Theorem necon3abid

Description: Deduction from equality to inequality. (Contributed by NM, 21-Mar-2007)

	Ref	Expression
Hypothesis	necon3abid.1	$$⊢\phi \to \left(A=B\leftrightarrow \psi \right)$$
Assertion	necon3abid	$$⊢\phi \to \left(A\ne B\leftrightarrow \neg \psi \right)$$

Proof

Step	Hyp	Ref	Expression
1		necon3abid.1	$$⊢\phi \to \left(A=B\leftrightarrow \psi \right)$$
2		df-ne	$$⊢A\ne B\leftrightarrow \neg A=B$$
3	1	notbid	$$⊢\phi \to \left(\neg A=B\leftrightarrow \neg \psi \right)$$
4	2 3	syl5bb	$$⊢\phi \to \left(A\ne B\leftrightarrow \neg \psi \right)$$