Theorem bj-nnor 13615

Description: Double negation of a disjunction in terms of implication. (Contributed by BJ, 9-Oct-2019.)

Assertion

Ref	Expression
bj-nnor	⊢ (¬ ¬ (𝜑 ∨ 𝜓) ↔ (¬ 𝜑 → ¬ ¬ 𝜓))

Proof of Theorem bj-nnor

Step	Hyp	Ref	Expression
1		ioran 742	. . 3 ⊢ (¬ (𝜑 ∨ 𝜓) ↔ (¬ 𝜑 ∧ ¬ 𝜓))
2	1	notbii 658	. 2 ⊢ (¬ ¬ (𝜑 ∨ 𝜓) ↔ ¬ (¬ 𝜑 ∧ ¬ 𝜓))
3		imnan 680	. 2 ⊢ ((¬ 𝜑 → ¬ ¬ 𝜓) ↔ ¬ (¬ 𝜑 ∧ ¬ 𝜓))
4	2, 3	bitr4i 186	1 ⊢ (¬ ¬ (𝜑 ∨ 𝜓) ↔ (¬ 𝜑 → ¬ ¬ 𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103   ↔ wb 104   ∨ wo 698

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605  ax-io 699

This theorem depends on definitions:  df-bi 116

This theorem is referenced by:  bj-nndcALT  13639