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Metamath Proof Explorer

Theorem necon3bii

Description: Inference from equality to inequality. (Contributed by NM, 23-Feb-2005)

Ref Expression
Hypothesis necon3bii.1 ⊢ ( 𝐴 = 𝐵 ↔ 𝐶 = 𝐷 )
Assertion necon3bii ( 𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷 )

Proof

Step Hyp Ref Expression
1 necon3bii.1 ⊢ ( 𝐴 = 𝐵 ↔ 𝐶 = 𝐷 )
2 1 necon3abii ⊢ ( 𝐴 ≠ 𝐵 ↔ ¬ 𝐶 = 𝐷 )
3 df-ne ⊢ ( 𝐶 ≠ 𝐷 ↔ ¬ 𝐶 = 𝐷 )
4 2 3 bitr4i ⊢ ( 𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷 )