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	|- ( ph -> A = B )
Assertion	neeq1d	|- ( ph -> ( A =/= C <-> B =/= C ) )

Proof

Step	Hyp	Ref	Expression
1		neeq1d.1	|- ( ph -> A = B )
2	1	eqeq1d	|- ( ph -> ( A = C <-> B = C ) )
3	2	necon3bid	|- ( ph -> ( A =/= C <-> B =/= C ) )