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Metamath Proof Explorer

Theorem necomd

Description: Deduction from commutative law for inequality. (Contributed by NM, 12-Feb-2008)

		Ref	Expression
	Hypothesis	necomd.1	$⊢ φ \to A \neq B$
	Assertion	necomd	$⊢ φ \to B \neq A$

Step	Hyp	Ref	Expression
1		necomd.1	$⊢ φ \to A \neq B$
2		necom	$⊢ A \neq B \leftrightarrow B \neq A$
3	1 2	sylib	$⊢ φ \to B \neq A$