Theorem ifpdfnan 40778

Description: Define nand as conditional logic operator. (Contributed by RP, 20-Apr-2020.)

Assertion

Ref	Expression
ifpdfnan	⊢ ((𝜑 ⊼ 𝜓) ↔ if-(𝜑, ¬ 𝜓, ⊤))

Proof of Theorem ifpdfnan

Step	Hyp	Ref	Expression
1		df-nan 1488	. 2 ⊢ ((𝜑 ⊼ 𝜓) ↔ ¬ (𝜑 ∧ 𝜓))
2		ifpdfan 40758	. . 3 ⊢ ((𝜑 ∧ 𝜓) ↔ if-(𝜑, 𝜓, ⊥))
3	2	notbii 323	. 2 ⊢ (¬ (𝜑 ∧ 𝜓) ↔ ¬ if-(𝜑, 𝜓, ⊥))
4		ifpnot23 40770	. . 3 ⊢ (¬ if-(𝜑, 𝜓, ⊥) ↔ if-(𝜑, ¬ 𝜓, ¬ ⊥))
5		notfal 1571	. . . 4 ⊢ (¬ ⊥ ↔ ⊤)
6		ifpbi3 40760	. . . 4 ⊢ ((¬ ⊥ ↔ ⊤) → (if-(𝜑, ¬ 𝜓, ¬ ⊥) ↔ if-(𝜑, ¬ 𝜓, ⊤)))
7	5, 6	ax-mp 5	. . 3 ⊢ (if-(𝜑, ¬ 𝜓, ¬ ⊥) ↔ if-(𝜑, ¬ 𝜓, ⊤))
8	4, 7	bitri 278	. 2 ⊢ (¬ if-(𝜑, 𝜓, ⊥) ↔ if-(𝜑, ¬ 𝜓, ⊤))
9	1, 3, 8	3bitri 300	1 ⊢ ((𝜑 ⊼ 𝜓) ↔ if-(𝜑, ¬ 𝜓, ⊤))

Syntax hints:  ¬ wn 3   ↔ wb 209   ∧ wa 399  if-wif 1063   ⊼ wnan 1487  ⊤wtru 1544  ⊥wfal 1555

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 210  df-an 400  df-or 848  df-ifp 1064  df-nan 1488  df-tru 1546  df-fal 1556

This theorem is referenced by: (None)