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 )