Theorem ifpdfnan 41093

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 1487	. 2 ⊢ ((𝜑 ⊼ 𝜓) ↔ ¬ (𝜑 ∧ 𝜓))
2		ifpdfan 41073	. . 3 ⊢ ((𝜑 ∧ 𝜓) ↔ if-(𝜑, 𝜓, ⊥))
3	2	notbii 320	. 2 ⊢ (¬ (𝜑 ∧ 𝜓) ↔ ¬ if-(𝜑, 𝜓, ⊥))
4		ifpnot23 41085	. . 3 ⊢ (¬ if-(𝜑, 𝜓, ⊥) ↔ if-(𝜑, ¬ 𝜓, ¬ ⊥))
5		notfal 1567	. . . 4 ⊢ (¬ ⊥ ↔ ⊤)
6		ifpbi3 41075	. . . 4 ⊢ ((¬ ⊥ ↔ ⊤) → (if-(𝜑, ¬ 𝜓, ¬ ⊥) ↔ if-(𝜑, ¬ 𝜓, ⊤)))
7	5, 6	ax-mp 5	. . 3 ⊢ (if-(𝜑, ¬ 𝜓, ¬ ⊥) ↔ if-(𝜑, ¬ 𝜓, ⊤))
8	4, 7	bitri 274	. 2 ⊢ (¬ if-(𝜑, 𝜓, ⊥) ↔ if-(𝜑, ¬ 𝜓, ⊤))
9	1, 3, 8	3bitri 297	1 ⊢ ((𝜑 ⊼ 𝜓) ↔ if-(𝜑, ¬ 𝜓, ⊤))

Syntax hints:  ¬ wn 3   ↔ wb 205   ∧ wa 396  if-wif 1060   ⊼ wnan 1486  ⊤wtru 1540  ⊥wfal 1551

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 206  df-an 397  df-or 845  df-ifp 1061  df-nan 1487  df-tru 1542  df-fal 1552

This theorem is referenced by: (None)