Theorem 2sb5

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

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

Proof

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