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Metamath Proof Explorer

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 ) )