Theorem neeq1d

Description: Deduction for inequality. (Contributed by NM, 25-Oct-1999) (Proof shortened by Wolf Lammen, 19-Nov-2019)

	Ref	Expression
Hypothesis	neeq1d.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion	neeq1d	⊢ ( 𝜑 → ( 𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐶 ) )

Proof

Step	Hyp	Ref	Expression
1		neeq1d.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
2	1	eqeq1d	⊢ ( 𝜑 → ( 𝐴 = 𝐶 ↔ 𝐵 = 𝐶 ) )
3	2	necon3bid	⊢ ( 𝜑 → ( 𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐶 ) )