Theorem nbbn 498

Description: Move negation outside of biconditional. Compare Theorem *5.18 of [WhiteheadRussell] p. 124.

Assertion

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

Proof of Theorem nbbn

Step	Hyp	Ref	Expression
1		bicom 398	. 2 ⊢ ((¬ φ ↔ ψ) ↔ (ψ ↔ ¬ φ))
2		bicom 398	. . . 4 ⊢ ((φ ↔ ψ) ↔ (ψ ↔ φ))
3		pm5.18 497	. . . 4 ⊢ ((ψ ↔ φ) ↔ ¬ (ψ ↔ ¬ φ))
4	2, 3	bitr 151	. . 3 ⊢ ((φ ↔ ψ) ↔ ¬ (ψ ↔ ¬ φ))
5	4	bicon2i 194	. 2 ⊢ ((ψ ↔ ¬ φ) ↔ ¬ (φ ↔ ψ))
6	1, 5	bitr 151	1 ⊢ ((¬ φ ↔ ψ) ↔ ¬ (φ ↔ ψ))

Syntax hints:  ¬ wn 1   ↔ wb 127

This theorem is referenced by:  xor 500  symdif2 1690  canth 2945

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198

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