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Metamath Proof Explorer

Theorem necom

Description: Commutation of inequality. (Contributed by NM, 14-May-1999)

Ref Expression
Assertion necom ( 𝐴 ≠ 𝐵 ↔ 𝐵 ≠ 𝐴 )

Proof

Step Hyp Ref Expression
1 eqcom ⊢ ( 𝐴 = 𝐵 ↔ 𝐵 = 𝐴 )
2 1 necon3bii ⊢ ( 𝐴 ≠ 𝐵 ↔ 𝐵 ≠ 𝐴 )