Theorem neeq12d

Description: Deduction for inequality. (Contributed by NM, 24-Jul-2012) (Proof shortened by Wolf Lammen, 25-Nov-2019)

	Ref	Expression
Hypotheses	neeq1d.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
	neeq12d.2	⊢ ( 𝜑 → 𝐶 = 𝐷 )
Assertion	neeq12d	⊢ ( 𝜑 → ( 𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐷 ) )

Proof

Step	Hyp	Ref	Expression
1		neeq1d.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
2		neeq12d.2	⊢ ( 𝜑 → 𝐶 = 𝐷 )
3	1 2	eqeq12d	⊢ ( 𝜑 → ( 𝐴 = 𝐶 ↔ 𝐵 = 𝐷 ) )
4	3	necon3bid	⊢ ( 𝜑 → ( 𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐷 ) )