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Metamath Proof Explorer

Theorem sbel2x

Description: Elimination of double substitution. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 5-Aug-1993) (Proof shortened by Wolf Lammen, 29-Sep-2018) (New usage is discouraged.)

Ref Expression
Assertion sbel2x 
|- ( ph <-> E. x E. y ( ( x = z /\ y = w ) /\ [ y / w ] [ x / z ] ph ) )

Proof

Step Hyp Ref Expression
1 nfv 
 |-  F/ y ph
2 nfv 
 |-  F/ x ph
3 1 2 2sb5rf 
 |-  ( ph <-> E. y E. x ( ( y = w /\ x = z ) /\ [ y / w ] [ x / z ] ph ) )
4 ancom 
 |-  ( ( y = w /\ x = z ) <-> ( x = z /\ y = w ) )
5 4 anbi1i 
 |-  ( ( ( y = w /\ x = z ) /\ [ y / w ] [ x / z ] ph ) <-> ( ( x = z /\ y = w ) /\ [ y / w ] [ x / z ] ph ) )
6 5 2exbii 
 |-  ( E. y E. x ( ( y = w /\ x = z ) /\ [ y / w ] [ x / z ] ph ) <-> E. y E. x ( ( x = z /\ y = w ) /\ [ y / w ] [ x / z ] ph ) )
7 excom 
 |-  ( E. y E. x ( ( x = z /\ y = w ) /\ [ y / w ] [ x / z ] ph ) <-> E. x E. y ( ( x = z /\ y = w ) /\ [ y / w ] [ x / z ] ph ) )
8 3 6 7 3bitri 
 |-  ( ph <-> E. x E. y ( ( x = z /\ y = w ) /\ [ y / w ] [ x / z ] ph ) )