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Theorem wl-2mintru2 35059
 Description: Using the recursion formula "(n+1)-mintru-(m+1)" ↔ if-(𝜑, "n-mintru-m" , "n-mintru-(m+1)" ) for "2-mintru-2" (meaning "2 out of 2 inputs are true") by plugging in n = 1, m = 1, and simplifying. See wl-1mintru1 35056 and wl-1mintru2 35057 to see that "1-mintru-1" / "1-mintru-2" evaluate to 𝜒 / ⊥ respectively. Negating a "n-mintru2" operation means 'at most one input is true', so all inputs exclude each other mutually. Such an exclusion is expressed by a NAND operation (𝜑 ⊼ 𝜓), not by a XOR. Applying this idea here (n = 2) yields the expected NAND in case of a pair of inputs. (Contributed by Wolf Lammen, 10-May-2024.)
Assertion
Ref Expression
wl-2mintru2 (if-(𝜓, 𝜒, ⊥) ↔ (𝜓𝜒))

Proof of Theorem wl-2mintru2
StepHypRef Expression
1 dfifp7 1065 . 2 (if-(𝜓, 𝜒, ⊥) ↔ ((⊥ → 𝜓) → (𝜓𝜒)))
2 falim 1555 . . 3 (⊥ → 𝜓)
32a1bi 366 . 2 ((𝜓𝜒) ↔ ((⊥ → 𝜓) → (𝜓𝜒)))
41, 3bitr4i 281 1 (if-(𝜓, 𝜒, ⊥) ↔ (𝜓𝜒))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 399  if-wif 1058  ⊥wfal 1550 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 845  df-ifp 1059  df-tru 1541  df-fal 1551 This theorem is referenced by: (None)
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