Theorem nanbi 1519

Description: Biconditional in terms of alternative denial. (Contributed by Jeff Hoffman, 19-Nov-2007.) (Proof shortened by Wolf Lammen, 27-Jun-2020.)

Assertion

Ref	Expression
nanbi	⊢ ((𝜑 ↔ 𝜓) ↔ ((𝜑 ⊼ 𝜓) ⊼ ((𝜑 ⊼ 𝜑) ⊼ (𝜓 ⊼ 𝜓))))

Proof of Theorem nanbi

Step	Hyp	Ref	Expression
1		dfbi3 1060	. . 3 ⊢ ((𝜑 ↔ 𝜓) ↔ ((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ ¬ 𝜓)))
2		df-or 859	. . 3 ⊢ (((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ ¬ 𝜓)) ↔ (¬ (𝜑 ∧ 𝜓) → (¬ 𝜑 ∧ ¬ 𝜓)))
3		df-nan 1511	. . . . 5 ⊢ ((𝜑 ⊼ 𝜓) ↔ ¬ (𝜑 ∧ 𝜓))
4	3	bicomi 226	. . . 4 ⊢ (¬ (𝜑 ∧ 𝜓) ↔ (𝜑 ⊼ 𝜓))
5		nannot 1518	. . . . 5 ⊢ (¬ 𝜑 ↔ (𝜑 ⊼ 𝜑))
6		nannot 1518	. . . . 5 ⊢ (¬ 𝜓 ↔ (𝜓 ⊼ 𝜓))
7	5, 6	anbi12i 637	. . . 4 ⊢ ((¬ 𝜑 ∧ ¬ 𝜓) ↔ ((𝜑 ⊼ 𝜑) ∧ (𝜓 ⊼ 𝜓)))
8	4, 7	imbi12i 352	. . 3 ⊢ ((¬ (𝜑 ∧ 𝜓) → (¬ 𝜑 ∧ ¬ 𝜓)) ↔ ((𝜑 ⊼ 𝜓) → ((𝜑 ⊼ 𝜑) ∧ (𝜓 ⊼ 𝜓))))
9	1, 2, 8	3bitri 299	. 2 ⊢ ((𝜑 ↔ 𝜓) ↔ ((𝜑 ⊼ 𝜓) → ((𝜑 ⊼ 𝜑) ∧ (𝜓 ⊼ 𝜓))))
10		nannan 1516	. 2 ⊢ (((𝜑 ⊼ 𝜓) ⊼ ((𝜑 ⊼ 𝜑) ⊼ (𝜓 ⊼ 𝜓))) ↔ ((𝜑 ⊼ 𝜓) → ((𝜑 ⊼ 𝜑) ∧ (𝜓 ⊼ 𝜓))))
11	9, 10	bitr4i 280	1 ⊢ ((𝜑 ↔ 𝜓) ↔ ((𝜑 ⊼ 𝜓) ⊼ ((𝜑 ⊼ 𝜑) ⊼ (𝜓 ⊼ 𝜓))))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208   ∧ wa 399   ∨ wo 858   ⊼ wnan 1510

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-nan 1511

This theorem is referenced by:  nic-dfim  1688  nic-dfneg  1689