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Metamath Proof Explorer

Theorem eqneqall

Description: A contradiction concerning equality implies anything. (Contributed by Alexander van der Vekens, 25-Jan-2018)

		Ref	Expression
	Assertion	eqneqall	⊢ ( 𝐴 = 𝐵 → ( 𝐴 ≠ 𝐵 → 𝜑 ) )

Step	Hyp	Ref	Expression
1		df-ne	⊢ ( 𝐴 ≠ 𝐵 ↔ ¬ 𝐴 = 𝐵 )
2		pm2.24	⊢ ( 𝐴 = 𝐵 → ( ¬ 𝐴 = 𝐵 → 𝜑 ) )
3	1 2	biimtrid	⊢ ( 𝐴 = 𝐵 → ( 𝐴 ≠ 𝐵 → 𝜑 ) )