Metamath Proof Explorer


Theorem neeq1i

Description: Inference for inequality. (Contributed by NM, 29-Apr-2005) (Proof shortened by Wolf Lammen, 19-Nov-2019)

Ref Expression
Hypothesis neeq1i.1 ⊢ 𝐴 = 𝐵
Assertion neeq1i ( 𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐶 )

Proof

Step Hyp Ref Expression
1 neeq1i.1 ⊢ 𝐴 = 𝐵
2 1 eqeq1i ⊢ ( 𝐴 = 𝐶 ↔ 𝐵 = 𝐶 )
3 2 necon3bii ⊢ ( 𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐶 )