Metamath Proof Explorer


Theorem ineqan12d

Description: Equality deduction for intersection of two classes. (Contributed by NM, 7-Feb-2007)

Ref Expression
Hypotheses ineq1d.1
|- ( ph -> A = B )
ineqan12d.2
|- ( ps -> C = D )
Assertion ineqan12d
|- ( ( ph /\ ps ) -> ( A i^i C ) = ( B i^i D ) )

Proof

Step Hyp Ref Expression
1 ineq1d.1
 |-  ( ph -> A = B )
2 ineqan12d.2
 |-  ( ps -> C = D )
3 ineq12
 |-  ( ( A = B /\ C = D ) -> ( A i^i C ) = ( B i^i D ) )
4 1 2 3 syl2an
 |-  ( ( ph /\ ps ) -> ( A i^i C ) = ( B i^i D ) )