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Metamath Proof Explorer

Theorem neeq12d

Description: Deduction for inequality. (Contributed by NM, 24-Jul-2012) (Proof shortened by Wolf Lammen, 25-Nov-2019)

Ref Expression
Hypotheses neeq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
neeq12d.2 ⊢ ( 𝜑 → 𝐶 = 𝐷 )
Assertion neeq12d ( 𝜑 → ( 𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐷 ) )

Proof

Step Hyp Ref Expression
1 neeq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 neeq12d.2 ⊢ ( 𝜑 → 𝐶 = 𝐷 )
3 1 2 eqeq12d ⊢ ( 𝜑 → ( 𝐴 = 𝐶 ↔ 𝐵 = 𝐷 ) )
4 3 necon3bid ⊢ ( 𝜑 → ( 𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐷 ) )