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Metamath Proof Explorer

Theorem sban

Description: Conjunction inside and outside of a substitution are equivalent. Compare 19.26 . (Contributed by NM, 14-May-1993) (Proof shortened by Steven Nguyen, 13-Aug-2023)

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

Proof

Step Hyp Ref Expression
1 simpl 
 |-  ( ( ph /\ ps ) -> ph )
2 1 sbimi 
 |-  ( [ y / x ] ( ph /\ ps ) -> [ y / x ] ph )
3 simpr 
 |-  ( ( ph /\ ps ) -> ps )
4 3 sbimi 
 |-  ( [ y / x ] ( ph /\ ps ) -> [ y / x ] ps )
5 2 4 jca 
 |-  ( [ y / x ] ( ph /\ ps ) -> ( [ y / x ] ph /\ [ y / x ] ps ) )
6 pm3.2 
 |-  ( ph -> ( ps -> ( ph /\ ps ) ) )
7 6 sb2imi 
 |-  ( [ y / x ] ph -> ( [ y / x ] ps -> [ y / x ] ( ph /\ ps ) ) )
8 7 imp 
 |-  ( ( [ y / x ] ph /\ [ y / x ] ps ) -> [ y / x ] ( ph /\ ps ) )
9 5 8 impbii 
 |-  ( [ y / x ] ( ph /\ ps ) <-> ( [ y / x ] ph /\ [ y / x ] ps ) )