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Theorem wl-1mintru1 35586
Description: Using the recursion formula:

"(n+1)-mintru-(m+1)" ↔ if-(𝜑, "n-mintru-m" , "n-mintru-(m+1)" )

for "1-mintru-1" (meaning "at least 1 out of 1 input is true") by plugging in n = 0, m = 0, and simplifying. The expressions "0-mintru-0" and "0-mintru-1" are base cases of the recursion, meaning "in a sequence of zero inputs, at least 0 / 1 input is true", respectively equvalent to / .

Negating an "n-mintru1" operation means: All n inputs 𝜑.. 𝜃 are false. This is also conveniently expressed as ¬ (𝜑.. 𝜃). Applying this idea here (n = 1) yields the obvious result that in an input sequence of size 1 only then all will be false, if its single input is. (Contributed by Wolf Lammen, 10-May-2024.)

Assertion
Ref Expression
wl-1mintru1 (if-(𝜒, ⊤, ⊥) ↔ 𝜒)

Proof of Theorem wl-1mintru1
StepHypRef Expression
1 tbtru 1547 . . . 4 (𝜒 ↔ (𝜒 ↔ ⊤))
21biimpi 215 . . 3 (𝜒 → (𝜒 ↔ ⊤))
3 nbfal 1554 . . . 4 𝜒 ↔ (𝜒 ↔ ⊥))
43biimpi 215 . . 3 𝜒 → (𝜒 ↔ ⊥))
52, 4casesifp 1075 . 2 (𝜒 ↔ if-(𝜒, ⊤, ⊥))
65bicomi 223 1 (if-(𝜒, ⊤, ⊥) ↔ 𝜒)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 205  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 206  df-an 396  df-or 844  df-ifp 1060  df-tru 1542  df-fal 1552
This theorem is referenced by: (None)
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