 183	. 2 |- (ph <-> E.xE.y(x = z /\ (y = w /\ [y / w][x / z]ph)))
7		anass 439	. . 3 |- (((x = z /\ y = w) /\ [y / w][x / z]ph) <-> (x = z /\ (y = w /\ [y / w][x / z]ph)))
8	7	2exbii 1052	. 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 176	1 |- (ph <-> E.xE.y((x = z /\ y = w) /\ [y / w][x / z]ph))

Syntax hints:   <-> wb 146   /\ wa 223   = wceq 956  E.wex 980  [wsbc 1170

This theorem is referenced by:  opabid 2810

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-10 966  ax-12 968  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-16 1210  ax-11o 1218

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 981  df-sb 1172

Copyright terms: Public domain