Metamath Proof Explorer

Theorem sb3an

Description: Threefold conjunction inside and outside of a substitution are equivalent. (Contributed by NM, 14-Dec-2006)

Ref Expression
Assertion sb3an 
|- ( [ y / x ] ( ph /\ ps /\ ch ) <-> ( [ y / x ] ph /\ [ y / x ] ps /\ [ y / x ] ch ) )

Proof

Step Hyp Ref Expression
1 df-3an 
 |-  ( ( ph /\ ps /\ ch ) <-> ( ( ph /\ ps ) /\ ch ) )
2 1 sbbii 
 |-  ( [ y / x ] ( ph /\ ps /\ ch ) <-> [ y / x ] ( ( ph /\ ps ) /\ ch ) )
3 sban 
 |-  ( [ y / x ] ( ( ph /\ ps ) /\ ch ) <-> ( [ y / x ] ( ph /\ ps ) /\ [ y / x ] ch ) )
4 sban 
 |-  ( [ y / x ] ( ph /\ ps ) <-> ( [ y / x ] ph /\ [ y / x ] ps ) )
5 4 anbi1i 
 |-  ( ( [ y / x ] ( ph /\ ps ) /\ [ y / x ] ch ) <-> ( ( [ y / x ] ph /\ [ y / x ] ps ) /\ [ y / x ] ch ) )
6 df-3an 
 |-  ( ( [ y / x ] ph /\ [ y / x ] ps /\ [ y / x ] ch ) <-> ( ( [ y / x ] ph /\ [ y / x ] ps ) /\ [ y / x ] ch ) )
7 5 6 bitr4i 
 |-  ( ( [ y / x ] ( ph /\ ps ) /\ [ y / x ] ch ) <-> ( [ y / x ] ph /\ [ y / x ] ps /\ [ y / x ] ch ) )
8 2 3 7 3bitri 
 |-  ( [ y / x ] ( ph /\ ps /\ ch ) <-> ( [ y / x ] ph /\ [ y / x ] ps /\ [ y / x ] ch ) )