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Metamath Proof Explorer

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 ) )