Theorem 2sb6 1331

Description: Equivalence for double substitution.

Assertion

Ref	Expression
2sb6	|- ([z / x][w / y]ph <-> A.xA.y((x = z /\ y = w) -> ph))

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

Proof of Theorem 2sb6

Step	Hyp	Ref	Expression
1		sb6 1262	. 2 |- ([z / x][w / y]ph <-> A.x(x = z -> [w / y]ph))
2		19.21v 1280	. . . 4 |- (A.y(x = z -> (y = w -> ph)) <-> (x = z -> A.y(y = w -> ph)))
3		impexp 347	. . . . 5 |- (((x = z /\ y = w) -> ph) <-> (x = z -> (y = w -> ph)))
4	3	albii 996	. . . 4 |- (A.y((x = z /\ y = w) -> ph) <-> A.y(x = z -> (y = w -> ph)))
5		sb6 1262	. . . . 5 |- ([w / y]ph <-> A.y(y = w -> ph))
6	5	imbi2i 185	. . . 4 |- ((x = z -> [w / y]ph) <-> (x = z -> A.y(y = w -> ph)))
7	2, 4, 6	3bitr4r 184	. . 3 |- ((x = z -> [w / y]ph) <-> A.y((x = z /\ y = w) -> ph))
8	7	albii 996	. 2 |- (A.x(x = z -> [w / y]ph) <-> A.xA.y((x = z /\ y = w) -> ph))
9	1, 8	bitr 173	1 |- ([z / x][w / y]ph <-> A.xA.y((x = z /\ y = w) -> ph))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223  A.wal 951   = wceq 953  [wsbc 1166

This theorem is referenced by:  2eu6 1447

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 960  ax-17 968  ax-4 970  ax-5o 972  ax-6o 975  ax-9o 1119  ax-16 1206  ax-11o 1213

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 978  df-sb 1168

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