Theorem necon3bid

Description: Deduction from equality to inequality. (Contributed by NM, 23-Feb-2005) (Proof shortened by Andrew Salmon, 25-May-2011)

	Ref	Expression
Hypothesis	necon3bid.1	⊢ ( 𝜑 → ( 𝐴 = 𝐵 ↔ 𝐶 = 𝐷 ) )
Assertion	necon3bid	⊢ ( 𝜑 → ( 𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷 ) )

Proof

Step	Hyp	Ref	Expression
1		necon3bid.1	⊢ ( 𝜑 → ( 𝐴 = 𝐵 ↔ 𝐶 = 𝐷 ) )
2		df-ne	⊢ ( 𝐴 ≠ 𝐵 ↔ ¬ 𝐴 = 𝐵 )
3	1	necon3bbid	⊢ ( 𝜑 → ( ¬ 𝐴 = 𝐵 ↔ 𝐶 ≠ 𝐷 ) )
4	2 3	syl5bb	⊢ ( 𝜑 → ( 𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷 ) )