Metamath Proof Explorer


Theorem disjeq2dv

Description: Equality deduction for disjoint collection. (Contributed by Mario Carneiro, 14-Nov-2016)

Ref Expression
Hypothesis disjeq2dv.1 φ x A B = C
Assertion disjeq2dv φ Disj x A B Disj x A C

Proof

Step Hyp Ref Expression
1 disjeq2dv.1 φ x A B = C
2 1 ralrimiva φ x A B = C
3 disjeq2 x A B = C Disj x A B Disj x A C
4 2 3 syl φ Disj x A B Disj x A C