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Metamath Proof Explorer

Theorem nbbn

Description: Move negation outside of biconditional. Compare Theorem *5.18 of WhiteheadRussell p. 124. (Contributed by NM, 27-Jun-2002) (Proof shortened by Wolf Lammen, 20-Sep-2013)

		Ref	Expression
	Assertion	nbbn	⊢ ( ( ¬ 𝜑 ↔ 𝜓 ) ↔ ¬ ( 𝜑 ↔ 𝜓 ) )

Proof

Step	Hyp	Ref	Expression
1		xor3	⊢ ( ¬ ( 𝜑 ↔ 𝜓 ) ↔ ( 𝜑 ↔ ¬ 𝜓 ) )
2		con2bi	⊢ ( ( 𝜑 ↔ ¬ 𝜓 ) ↔ ( 𝜓 ↔ ¬ 𝜑 ) )
3		bicom	⊢ ( ( 𝜓 ↔ ¬ 𝜑 ) ↔ ( ¬ 𝜑 ↔ 𝜓 ) )
4	1 2 3	3bitrri	⊢ ( ( ¬ 𝜑 ↔ 𝜓 ) ↔ ¬ ( 𝜑 ↔ 𝜓 ) )