Theorem 2sb5 1859

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 1767	. 2 |- ( [ z / x ] [ w / y ] ph <-> E. x ( x = z /\ [ w / y ] ph ) )
2		19.42v 1786	. . . 4 |- ( E. y ( x = z /\ ( y = w /\ ph ) ) <-> ( x = z /\ E. y ( y = w /\ ph ) ) )
3		anass 381	. . . . 5 |- ( ( ( x = z /\ y = w ) /\ ph ) <-> ( x = z /\ ( y = w /\ ph ) ) )
4	3	exbii 1496	. . . 4 |- ( E. y ( ( x = z /\ y = w ) /\ ph ) <-> E. y ( x = z /\ ( y = w /\ ph ) ) )
5		sb5 1767	. . . . 5 |- ( [ w / y ] ph <-> E. y ( y = w /\ ph ) )
6	5	anbi2i 430	. . . 4 |- ( ( x = z /\ [ w / y ] ph ) <-> ( x = z /\ E. y ( y = w /\ ph ) ) )
7	2, 4, 6	3bitr4ri 202	. . 3 |- ( ( x = z /\ [ w / y ] ph ) <-> E. y ( ( x = z /\ y = w ) /\ ph ) )
8	7	exbii 1496	. 2 |- ( E. x ( x = z /\ [ w / y ] ph ) <-> E. x E. y ( ( x = z /\ y = w ) /\ ph ) )
9	1, 8	bitri 173	1 |- ( [ z / x ] [ w / y ] ph <-> E. x E. y ( ( x = z /\ y = w ) /\ ph ) )

Syntax hints:   /\ wa 97   <-> wb 98   E.wex 1381   [wsb 1645

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1336  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-11 1397  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427

This theorem depends on definitions:  df-bi 110  df-sb 1646

This theorem is referenced by:  opelopabsbALT  3996