Theorem 2sb5 1956

Description: Equivalence for double substitution. (Contributed by NM, 3-Feb-2005.)

Assertion

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

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

Allowed substitution hints:   ph( x, y, z, w)

Proof of Theorem 2sb5

Step	Hyp	Ref	Expression
1		sb5 1859	. 2 |- ( [ z / x ] [ w / y ] ph <-> E. x ( x = z /\ [ w / y ] ph ) )
2		19.42v 1878	. . . 4 |- ( E. y ( x = z /\ ( y = w /\ ph ) ) <-> ( x = z /\ E. y ( y = w /\ ph ) ) )
3		anass 398	. . . . 5 |- ( ( ( x = z /\ y = w ) /\ ph ) <-> ( x = z /\ ( y = w /\ ph ) ) )
4	3	exbii 1584	. . . 4 |- ( E. y ( ( x = z /\ y = w ) /\ ph ) <-> E. y ( x = z /\ ( y = w /\ ph ) ) )
5		sb5 1859	. . . . 5 |- ( [ w / y ] ph <-> E. y ( y = w /\ ph ) )
6	5	anbi2i 452	. . . 4 |- ( ( x = z /\ [ w / y ] ph ) <-> ( x = z /\ E. y ( y = w /\ ph ) ) )
7	2, 4, 6	3bitr4ri 212	. . 3 |- ( ( x = z /\ [ w / y ] ph ) <-> E. y ( ( x = z /\ y = w ) /\ ph ) )
8	7	exbii 1584	. 2 |- ( E. x ( x = z /\ [ w / y ] ph ) <-> E. x E. y ( ( x = z /\ y = w ) /\ ph ) )
9	1, 8	bitri 183	1 |- ( [ z / x ] [ w / y ] ph <-> E. x E. y ( ( x = z /\ y = w ) /\ ph ) )

Syntax hints:   /\ wa 103   <-> wb 104   E.wex 1468   [wsb 1735

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1423  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-11 1484  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514

This theorem depends on definitions:  df-bi 116  df-sb 1736

This theorem is referenced by:  opelopabsbALT  4176