Metamath Proof Explorer


Theorem rneqd

Description: Equality deduction for range. (Contributed by NM, 4-Mar-2004)

Ref Expression
Hypothesis rneqd.1 φ A = B
Assertion rneqd φ ran A = ran B

Proof

Step Hyp Ref Expression
1 rneqd.1 φ A = B
2 rneq A = B ran A = ran B
3 1 2 syl φ ran A = ran B