Theorem sbel2x 1991

Description: Elimination of double substitution. (Contributed by NM, 5-Aug-1993.)

Assertion

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

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

Allowed substitution hints:   ph( z, w)

Proof of Theorem sbel2x

Step	Hyp	Ref	Expression
1		sbelx 1990	. . . . 5 |- ( [ x / z ] ph <-> E. y ( y = w /\ [ y / w ] [ x / z ] ph ) )
2	1	anbi2i 454	. . . 4 |- ( ( x = z /\ [ x / z ] ph ) <-> ( x = z /\ E. y ( y = w /\ [ y / w ] [ x / z ] ph ) ) )
3	2	exbii 1598	. . 3 |- ( E. x ( x = z /\ [ x / z ] ph ) <-> E. x ( x = z /\ E. y ( y = w /\ [ y / w ] [ x / z ] ph ) ) )
4		sbelx 1990	. . 3 |- ( ph <-> E. x ( x = z /\ [ x / z ] ph ) )
5		exdistr 1902	. . 3 |- ( E. x E. 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	3bitr4i 211	. 2 |- ( ph <-> E. x E. y ( x = z /\ ( y = w /\ [ y / w ] [ x / z ] ph ) ) )
7		anass 399	. . 3 |- ( ( ( x = z /\ y = w ) /\ [ y / w ] [ x / z ] ph ) <-> ( x = z /\ ( y = w /\ [ y / w ] [ x / z ] ph ) ) )
8	7	2exbii 1599	. 2 |- ( E. x E. y ( ( x = z /\ y = w ) /\ [ y / w ] [ x / z ] ph ) <-> E. x E. y ( x = z /\ ( y = w /\ [ y / w ] [ x / z ] ph ) ) )
9	6, 8	bitr4i 186	1 |- ( ph <-> E. x E. y ( ( x = z /\ y = w ) /\ [ y / w ] [ x / z ] ph ) )

Syntax hints:   /\ wa 103   <-> wb 104   E.wex 1485   [wsb 1755

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 1440  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-11 1499  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527

This theorem depends on definitions:  df-bi 116  df-sb 1756

This theorem is referenced by: (None)