Theorem sbel2x 1998

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 1997	. . . . 5 |- ( [ x / z ] ph <-> E. y ( y = w /\ [ y / w ] [ x / z ] ph ) )
2	1	anbi2i 457	. . . 4 |- ( ( x = z /\ [ x / z ] ph ) <-> ( x = z /\ E. y ( y = w /\ [ y / w ] [ x / z ] ph ) ) )
3	2	exbii 1605	. . 3 |- ( E. x ( x = z /\ [ x / z ] ph ) <-> E. x ( x = z /\ E. y ( y = w /\ [ y / w ] [ x / z ] ph ) ) )
4		sbelx 1997	. . 3 |- ( ph <-> E. x ( x = z /\ [ x / z ] ph ) )
5		exdistr 1909	. . 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 212	. 2 |- ( ph <-> E. x E. y ( x = z /\ ( y = w /\ [ y / w ] [ x / z ] ph ) ) )
7		anass 401	. . 3 |- ( ( ( x = z /\ y = w ) /\ [ y / w ] [ x / z ] ph ) <-> ( x = z /\ ( y = w /\ [ y / w ] [ x / z ] ph ) ) )
8	7	2exbii 1606	. 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 187	1 |- ( ph <-> E. x E. y ( ( x = z /\ y = w ) /\ [ y / w ] [ x / z ] ph ) )

Syntax hints:   /\ wa 104   <-> wb 105   E.wex 1492   [wsb 1762

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1447  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-11 1506  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534

This theorem depends on definitions:  df-bi 117  df-sb 1763

This theorem is referenced by: (None)