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