Theorem wl-1xor 35164

Description: In the recursive scheme

"(n+1)-xor" ↔ if-(𝜑, ¬ "n-xor" , "n-xor" )

we set n = 0 to formally arrive at an expression for "1-xor". The base case "0-xor" is replaced with ⊥, as a sequence of 0 inputs never has an odd number being part of it. (Contributed by Wolf Lammen, 11-May-2024.)

Assertion

Ref	Expression
wl-1xor	⊢ (if-(𝜓, ¬ ⊥, ⊥) ↔ 𝜓)

Proof of Theorem wl-1xor

Step	Hyp	Ref	Expression
1		tbtru 1547	. . . . 5 ⊢ (𝜓 ↔ (𝜓 ↔ ⊤))
2	1	biimpi 219	. . . 4 ⊢ (𝜓 → (𝜓 ↔ ⊤))
3		notfal 1567	. . . 4 ⊢ (¬ ⊥ ↔ ⊤)
4	2, 3	bitr4di 293	. . 3 ⊢ (𝜓 → (𝜓 ↔ ¬ ⊥))
5		nbfal 1554	. . . 4 ⊢ (¬ 𝜓 ↔ (𝜓 ↔ ⊥))
6	5	biimpi 219	. . 3 ⊢ (¬ 𝜓 → (𝜓 ↔ ⊥))
7	4, 6	casesifp 1075	. 2 ⊢ (𝜓 ↔ if-(𝜓, ¬ ⊥, ⊥))
8	7	bicomi 227	1 ⊢ (if-(𝜓, ¬ ⊥, ⊥) ↔ 𝜓)

Syntax hints:  ¬ wn 3   ↔ wb 209  if-wif 1059  ⊤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 210  df-an 401  df-or 846  df-ifp 1060  df-tru 1542  df-fal 1552

This theorem is referenced by: (None)