Theorem nannan 1451

Description: Lemma for handling nested 'nand's. (Contributed by Jeff Hoffman, 19-Nov-2007.) (Proof shortened by Wolf Lammen, 7-Mar-2020.)

Assertion

Ref	Expression
nannan	⊢ ((𝜑 ⊼ (𝜒 ⊼ 𝜓)) ↔ (𝜑 → (𝜒 ∧ 𝜓)))

Proof of Theorem nannan

Step	Hyp	Ref	Expression
1		imnan 438	. 2 ⊢ ((𝜑 → ¬ (𝜒 ⊼ 𝜓)) ↔ ¬ (𝜑 ∧ (𝜒 ⊼ 𝜓)))
2		nanan 1449	. . 3 ⊢ ((𝜒 ∧ 𝜓) ↔ ¬ (𝜒 ⊼ 𝜓))
3	2	imbi2i 326	. 2 ⊢ ((𝜑 → (𝜒 ∧ 𝜓)) ↔ (𝜑 → ¬ (𝜒 ⊼ 𝜓)))
4		df-nan 1448	. 2 ⊢ ((𝜑 ⊼ (𝜒 ⊼ 𝜓)) ↔ ¬ (𝜑 ∧ (𝜒 ⊼ 𝜓)))
5	1, 3, 4	3bitr4ri 293	1 ⊢ ((𝜑 ⊼ (𝜒 ⊼ 𝜓)) ↔ (𝜑 → (𝜒 ∧ 𝜓)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 196   ∧ wa 384   ⊼ wnan 1447

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 197  df-an 386  df-nan 1448

This theorem is referenced by:  nanim  1452  nanbi  1454  nic-mp  1596  nic-ax  1598  waj-ax  32413  lukshef-ax2  32414  arg-ax  32415  rp-fakenanass  37860