Metamath Proof Explorer


Theorem iineq2d

Description: Equality deduction for indexed intersection. (Contributed by NM, 7-Dec-2011)

Ref Expression
Hypotheses iineq2d.1
|- F/ x ph
iineq2d.2
|- ( ( ph /\ x e. A ) -> B = C )
Assertion iineq2d
|- ( ph -> |^|_ x e. A B = |^|_ x e. A C )

Proof

Step Hyp Ref Expression
1 iineq2d.1
 |-  F/ x ph
2 iineq2d.2
 |-  ( ( ph /\ x e. A ) -> B = C )
3 2 ex
 |-  ( ph -> ( x e. A -> B = C ) )
4 1 3 ralrimi
 |-  ( ph -> A. x e. A B = C )
5 iineq2
 |-  ( A. x e. A B = C -> |^|_ x e. A B = |^|_ x e. A C )
6 4 5 syl
 |-  ( ph -> |^|_ x e. A B = |^|_ x e. A C )