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