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Metamath Proof Explorer

Theorem sbc3an

Description: Distribution of class substitution over triple conjunction. (Contributed by NM, 14-Dec-2006) (Revised by NM, 17-Aug-2018)

Ref Expression
Assertion sbc3an 
|- ( [. A / x ]. ( ph /\ ps /\ ch ) <-> ( [. A / x ]. ph /\ [. A / x ]. ps /\ [. A / x ]. ch ) )

Proof

Step Hyp Ref Expression
1 sbcan 
 |-  ( [. A / x ]. ( ( ph /\ ps ) /\ ch ) <-> ( [. A / x ]. ( ph /\ ps ) /\ [. A / x ]. ch ) )
2 sbcan 
 |-  ( [. A / x ]. ( ph /\ ps ) <-> ( [. A / x ]. ph /\ [. A / x ]. ps ) )
3 1 2 bianbi 
 |-  ( [. A / x ]. ( ( ph /\ ps ) /\ ch ) <-> ( ( [. A / x ]. ph /\ [. A / x ]. ps ) /\ [. A / x ]. ch ) )
4 df-3an 
 |-  ( ( ph /\ ps /\ ch ) <-> ( ( ph /\ ps ) /\ ch ) )
5 4 sbcbii 
 |-  ( [. A / x ]. ( ph /\ ps /\ ch ) <-> [. A / x ]. ( ( ph /\ ps ) /\ ch ) )
6 df-3an 
 |-  ( ( [. A / x ]. ph /\ [. A / x ]. ps /\ [. A / x ]. ch ) <-> ( ( [. A / x ]. ph /\ [. A / x ]. ps ) /\ [. A / x ]. ch ) )
7 3 5 6 3bitr4i 
 |-  ( [. A / x ]. ( ph /\ ps /\ ch ) <-> ( [. A / x ]. ph /\ [. A / x ]. ps /\ [. A / x ]. ch ) )