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Metamath Proof Explorer

Theorem neeq2d

Description: Deduction for inequality. (Contributed by NM, 25-Oct-1999) (Proof shortened by Wolf Lammen, 19-Nov-2019)

Ref Expression
Hypothesis neeq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion neeq2d ( 𝜑 → ( 𝐶 ≠ 𝐴 ↔ 𝐶 ≠ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 neeq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 1 eqeq2d ⊢ ( 𝜑 → ( 𝐶 = 𝐴 ↔ 𝐶 = 𝐵 ) )
3 2 necon3bid ⊢ ( 𝜑 → ( 𝐶 ≠ 𝐴 ↔ 𝐶 ≠ 𝐵 ) )