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Metamath Proof Explorer

Theorem necon3abid

Description: Deduction from equality to inequality. (Contributed by NM, 21-Mar-2007)

Ref Expression
Hypothesis necon3abid.1 ⊢ ( 𝜑 → ( 𝐴 = 𝐵 ↔ 𝜓 ) )
Assertion necon3abid ( 𝜑 → ( 𝐴 ≠ 𝐵 ↔ ¬ 𝜓 ) )

Proof

Step Hyp Ref Expression
1 necon3abid.1 ⊢ ( 𝜑 → ( 𝐴 = 𝐵 ↔ 𝜓 ) )
2 df-ne ⊢ ( 𝐴 ≠ 𝐵 ↔ ¬ 𝐴 = 𝐵 )
3 1 notbid ⊢ ( 𝜑 → ( ¬ 𝐴 = 𝐵 ↔ ¬ 𝜓 ) )
4 2 3 bitrid ⊢ ( 𝜑 → ( 𝐴 ≠ 𝐵 ↔ ¬ 𝜓 ) )