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	|- ( ph -> A = B )
	neeq12d.2	|- ( ph -> C = D )
Assertion	neeq12d	|- ( ph -> ( A =/= C <-> B =/= D ) )

Proof

Step	Hyp	Ref	Expression
1		neeq1d.1	|- ( ph -> A = B )
2		neeq12d.2	|- ( ph -> C = D )
3	1 2	eqeq12d	|- ( ph -> ( A = C <-> B = D ) )
4	3	necon3bid	|- ( ph -> ( A =/= C <-> B =/= D ) )