Theorem 2sb5 1335

Description: Equivalence for double substitution.

Assertion

Ref	Expression
2sb5	|- ([z / x][w / y]ph <-> E.xE.y((x = z /\ y = w) /\ ph))

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

Proof of Theorem 2sb5

Step	Hyp	Ref	Expression
1		sb5 1268	. 2 |- ([z / x][w / y]ph <-> E.x(x = z /\ [w / y]ph))
2		19.42v 1308	. . . 4 |- (E.y(x = z /\ (y = w /\ ph)) <-> (x = z /\ E.y(y = w /\ ph)))
3		anass 439	. . . . 5 |- (((x = z /\ y = w) /\ ph) <-> (x = z /\ (y = w /\ ph)))
4	3	exbii 1051	. . . 4 |- (E.y((x = z /\ y = w) /\ ph) <-> E.y(x = z /\ (y = w /\ ph)))
5		sb5 1268	. . . . 5 |- ([w / y]ph <-> E.y(y = w /\ ph))
6	5	anbi2i 480	. . . 4 |- ((x = z /\ [w / y]ph) <-> (x = z /\ E.y(y = w /\ ph)))
7	2, 4, 6	3bitr4r 184	. . 3 |- ((x = z /\ [w / y]ph) <-> E.y((x = z /\ y = w) /\ ph))
8	7	exbii 1051	. 2 |- (E.x(x = z /\ [w / y]ph) <-> E.xE.y((x = z /\ y = w) /\ ph))
9	1, 8	bitr 173	1 |- ([z / x][w / y]ph <-> E.xE.y((x = z /\ y = w) /\ ph))

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

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 963  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  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

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