Theorem 2sb6

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

	Ref	Expression
Assertion	2sb6	|- ( [ z / x ] [ w / y ] ph <-> A. x A. y ( ( x = z /\ y = w ) -> ph ) )

Proof

Step	Hyp	Ref	Expression
1		sb6	|- ( [ z / x ] [ w / y ] ph <-> A. x ( x = z -> [ w / y ] ph ) )
2		19.21v	|- ( A. y ( x = z -> ( y = w -> ph ) ) <-> ( x = z -> A. y ( y = w -> ph ) ) )
3		impexp	|- ( ( ( x = z /\ y = w ) -> ph ) <-> ( x = z -> ( y = w -> ph ) ) )
4	3	albii	|- ( A. y ( ( x = z /\ y = w ) -> ph ) <-> A. y ( x = z -> ( y = w -> ph ) ) )
5		sb6	|- ( [ w / y ] ph <-> A. y ( y = w -> ph ) )
6	5	imbi2i	|- ( ( x = z -> [ w / y ] ph ) <-> ( x = z -> A. y ( y = w -> ph ) ) )
7	2 4 6	3bitr4ri	|- ( ( x = z -> [ w / y ] ph ) <-> A. y ( ( x = z /\ y = w ) -> ph ) )
8	7	albii	|- ( A. x ( x = z -> [ w / y ] ph ) <-> A. x A. y ( ( x = z /\ y = w ) -> ph ) )
9	1 8	bitri	|- ( [ z / x ] [ w / y ] ph <-> A. x A. y ( ( x = z /\ y = w ) -> ph ) )